NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

Edited By Ramraj Saini | Updated on Mar 30, 2025 06:30 PM IST

In our daily lives, we see the motion of vehicles, speedometers, and changing temperatures, which involve changing quantities. These changes are studied using limits and derivatives. So, what is a limit? A limit helps us understand the behavior of a function as it approaches a particular value. A derivative tells us the rate of change of a function. For example, speed is the derivative of distance concerning time. Derivatives are based on limits. They form the foundation of calculus and are used in physics, engineering, and economics to analyze motion, growth, and optimization problems in real-world scenarios.

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Limits and Derivatives Class 11 Questions And Answers PDF Free Download
Limits and Derivatives Class 11 Solutions - Important Formulae
Limits and Derivatives Class 11 NCERT Solutions
NCERT Solutions for Class 11 Mathematics - Chapter Wise
Importance of Solving NCERT Questions for Class 11 Chapter 12
NCERT Solutions for Class 11- Subject Wise
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NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

Students can get all NCERT solutions in one place and practice them to develop an in-depth understanding of the concepts. Also, after practicing all the questions, students can try the advanced questions of NCERT Exemplar and can refer to the solutions created by our experts in NCERT Exemplar Solutions For Class 11 Mathematics Chapter Limits And Derivative.

Limits and Derivatives Class 11 Questions And Answers PDF Free Download

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Limits and Derivatives Class 11 Solutions - Important Formulae

Limits

The limit of a function $f (x)$ as $x$ approaches $a$ , denoted as $lim_{x \to a} f (x)$ , exists when both the left-hand limit and the right-hand limit exist and are equal:

$lim_{x \to a} f (x)$ exists if $lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x)$

Left-Hand Limit:

$lim_{x \to a^{-}} f (x) = lim_{h \to 0} f (a - h)$

Right-Hand Limit:

$lim_{x \to a^{+}} f (x) = lim_{h \to 0} f (a + h)$

Properties of Limits:

If $lim_{x \to a} f (x)$ and $lim_{x \to a} g (x)$ both exist:
Sum Rule:

$lim_{x \to a} [f (x) \pm g (x)] = lim_{x \to a} f (x) \pm lim_{x \to a} g (x)$

Constant Rule:

$lim_{x \to a} [k \cdot f (x)] = k \cdot lim_{x \to a} f (x)$

Product Rule:

$lim_{x \to a} [f (x) \cdot g (x)] = lim_{x \to a} f (x) \cdot lim_{x \to a} g (x)$

Quotient Rule:

$lim_{x \to a} [\frac{f (x)}{g (x)}] = \frac{lim_{x \to a} f (x)}{lim_{x \to a} g (x)},$ provided $lim_{x \to a} g (x) \neq 0$

Standard Limits:

$lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}$
$lim_{x \to 0} \frac{\sin x}{x} = 1$
$lim_{x \to 0} \frac{\tan x}{x} = 1$
$lim_{x \to 0} \frac{a^{x} - 1}{x} = \ln a$
$lim_{x \to 0} \frac{e^{x} - 1}{x} = 1$
$lim_{x \to 0} \frac{\ln (1 + x)}{x} = 1$

Derivatives:

The derivative of a function $f$ at point $x$ is given by:

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

Properties of Derivatives:

If $f (x)$ and $g (x)$ are differentiable:
Sum Rule:

$\frac{d}{d x} [f (x) + g (x)] = \frac{d}{d x} f (x) + \frac{d}{d x} g (x)$

Difference Rule:

$\frac{d}{d x} [f (x) - g (x)] = \frac{d}{d x} f (x) - \frac{d}{d x} g (x)$

Product Rule:

$\frac{d}{d x} [f (x) \cdot g (x)] = f^{'} (x) g (x) + f (x) g^{'} (x)$

Quotient Rule:

$\frac{d}{d x} [\frac{f (x)}{g (x)}] = \frac{f^{'} (x) g (x) - f (x) g^{'} (x)}{[g (x)]^{2}}$

Standard Derivatives:

$\frac{d}{d x} (x^{n}) = n x^{n - 1}$
$\frac{d}{d x} (\sin x) = \cos x$
$\frac{d}{d x} (\cos x) = - \sin x$
$\frac{d}{d x} (\tan x) = \sec^{2} x$
$\frac{d}{d x} (\cot x) = - \csc^{2} x$
$\frac{d}{d x} (\sec x) = \sec x \cdot \tan x$
$\frac{d}{d x} (\csc x) = - \csc x \cdot \cot x$
$\frac{d}{d x} (a^{x}) = a^{x} \cdot \ln a$
$\frac{d}{d x} (e^{x}) = e^{x}$
$\frac{d}{d x} (\ln x) = \frac{1}{x}$

Limits and Derivatives Class 11 NCERT Solutions

Class 11 Maths chapter 12 solutions Exercise: 12.1

Page number: 237-239

Total questions: 32

Limits and derivatives class 11 questions and answers - Exercise: 12.1

Question:1 Evaluate the following limits $lim_{x \to 3} x + 3$

Answer:

$lim_{x \to 3} x + 3$

$\Rightarrow lim_{x \to 3} 3 + 3$

$\Rightarrow 6$

Question:2 Evaluate the following limits $lim_{x \to π} (x - 22 / 7)$

Answer:

Below, you can find the solution:

$lim_{x \to π} (x - 22 / 7) = π - \frac{22}{7}$

Question:3 Evaluate the following limits $lim_{r \to 1} π r^{2}$

Answer:

The limit

$lim_{r \to 1} π r^{2} = π (1)^{2} = π$

Hence, the answer is $π$

Question:4 Evaluate the following limits $lim_{x \to 4} \frac{4 x + 3}{x - 2}$

Answer:

The limit

$lim_{x \to 4} \frac{4 x + 3}{x - 2}$

$\Rightarrow \frac{4 (4) + 3}{(4) - 2}$

$\Rightarrow \frac{19}{2}$

Question:5 Evaluate the following limits $lim_{x \to - 1} \frac{x^{10} + x^{5} + 1}{x - 1}$

Answer:

The limit

$lim_{x \to 4} \frac{x^{10} + x^{5} + 1}{x - 1}$

$\Rightarrow \frac{(- 1)^{10} + (- 1)^{5} + 1}{(- 1) - 1}$

$\Rightarrow \frac{1 - 1 + 1}{- 2}$

$\Rightarrow - \frac{1}{2}$

Question:6 Evaluate the following limits $lim_{x \to 0} \frac{(x + 1)^{5} - 1}{x}$

Answer:

The limit

$lim_{x \to 0} \frac{(x + 1)^{5} - 1}{x}$

Let's put

$x + 1 = y$

Since we have changed the function, its limit will also change,

$x \to 0, y \to 0 + 1 = 1$

So our function has became

$lim_{y \to 1} \frac{y^{5} - 1}{y - 1}$

Now, as we know the property

$lim_{x \to 1} \frac{x^{5} - a^{n}}{x - a} = n a^{n - 1}$

$lim_{y \to 1} \frac{y^{5} - 1}{y - 1} = 5 (1)^{5} = 5$

Hence,

$lim_{x \to 0} \frac{(x + 1)^{5} - 1}{x} = 5$

Question:7 Evaluate the following limits $lim_{x \to 2} \frac{3 x^{2} - x - 10}{x^{2} - 4}$

Answer:

The limit

$lim_{x \to 2} \frac{3 x^{2} - x - 10}{x^{2} - 4}$

$\Rightarrow lim_{x \to 2} \frac{(x - 2) (3 x + 5)}{(x - 2) (x + 2)}$

$\Rightarrow lim_{x \to 2} \frac{(3 x + 5)}{(x + 2)}$

$\Rightarrow \frac{(3 (2) + 5)}{((2) + 2)}$

$\Rightarrow \frac{11}{4}$

Question:8 Evaluate the following limits $lim_{x \to 3} \frac{x^{4} - 81}{2 x^{2} - 5 x - 3}$

Answer:

The limit

$lim_{x \to 3} \frac{x^{4} - 81}{2 x^{2} - 5 x - 3}$

At x = 2, both the numerator and denominator become zero, so let's factorise the function

$lim_{x \to 3} \frac{(x - 3) (x + 3) (x^{2} + 9)}{(x - 3) (2 x + 1)}$

$lim_{x \to 3} \frac{(x + 3) (x^{2} + 9)}{(2 x + 1)}$

Now we can put the limit directly, so

$lim_{x \to 3} \frac{(x + 3) (x^{2} + 9)}{(2 x + 1)}$

$\Rightarrow \frac{((3) + 3) ((3)^{2} + 9)}{(2 (3) + 1)}$

$\Rightarrow \frac{6 \times 18}{7}$

$\Rightarrow \frac{108}{7}$

Question:9 Evaluate the following limits $lim_{x \to 0} \frac{a x + b}{c x + 1}$

Answer:

The limit,

$lim_{x \to 0} \frac{a x + b}{c x + 1}$

$\Rightarrow \frac{a (0) + b}{c (0) + 1}$

$\Rightarrow b$

Question:10 Evaluate the following limits $lim_{z \to 1} \frac{z^{1 / 3} - 1}{z^{1 / 6} - 1}$

Answer:

The limit

$lim_{z \to 1} \frac{z^{1 / 3} - 1}{z^{1 / 6} - 1}$

Here, on directly putting the limit, both the numerator and the denominator become zero so we factorize the function and then put the limit.

$lim_{z \to 1} \frac{z^{1 / 3} - 1}{z^{1 / 6} - 1} = lim_{z \to 1} \frac{z^{(1 / 6)^{2}} - 1^{2}}{z^{1 / 6} - 1}$

$= lim_{z \to 1} \frac{(z^{(1 / 6)} - 1) (z^{(1 / 6)} + 1)}{z^{1 / 6} - 1}$

$= lim_{z \to 1} (z^{1 / 6} + 1)$

$= (1^{1 / 6} + 1)$

$= 1 + 1$

$= 2$

Question:11 Evaluate the following limits $lim_{x \to 1} \frac{a x^{2} + b x + c}{c x^{2} + b x + a}, a + b + c \neq 0$

Answer:

The limit:

$lim_{x \to 1} \frac{a x^{2} + b x + c}{c x^{2} + b x + a}, a + b + c \neq 0$

Since the Denominator is not zero on directly putting the limit, we can directly put the limits, so,

$lim_{x \to 1} \frac{a x^{2} + b x + c}{c x^{2} + b x + a}, a + b + c \neq 0$

$= \frac{a (1)^{2} + b (1) + c}{c (1)^{2} + b (1) + a},$

$= \frac{a + b + c}{a + b + c},$

$= 1$

Question:12 Evaluate the following limits $lim_{x \to - 2} \frac{\frac{1}{x} + \frac{x}{2}}{x + 2}$

Answer:

$lim_{x \to - 2} \frac{\frac{1}{x} + \frac{x}{2}}{x + 2}$

Here, since the denominator becomes zero on putting the limit directly, we first simplify the function and then put the limit,

$lim_{x \to - 2} \frac{\frac{1}{x} + \frac{x}{2}}{x + 2}$

$= lim_{x \to - 2} \frac{\frac{x + 2}{2 x}}{x + 2}$

$= lim_{x \to - 2} \frac{1}{2 x}$

$= \frac{1}{2 (- 2)}$

$= - \frac{1}{4}$

Question:13 Evaluate the following limits $lim_{x \to 0} \frac{\sin a x}{b x}$

Answer:

The limit

$lim_{x \to 0} \frac{\sin a x}{b x}$

Here, on directly putting the limits, the function becomes $\frac{0}{0}$ form. So we try to make the function in the form of $\frac{s i n x}{x}$ . so,

$lim_{x \to 0} \frac{\sin a x}{b x}$

$= lim_{x \to 0} \frac{\sin a x (a x)}{b x (a x)}$

$= lim_{x \to 0} \frac{\sin a x}{a x} \frac{a}{b}$

As $lim_{x \to 0} \frac{s i n x}{x} = 1$

$= 1 \cdot \frac{a}{b}$

$= \frac{a}{b}$

Question:14 Evaluate the following limits $lim_{x \to 0} \frac{\sin a x}{\sin b x}, a, b \neq 0$

Answer:

The limit,

$lim_{x \to 0} \frac{\sin a x}{\sin b x}, a, b \neq 0$

On putting the limit directly, the function takes the zero by zero form. So, we convert it in the form of $\frac{s i n a}{a}$ .and then put the limit,

$\Rightarrow lim_{x \to 0} \frac{\frac{s i n a x}{a x}}{\frac{s i n b x}{b x}} \cdot \frac{a x}{b x}$

$= \frac{lim_{a x \to 0} \frac{s i n a x}{a x}}{lim_{b x \to 0} \frac{s i n b x}{b x}} \cdot \frac{a}{b}$

$= \frac{a}{b}$

Question:15 Evaluate the following limits $lim_{x \to π} \frac{\sin (π - x)}{π (π - x)}$

Answer:

The limit

$lim_{x \to π} \frac{\sin (π - x)}{π (π - x)}$

$\Rightarrow lim_{x \to π} \frac{\sin (π - x)}{(π - x)} \times \frac{1}{π}$

$= 1 \times \frac{1}{π}$

$= \frac{1}{π}$

Question:16 Evaluate the following limits $lim_{x \to 0} \frac{\cos x}{π - x}$

Answer:

The limit

$lim_{x \to 0} \frac{\cos x}{π - x}$

The function behaves well on directly putting the limit, so we put the limit directly. So.

$lim_{x \to 0} \frac{\cos x}{π - x}$

$= \frac{\cos (0)}{π - (0)}$

$= \frac{1}{π}$

Question:17 Evaluate the following limits $lim_{x \to 0}$ $\frac{\cos 2 x - 1}{\cos x - 1}$

Answer:

The limit:

$lim_{x \to 0}$ $\frac{\cos 2 x - 1}{\cos x - 1}$

The function takes the zero-by-zero form when the limit is put directly, so we simplify the function and then put the limit

$lim_{x \to 0}$ $\frac{\cos 2 x - 1}{\cos x - 1}$

$lim_{x \to 0}$ $\frac{- 2 (s i n^{2} x)}{- 2 (s i n^{2} (\frac{x}{2}))}$

$= lim_{x \to 0}$ $\frac{\frac{(s i n^{2} x)}{x^{2}}}{\frac{(s i n^{2} (\frac{x}{2}))}{(\frac{x}{2})^{2}}} \times \frac{x^{2}}{(\frac{x}{2})^{2}}$

$= \frac{1^{2}}{1^{2}} \times 4$

$= 4$

Question:18 Evaluate the following limits $lim_{x \to 0} \frac{a x + x \cos x}{b \sin x}$

Answer:

$lim_{x \to 0} \frac{a x + x \cos x}{b \sin x}$

The function takes the form zero by zero when we put the limit directly in the function Since theunction consists of the sin function and cos function, we try to make the function in the form of $\frac{s i n x}{x}$ as we know that it tends to 1 when x tends to 0.

So,

$lim_{x \to 0} \frac{a x + x \cos x}{b \sin x}$

$= \frac{1}{b} lim_{x \to 0} \frac{x (a + \cos x)}{\sin x}$

$= \frac{1}{b} lim_{x \to 0} \frac{x}{\sin x} \times (a + \cos x)$

$= \frac{1}{b} \times 1 \times (a + \cos (0))$

$= \frac{a + 1}{b}$

Question:19 Evaluate the following limits $lim_{x \to 0} x \sec x$

Answer:

$lim_{x \to 0} x \sec x$

As the function doesn't create any abnormality on putting the limit directly, we can put the limit directly. So,

$lim_{x \to 0} x \sec x$

$= (0) \times 1$

$= 0$ .

Question:20 Evaluate the following limits $lim_{x \to 0} \frac{\sin a x + b x}{a x + \sin b x} a, b, a + b \neq 0$

Answer:

$lim_{x \to 0} \frac{\sin a x + b x}{a x + \sin b x} a, b, a + b \neq 0$

The function takes the zero-by-zero form when we put the limit into the function directly, so we try to eliminate this case by simplifying the function. So

$lim_{x \to 0} \frac{\sin a x + b x}{a x + \sin b x} a, b, a + b \neq 0$

$= lim_{x \to 0} \frac{\frac{\sin a x}{a x} \cdot a x + b x}{a x + \frac{\sin b x}{b x} \cdot b x}$

$= lim_{x \to 0} \frac{\frac{\sin a x}{a x} \cdot a + b}{a + \frac{\sin b x}{b x} \cdot b}$

$= \frac{1 \cdot a + b}{a + 1 \cdot b}$

$= \frac{a + b}{a + b}$

$= 1$

Question:21 Evaluate the following limits $lim_{x \to 0} (\csc x - \cot x)$

Answer:

$lim_{x \to 0} (\csc x - \cot x)$

On putting the limit directly, the function takes infinity by infinity form, so we simplify the function and then put the limit

$lim_{x \to 0} (\csc x - \cot x)$

$= lim_{x \to 0} (\frac{1}{s i n x} - \frac{c o s x}{s i n x})$

$= lim_{x \to 0} (\frac{1 - c o s x}{s i n x})$

$= lim_{x \to 0} (\frac{2 s i n^{2} (\frac{x}{2})}{s i n x})$

$= lim_{x \to 0} (\frac{2 s i n^{2} (\frac{x}{2})}{(\frac{x}{2})^{2}}) (\frac{(\frac{x}{2})^{2}}{s i n x})$

$= lim_{x \to 0} \frac{2}{4} (\frac{s i n^{2} (\frac{x}{2})}{(\frac{x}{2})^{2}}) (\frac{(x)}{s i n x}) \cdot x$

$= \frac{2}{4} \times (1)^{2} \times 0$

$= 0$

Question:22 Evaluate the following limits $lim_{x \to π / 2} \frac{\tan 2 x}{x - π / 2}$

Answer:

$lim_{x \to π / 2} \frac{\tan 2 x}{x - π / 2}$

The function takes zero by zero form when the limit is put directly, so we simplify the function and then put the limits,

Let's put

$y = x - \frac{π}{2}$

Since we are changing the variable, the limit will also change.

$x \to \frac{π}{2}, y = x - \frac{π}{2} \to \frac{π}{2} - \frac{π}{2} = 0$

So, the function in the new variable becomes,

$lim_{y \to 0} \frac{\tan 2 (y + \frac{π}{2})}{y + π / 2 - π / 2}$

$= lim_{y \to 0} \frac{\tan (2 y + π)}{y}$

As we know that property $t a n (π + x) = t a n x$

$= lim_{y \to 0} \frac{\tan (2 y)}{y}$

$= lim_{y \to 0} \frac{s i n 2 y}{2 y} \cdot \frac{2}{c o s 2 y}$

$= 1 \times 2$

$= 2$

Question:23 Find $lim_{x \to 0} f (x) lim_{x \to 1} f (x) w h e r e f (x) = {\begin{matrix} 2 x + 3 & x \leq 0 \\ 3 (x + 1) & x > 0 \end{matrix}$

Answer:

Given Function

$f (x) = {\begin{matrix} 2 x + 3 & x \leq 0 \\ 3 (x + 1) & x > 0 \end{matrix}$

Now,

Limit at x = 0 :

$a t x = 0^{-}$

: $lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (2 x + 3) = 2 (0) + 3 = 3$

$a t x = 0^{+}$

$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} 3 (x + 1) = 3 (0 + 1) = 3$

Hence limit at x = 0 is 3.

Limit at x = 1

$a t x = 1^{+}$

$lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} 3 (x + 1) = 3 (1 + 1) = 6$

$a t x = 1^{-}$

$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} 3 (x + 1) = 3 (1 + 1) = 6$

Hence limit at x = 1 is 6.

Question:24 Find $lim_{x \to 1} f (x), w h e r e f (x) = {\begin{matrix} x^{2} - 1 & x \neq 0 \\ - x^{2} - 1 & x > 1 \end{matrix}$

Answer:

$lim_{x \to 1} f (x), w h e r e f (x) = {\begin{matrix} x^{2} - 1 & x \neq 0 \\ - x^{2} - 1 & x > 1 \end{matrix}$

Limit at $x = 1^{+}$

$lim_{x \to 1^{+}} f (x) = lim_{x \to 1} (- x^{2} - 1) = - (1)^{2} - 1 = - 2$

Limit at $x = 1^{-}$

$lim_{x \to 1^{-}} f (x) = lim_{x \to 1} (x^{2} - 1) = (1)^{2} - 1 = 0$

As we can see Limit at $x = 1^{+}$ is not equal to the Limit at $x = 1^{-}$ , The limit of this function at x = 1 does not exist.

Question:25 Evaluate $lim_{x \to 0} f (x), w h e r e f (x) = {\begin{matrix} \frac{| x |}{x} & x \neq 0 \\ 0 & x = 0 \end{matrix}$

Answer:

$lim_{x \to 0} f (x), w h e r e f (x) = {\begin{matrix} \frac{| x |}{x} & x \neq 0 \\ 0 & x = 0 \end{matrix}$

The right-hand Limit or Limit at $x = 0^{+}$

$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} \frac{| x |}{x} = \frac{x}{x} = 1$

The left-hand limit or Limit at $x = 0^{-}$

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} \frac{| x |}{x} = \frac{- x}{x} = - 1$

Since The left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exist.

Question:26 Evaluate $lim_{x \to 0} f (x), w h e r e f (x) = {\begin{matrix} \frac{x}{| x |} & x \neq 0 \\ 0 & x = 0 \end{matrix}$

Answer:

$lim_{x \to 0} f (x), w h e r e f (x) = {\begin{matrix} \frac{x}{| x |} & x \neq 0 \\ 0 & x = 0 \end{matrix}$

The right-hand Limit or Limit at $x = 0^{+}$

$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} \frac{x}{| x |} = \frac{x}{x} = 1$

The left-hand limit or Limit at $x = 0^{-}$

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} \frac{x}{| x |} = \frac{x}{- x} = - 1$

Since The left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exist.

Question:27 Find $lim_{x \to 5} f (x), w h e r e f (x) = | x | - 5$

Answer:

$lim_{x \to 5} f (x), w h e r e f (x) = | x | - 5$

The right-hand Limit or Limit at $x = 5^{+}$

$lim_{x \to 5^{+}} f (x) = lim_{x \to 5^{+}} | x | - 5 = 5 - 5 = 0$

The left-hand limit or Limit at $x = 5^{-}$

$lim_{x \to 5^{-}} f (x) = lim_{x \to 5^{-}} | x | - 5 = 5 - 5 = 0$

Since The left-hand limit and right-hand limit are equal, the limit of this function at x = 5 is 0.

Question:28 Suppose $f (x) = {\begin{matrix} a + b x & x < 1 \\ 4 & x = 1 \\ b - a x & x > 1 \end{matrix}$ f (x) = f (1) what are possible values of a and b?

Answer:

Given,

$f (x) = {\begin{matrix} a + b x & x < 1 \\ 4 & x = 1 \\ b - a x & x > 1 \end{matrix}$

And

$lim_{x \to 1} f (x) = f (1)$

Since the limit exists,

left-hand limit = Right-hand limit = f(1).

Left-hand limit = f(1)

$lim_{x \to 1^{-}} f (x) = lim_{x \to 1} (a + b x) = a + b (1) = a + b = 4$

Right-hand limit

$lim_{x \to 1^{+}} f (x) = lim_{x \to 1} (b - a x) = b - a (1) = b - a = 4$

From both equations, we get that,

$a = 0$ and $b = 4$

Hence, the possible values of a and b are 0 and 4, respectively.

Question:29 Let a1, a2, . . ., an be fixed real numbers and define a function $f (x) = (x - a_{1}) (x - a_{2}) . . . (x - a_{n}) .$ What is $lim_{x \to a_{1}}$ f (x) ? For some $a \neq a_{1}, a_{2} . . . . a_{n}$ , compute l $lim_{x \to a} f (x)$

Answer:

Given,

$f (x) = (x - a_{1}) (x - a_{2}) . . . (x - a_{n}) .$

Now,

$lim_{x \to a_{1}} f (x) = lim_{x \to a_{1}} [(x - a_{1}) (x - a_{2}) . . . (x - a_{n})] . = [lim_{x \to a_{1}} (x - a_{1})] [lim_{x \to a_{1}} (x - a_{2})] [lim_{x \to a_{1}} (x - a_{n})] = 0$

Hence,

$lim_{x \to a_{1}} f (x) = 0$

Now,

$lim_{x \to a} f (x) = lim_{x \to a} (x - a_{1}) (x - a_{2}) . . . (x - a_{n})$

$lim_{x \to a} f (x) = (a - a_{1}) (a - a_{2}) (a - a_{3})$

Hence

$lim_{x \to a} f (x) = (a - a_{1}) (a - a_{2}) (a - a_{3})$ .

Question:30If $f (x) = {\begin{matrix} | x | + 1 & x < 0 \\ 0 & x = 0 \\ | x | - 1 & x > 0 \end{matrix}$ For what value (s) of a does $lim_{x \to a} f (x)$ exists ?

Answer:

$f (x) = {\begin{matrix} | x | + 1 & x < 0 \\ 0 & x = 0 \\ | x | - 1 & x > 0 \end{matrix}$

The limit at x = a exists when the right-hand limit is equal to the left-hand limit. So,

Case 1: when a = 0

The right-hand Limit or Limit at $x = 0^{+}$

$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} | x | - 1 = 1 - 1 = 0$

The left-hand limit or Limit at $x = 0^{-}$

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} | x | + 1 = 0 + 1 = 1$

Since the Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exist.

Case 2: When a < 0

The right-hand Limit or Limit at $x = a^{+}$

$lim_{x \to a^{+}} f (x) = lim_{x \to a^{+}} | x | - 1 = a - 1$

The left-hand limit or Limit at $x = a^{-}$

$lim_{x \to a^{-}} f (x) = lim_{x \to a^{-}} | x | - 1 = a - 1$

Since LHL = RHL, the Limit exists at x = a and is equal to a-1.

Case 3: When a > 0

The right-hand Limit or Limit at $x = a^{+}$

$lim_{x \to a^{+}} f (x) = lim_{x \to a^{+}} | x | + 1 = a + 1$

The left-hand limit or Limit at $x = a^{-}$

$lim_{x \to a^{-}} f (x) = lim_{x \to a^{-}} | x | + 1 = a + 1$

Since LHL = RHL, Limit exists at x = a and is equal to a+1

Hence,

The limit exists at all points except at x=0.

Question:31If the function f(x) satisfies $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$ , evaluate $lim_{x \to 1} f (x)$

Answer:

Given

$lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$

Now,

$lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = \frac{lim_{x \to 1} (f (x) - 2)}{lim_{x \to 1} (x^{2} - 1)} = π$

$lim_{x \to 1} (f (x) - 2) = π lim_{x \to 1} (x^{2} - 1)$

$lim_{x \to 1} (f (x) - 2) = π (1 - 1)$

$lim_{x \to 1} (f (x) - 2) = 0$

$lim_{x \to 1} f (x) = 2$

Question:32 If $f (x) = [\begin{array}{cl} m x^{2} + n & ; x < 0 \\ n x + m & ; x \leq 0 \leq 1 t \\ n x^{3} + m & ; x > 1 \end{array}]$

Answer:

Given,

$f (x) = [\begin{array}{cl} m x^{2} + n & ; x < 0 \\ n x + m & ; x \leq 0 \leq 1 t \\ n x^{3} + m & ; x > 1 \end{array}]$

Case 1: Limit at x = 0

The right-hand Limit or Limit at $x = 0^{+}$

$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} n x + m = n (0) + m = m$

The left-hand limit or Limit at $x = 0^{-}$

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} m x^{2} + n = m (0)^{2} + n = n$

Hence Limit will exist at x = 0 when m = n.

Case 2: Limit at x = 1

The right-hand Limit or Limit at $x = 1^{+}$

$lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} n x^{3} + m = n (1)^{3} + m = n + m$

The left-hand limit or Limit at $x = 1^{-}$

$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} n x + m = n (1) + m = n + m$

Hence Limit at 1 exists at all integers.

Class 11 Maths chapter 12 solutions Exercise: 12.2

Page number: 248-249

Total questions: 11

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Class 11 Maths Chapter 12 NCERT solutions - Exercise: 12.2

Question:1 Find the derivative of $x^{2} - 2 a t x = 10$

Answer:

F(x)= $x^{2} - 2$

Now, As we know, The derivative of any function at x is

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

The derivative of f(x) at x = 10:

$f^{'} (10) = lim_{h \to 0} \frac{f (10 + h) - f (10)}{h}$

$f^{'} (10) = lim_{h \to 0} \frac{(10 + h)^{2} - 2 - ((10)^{2} - 2)}{h}$

$f^{'} (10) = lim_{h \to 0} \frac{100 + 20 h + h^{2} - 2 - 100 + 2}{h}$

$f^{'} (10) = lim_{h \to 0} \frac{20 h + h^{2}}{h}$

$f^{'} (10) = lim_{h \to 0} 20 + h$

$f^{'} (10) = 20 + 0$

$f^{'} (10) = 20$

Question:2 Find the derivative of x at x = 1.

Answer:

Given

f(x)= x

Now, As we know, The derivative of any function at x is

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

The derivative of f(x) at x = 1:

$f^{'} (1) = lim_{h \to 0} \frac{f (1 + h) - f (1)}{h}$

$f^{'} (1) = lim_{h \to 0} \frac{(1 + h) - (1)}{h}$

$f^{'} (1) = lim_{h \to 0} \frac{(h)}{h}$

$f^{'} (1) = 1$ (Answer)

Question:3 Find the derivative of 99x at x = l00.

Answer:

f(x)= 99x

Now, as we know, the derivative of any function at x is

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

The derivative of f(x) at x = 100:

$f^{'} (100) = lim_{h \to 0} \frac{f (100 + h) - f (100)}{h}$

$f^{'} (100) = lim_{h \to 0} \frac{99 (100 + h) - 99 (100)}{h}$

$f^{'} (100) = lim_{h \to 0} \frac{99 h}{h}$

$f^{'} (100) = 99$

Question:4 (i) Find the derivative of the following functions from the first principle. $x^{3} - 27$

Answer:

Given

f(x)= $x^{3} - 27$

Now, as we know, the derivative of any function at x is

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{(x + h)^{3} - 27 - ((x)^{3} - 27)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{x^{3} + h^{3} + 3 x^{2} h + 3 h^{2} x - 27 + x^{3} + 27}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{h^{3} + 3 x^{2} h + 3 h^{2} x}{h}$

$f^{'} (x) = lim_{h \to 0} h^{2} + 3 x^{2} + 3 h x$

$f^{'} (x) = 3 x^{2}$

Question:4.(ii) Find the derivative of the following function from the first principle. $(x - 1) (x - 2)$

Answer:

f(x)= $(x - 1) (x - 2)$

Now, as we know, the derivative of any function at x is

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{(x + h - 1) (x + h - 2) - (x - 1) (x - 2)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{x^{2} + x h - 2 x + h x + h^{2} - 2 h - x - h + 2 - x^{2} + 2 x + x - 2}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{2 h x + h^{2} - 3 h}{h}$

$f^{'} (x) = lim_{h \to 0} 2 x + h - 3$

$f^{'} (x) = 2 x - 3$ (Answer)

Question:4(iii) Find the derivative of the following functions from the first principle. $1 / x^{2}$

Answer:

f(x)= $1 / x^{2}$

Now, as we know, the derivative of any function at x is

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{1 / (x + h)^{2} - 1 / (x^{2})}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{\frac{x^{2} - (x + h)^{2}}{(x + h)^{2} x^{2}}}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{x^{2} - x^{2} - 2 x h - h^{2}}{h (x + h)^{2} x^{2}}$

$f^{'} (x) = lim_{h \to 0} \frac{- 2 x h - h^{2}}{h (x + h)^{2} x^{2}}$

$f^{'} (x) = lim_{h \to 0} \frac{- 2 x - h}{(x + h)^{2} x^{2}}$

$f^{'} (x) = \frac{- 2 x - 0}{(x + 0)^{2} x^{2}}$

$f^{'} (x) = \frac{- 2}{x^{3}}$ (Answer)

Question:4(iv) Find the derivative of the following functions from the first principle. $\frac{x + 1}{x - 1}$

Answer:

Given:

$f (x) = \frac{x + 1}{x - 1}$

Now, as we know, the derivative of any function at x is

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{\frac{x + h + 1}{x + h - 1} - \frac{x + 1}{x - 1}}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{\frac{(x + h + 1) (x - 1) - (x + 1) (x + h - 1)}{(x - 1) (x + h - 1)}}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{x^{2} - x + h x - h + x - 1 - x^{2} - x h + x - x - h + 1}{(x - 1) (x + h - 1) h}$

$f^{'} (x) = lim_{h \to 0} \frac{- 2 h}{(x - 1) (x + h - 1) h}$

$f^{'} (x) = lim_{h \to 0} \frac{- 2}{(x - 1) (x + h - 1)}$

$f^{'} (x) = \frac{- 2}{(x - 1) (x + 0 - 1)}$

$f^{'} (x) = \frac{- 2}{(x - 1)^{2}}$

Question:5 For the function $f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + . . . . + \frac{x^{2}}{2} + x + 1$ Prove that f '(1) =100 f '(0).

Answer:

$f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + . . . . + \frac{x^{2}}{2} + x + 1$

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

Applying that property, we get

$f^{'} (x) = 100 \frac{x^{99}}{100} + 99 \frac{x^{98}}{99} + . . . . + 2 \frac{x}{2} + 1 + 0$

$f^{'} (x) = x^{99} + x^{98} + . . . . . . x + 1$

Now.

$f^{'} (0) = 0^{99} + 0^{98} + . . . . . .0 + 1 = 1$

$f^{'} (1) = 1^{99} + 1^{98} + . . . . . .1 + 1 = 100$

So,

Here

$1 \times 100 = 100$

$f^{'} (0) \times 100 = f^{'} (1)$

Hence Proved.

Question:6 Find the derivative of $x^{n} + a x^{n - 1} + a^{2} x^{n - 2} + . . . . + a^{n - 1} x + a^{n}$ for some fixed real number a.

Answer:

Given

$f (x) = x^{n} + a x^{n - 1} + a^{2} x^{n - 2} + . . . . + a^{n - 1} x + a^{n}$

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

Applying that property, we get

$f^{'} (x) = n x^{n - 1} + a (n - 1) x^{n - 2} + a^{2} (n - 2) x^{n - 3} + . . . . + a^{n - 1} 1 + 0$

$f^{'} (x) = n x^{n - 1} + a (n - 1) x^{n - 2} + a^{2} (n - 2) x^{n - 3} + . . . . + a^{n - 1}$

Question:7(i) For some constants a and b, find the derivative of $(x - a) (x - b)$

Answer:

Given

$f (x) = (x - a) (x - b) = x^{2} - a x - b x + a b$

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

And the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying that property, we get

$f^{'} (x) = 2 x - a - b$

Question:7(ii) For some constants a and b, find the derivative of $(a x^{2} + b)^{2}$

Answer:

Given

$f (x) = (a x^{2} + b)^{2} = a^{2} x^{4} + 2 a b x^{2} + b^{2}$

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

And the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying those properties, we get

$f^{'} (x) = 4 a^{2} x^{3} + 2 (2) a b x + 0$

$f^{'} (x) = 4 a^{2} x^{3} + 4 a b x$

$f^{'} (x) = 4 a x (a x^{2} + b)$

Question:7(iii) For some constants a and b, find the derivative of $\frac{x - a}{x - b}$

Answer:

Given,

$f (x) = \frac{x - a}{x - b}$

Now, as we know the quotient rule of derivative,

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} \frac{d y_{1}}{d x} - y_{1} \frac{d y_{2}}{d x}}{y_{2}^{2}}$

So,, applying this rule, we get

$\frac{d (\frac{x - a}{x - b})}{d x} = \frac{(x - b) \frac{d (x - a)}{d x} - (x - a) \frac{d (x - b)}{d x}}{(x - b)^{2}}$

$\frac{d (\frac{x - a}{x - b})}{d x} = \frac{(x - b) - (x - a)}{(x - b)^{2}}$

$\frac{d (\frac{x - a}{x - b})}{d x} = \frac{a - b}{(x - b)^{2}}$

Hence

$f^{'} (x) = \frac{a - b}{(x - b)^{2}}$

Question:8 Find the derivative of $\frac{x^{n} - a^{n}}{x - a}$ for some constant a.

Answer:

Given,

$f (x) = \frac{x^{n} - a^{n}}{x - a}$

Now, as we know the quotient rule of derivative,

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} \frac{d y_{1}}{d x} - y_{1} \frac{d y_{2}}{d x}}{y_{2}^{2}}$

So, applying this rule, we get

$\frac{d (\frac{x^{n} - a^{n}}{x - a})}{d x} = \frac{(x - a) \frac{d (x^{n} - a^{n})}{d x} - (x^{n} - a^{n}) \frac{d (x - a)}{d x}}{(x - a)^{2}}$

$\frac{d (\frac{x^{n} - a^{n}}{x - a})}{d x} = \frac{(x - a) n x^{n - 1} - (x^{n} - a^{n})}{(x - a)^{2}}$

$\frac{d (\frac{x^{n} - a^{n}}{x - a})}{d x} = \frac{n x^{n} - a n x^{n - 1} - x^{n} + a^{n}}{(x - a)^{2}}$

Hence

$f^{'} (x) = \frac{n x^{n} - a n x^{n - 1} - x^{n} + a^{n}}{(x - a)^{2}}$

Question:9(i) Find the derivative of $2 x - 3 / 4$

Answer:

Given:

$f (x) = 2 x - 3 / 4$

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

And the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying that property, we get

$f^{'} (x) = 2 - 0$

$f^{'} (x) = 2$

Question:9(ii) Find the derivative of $(5 x^{3} + 3 x - 1) (x - 1)$

Answer:

Given.

$f (x) = (5 x^{3} + 3 x - 1) (x - 1) = 5 x^{4} + 3 x^{2} - x - 5 x^{3} - 3 x + 1$

$f (x) = 5 x^{4} - 5 x^{3} + 3 x^{2} - 4 x + 1$

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

And the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying that property, we get

$f^{'} (x) = 5 (4) x^{3} - 5 (3) x^{2} + 3 (2) x - 4 + 0$

$f^{'} (x) = 20 x^{3} - 15 x^{2} + 6 x - 4$

Question:9(iii) Find the derivative of $x^{- 3} (5 + 3 x)$

Answer:

Given

$f (x) = x^{- 3} (5 + 3 x) = 5 x^{- 3} + 3 x^{- 2}$

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

And the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying that property, we get

$f^{'} (x) = (- 3) 5 x^{- 4} + 3 (- 2) x^{- 3}$

$f^{'} (x) = - 15 x^{- 4} - 6 x^{- 3}$

Question:9(iv) Find the derivative of $x^{5} (3 - 6 x^{- 9})$

Answer:

Given

$f (x) = x^{5} (3 - 6 x^{- 9}) = 3 x^{5} - 6 x^{- 4}$

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

And the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying that property, we get

$f^{'} (x) = (5) 3 x^{4} - 6 (- 4) x^{- 5}$

$f^{'} (x) = 15 x^{4} + 24 x^{- 5}$

Question:9(v) Find the derivative of $x^{- 4} (3 - 4 x^{- 5})$

Answer:

Given

$f (x) = x^{- 4} (3 - 4 x^{- 5}) = 3 x^{- 4} - 4 x^{- 9}$

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

And the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying that property, we get

$f^{'} (x) = (- 4) 3 x^{- 5} - (- 9) 4 x^{- 10}$

$f^{'} (x) = - 12 x^{- 5} + 36 x^{- 10}$

Question:9(vi) Find the derivative of $\frac{2}{x + 1} - \frac{x^{2}}{3 x - 1}$

Answer:

Given

$f (x) = \frac{2}{x + 1} - \frac{x^{2}}{3 x - 1}$

As we know, the quotient rule of the derivative is:

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} \frac{d y_{1}}{d x} - y_{1} \frac{d y_{2}}{d x}}{y_{2}^{2}}$

And the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

So, applying this rule, we get

$\frac{d (\frac{2}{x + 1} - \frac{x^{2}}{3 x - 1})}{d x} = \frac{(x + 1) \frac{d (2)}{d x} - 2 \frac{d (x + 1)}{d x}}{(x + 1)^{2}} - \frac{(3 x - 1) \frac{d (x^{2})}{d x} - x^{2} \frac{d (3 x - 1)}{d x}}{(3 x - 1)^{2}}$

$\frac{d (\frac{2}{x + 1} - \frac{x^{2}}{3 x - 1})}{d x} = \frac{- 2}{(x + 1)^{2}} - \frac{(3 x - 1) 2 x - x^{2} 3}{(3 x - 1)^{2}}$

$\frac{d (\frac{2}{x + 1} - \frac{x^{2}}{3 x - 1})}{d x} = \frac{- 2}{(x + 1)^{2}} - \frac{6 x^{2} - 2 x - 3 x^{2}}{(3 x - 1)^{2}}$

$\frac{d (\frac{2}{x + 1} - \frac{x^{2}}{3 x - 1})}{d x} = \frac{- 2}{(x + 1)^{2}} - \frac{3 x^{2} - 2 x}{(3 x - 1)^{2}}$

Hence

$f^{'} (x) = \frac{- 2}{(x + 1)^{2}} - \frac{3 x^{2} - 2 x}{(3 x - 1)^{2}}$

Question:10 Find the derivative of $\cos x$ from the first principle.

Answer:

Given,

f(x)= $\cos x$

Now, as we know, the derivative of any function at x is

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{\cos (x + h) - \cos (x)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{\cos (x) \cos (h) - \sin (x) \sin (h) - \cos (x)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{\cos (x) \cos (h) - \cos (x)}{h} - \frac{\sin (x) \sin (h)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{\cos (x) (\cos (h) - 1)}{h} - \frac{\sin (x) \sin (h)}{h}$

$f^{'} (x) = lim_{h \to 0} \cos (x) \frac{- 2 s i n^{2} (h / 2)}{h} \cdot - \sin (x) \frac{s i n h}{h}$

$f^{'} (x) = \cos (x) (0) - s i n x (1)$

$f^{'} (x) = - \sin (x)$

Question:11(i) Find the derivative of the following functions: $\sin x \cos x$

Answer:

Given,

f(x)= $\sin x \cos x$

Now, as we know the product rule of derivative,

$\frac{d (y_{1} y_{2})}{d x} = y_{1} \frac{d y_{2}}{d x} + y_{2} \frac{d y_{1}}{d x}$

So, applying the rule here,

$\frac{d (\sin x \cos x)}{d x} = \sin x \frac{d \cos x}{d x} + \cos x \frac{d \sin x}{d x}$

$\frac{d (\sin x \cos x)}{d x} = \sin x (- \sin x) + \cos x (\cos x)$

$\frac{d (\sin x \cos x)}{d x} = - \sin^{2} x + \cos^{2} x$

$\frac{d (\sin x \cos x)}{d x} = \cos 2 x$

Question:11(ii) Find the derivative of the following functions: $\sec x$

Answer:

Given

$f (x) = \sec x = \frac{1}{\cos x}$

Now, as we know the quotient rule of derivative,

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} \frac{d y_{1}}{d x} - y_{1} \frac{d y_{2}}{d x}}{y_{2}^{2}}$

So, applying this rule, we get

$\frac{d (\frac{1}{\cos x})}{d x} = \frac{\cos x \frac{d (1)}{d x} - 1 \frac{d (\cos x)}{d x}}{\cos^{2} x}$

$\frac{d (\frac{1}{\cos x})}{d x} = \frac{- 1 (- \sin x)}{\cos^{2} x}$

$\frac{d (\frac{1}{\cos x})}{d x} = \frac{\sin x}{\cos^{2} x} = \frac{\sin x}{\cos x} \frac{1}{\cos x}$

$\frac{d (\sec x)}{d x} = \tan x \sec x$

Question:11 (iii) Find the derivative of the following functions: $5 \sec x + 4 \cos x$

Answer:

Given

$f (x) = 5 \sec x + 4 \cos x$

As we know, the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying the property, we get

$\frac{d (5 \sec x + 4 \cos x)}{d x} = \frac{d (5 \sec x)}{d x} + \frac{d (4 \cos x)}{d x}$

$\frac{d (5 \sec x + 4 \cos x)}{d x} = 5 \tan x \sec x + 4 (- \sin x)$

$\frac{d (5 \sec x + 4 \cos x)}{d x} = 5 \tan x \sec x - 4 \sin x$

Question:11(iv) Find the derivative of the following functions: $\csc x$

Answer:

Given :

$f (x) = \csc x = \frac{1}{\sin x}$

Now, as we know the quotient rule of derivative,

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} \frac{d y_{1}}{d x} - y_{1} \frac{d y_{2}}{d x}}{y_{2}^{2}}$

So, applying this rule, we get

$\frac{d (\frac{1}{\sin x})}{d x} = \frac{(\sin x) \frac{d (1)}{d x} - 1 \frac{d (\sin x)}{d x}}{(\sin x)^{2}}$

$\frac{d (\frac{1}{\sin x})}{d x} = \frac{- 1 (\cos x)}{(\sin x)^{2}}$

$\frac{d (\frac{1}{\sin x})}{d x} = - \frac{(\cos x)}{(\sin x)} \frac{1}{\sin x}$

$\frac{d (\csc x)}{d x} = - \cot x \csc x$

Question:11(v) Find the derivative of the following functions: $3 \cot x + 5 \csc x$

Answer:

Given,

$f (x) = 3 \cot x + 5 \csc x$

As we know, the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying the property,

$\frac{d (3 \cot x + 5 \csc x)}{d x} = \frac{d (3 \cot x)}{d x} + \frac{d (5 \csc x)}{d x}$

$\frac{d (3 \cot x + 5 \csc x)}{d x} = 3 \frac{d (\frac{\cos x}{\sin x})}{d x} + \frac{d (5 \csc x)}{d x}$

$\frac{d (3 \cot x + 5 \csc x)}{d x} = - 5 \csc x \cot x + 3 \frac{d (\frac{\cos x}{\sin x})}{d x}$

Now, as we know the quotient rule of the derivative,

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} \frac{d y_{1}}{d x} - y_{1} \frac{d y_{2}}{d x}}{y_{2}^{2}}$

So, applying this rule, we get

$\frac{d (3 \cot x + 5 \csc x)}{d x} = - 5 \csc x \cot x + 3 [\frac{\sin x \frac{d (\cos x)}{d x} - \cos x (\frac{d (\sin x)}{d x})}{\sin^{2} x}]$

$\frac{d (3 \cot x + 5 \csc x)}{d x} = - 5 \csc x \cot x + 3 [\frac{\sin x (- \sin x) - \cos x (\cos x)}{\sin^{2} x}]$

$\frac{d (3 \cot x + 5 \csc x)}{d x} = - 5 \csc x \cot x + 3 [\frac{- \sin^{2} x - \cos^{2} x}{\sin^{2} x}]$

$\frac{d (3 \cot x + 5 \csc x)}{d x} = - 5 \csc x \cot x - 3 [\frac{\sin^{2} x + \cos^{2} x}{\sin^{2} x}]$

$\frac{d (3 \cot x + 5 \csc x)}{d x} = - 5 \csc x \cot x - 3 [\frac{1}{\sin^{2} x}]$

$\frac{d (3 \cot x + 5 \csc x)}{d x} = - 5 \csc x \cot x - 3 \csc^{2} x$

Question:11(vi) Find the derivative of the following functions: $5 \sin x - 6 \cos x + 7$

Answer:

Given,

$f (x) = 5 \sin x - 6 \cos x + 7$

Now, as we know the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

So, applying the property,

$f^{'} (x) = 5 \cos x - 6 (- \sin x) + 0$

$f^{'} (x) = 5 \cos x + 6 (\sin x)$

$f^{'} (x) = 5 \cos x + 6 \sin x$

Question:11(vii) Find the derivative of the following functions: $2 \tan x - 7 \sec x$

Answer:

Given

$f (x) = 2 \tan x - 7 \sec x$

As we know, the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying this property,

$\frac{d (2 \tan x + 7 \sec x)}{d x} = 2 \frac{d \tan x}{d x} + 7 \frac{d \sec x}{d x}$

$\frac{d (2 \tan x + 7 \sec x)}{d x} = 2 \sec^{2} x + 7 (- \sec x \tan x)$

$\frac{d (2 \tan x + 7 \sec x)}{d x} = 2 \sec^{2} x - 7 \sec x \tan x$

NCERT Limits and Derivatives Class 11 Solutions: Exercise: Miscellaneous Exercise

Page Number: 253-254

Total Questions: 30

Question:1(i) Find the derivative of the following functions from the first principle: -x

Answer:

Given.

f(x)=-x

Now, as we know, the derivative of any function at x is

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{- (x + h) - (- x)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{- h}{h}$

$f^{'} (x) = - 1$

Question 1 (ii) Find the derivative of the following functions from the first principle: $(- x)^{- 1}$

Answer:

Given.

f(x)= $(- x)^{- 1}$

Now, as we know, the derivative of any function at x is

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{- (x + h)^{- 1} - (- x)^{- 1}}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{- \frac{1}{x + h} + \frac{1}{x}}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{\frac{- x + x + h}{x (x + h)}}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{\frac{h}{(x + h) (x)}}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{1}{x (x + h)}$

$f^{'} (x) = \frac{1}{x^{2}}$

Question:1(iii) Find the derivative of the following functions from the first principle: $\sin (x + 1)$

Answer:

Given.

$f (x) = \sin (x + 1)$

Now, as we know, the derivative of any function at x is

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{\sin (x + h + 1) - \sin (x + 1)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{2 \cos (\frac{x + h + 1 + x + 1}{2}) \sin (\frac{x + h + 1 - x - 1}{2})}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{2 \cos (\frac{2 x + h + 2}{2}) \sin (\frac{h}{2})}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{\cos (\frac{2 x + h + 2}{2}) \sin (\frac{h}{2})}{\frac{h}{2}}$

$f^{'} (x) = \cos (\frac{2 x + 0 + 2}{2}) \times 1$

$f^{'} (x) = \cos (x + 1)$

Question:1(iv) Find the derivative of the following functions from the first principle: $\cos (x - π / 8)$

Answer:

Given.

$f (x) = \cos (x - π / 8)$

Now, as we know, the derivative of any function at x is

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{\cos (x + h - π / 8) - c o s (x - π / 8)}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{- 2 \sin (\frac{x + h - π / 8 + x - π / 8}{2}) \sin (\frac{x + h - π / 8 - x + π / 8}{2})}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{- 2 \sin (\frac{2 x + h - π / 4}{2}) \sin (\frac{h}{2})}{h}$

$f^{'} (x) = lim_{h \to 0} \frac{- \sin (\frac{2 x + h - π / 4}{2}) \sin (\frac{h}{2})}{\frac{h}{2}}$

$f^{'} (x) = \sin (\frac{2 x + 0 - π / 4}{2}) \times 1$

$f^{'} (x) = - \sin (x - π / 8)$

Question:2 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (x+a )

Answer:

Given

f(x)= x + a

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

Applying that property, we get

$f^{'} (x) = 1 + 0$

$f^{'} (x) = 1$

Question:3 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $(p x + q) (\frac{r}{x} + s)$

Answer:

Given

$f (x) = (p x + q) (\frac{r}{x} + s)$

$f (x) = p r + p s x + \frac{q r}{x} + q s$

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

Applying that property, we get

$f^{'} (x) = 0 + p s + \frac{- q r}{x^{2}} + 0$

$f^{'} (x) = p s + q (\frac{- r}{x^{2}})$

$f^{'} (x) = p s - (\frac{q r}{x^{2}})$

Question:4 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $(a x + b) (c x + d)^{2}$

Answer:

Given,

$f (x) = (a x + b) (c x + d)^{2}$

$f (x) = (a x + b) (c^{2} x^{2} + 2 c d x + d^{2})$

$f (x) = a c^{2} x^{3} + 2 a c d x^{2} + a d^{2} x + b c^{2} x^{2} + 2 b c d x + b d^{2}$

Now,

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

And the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying that property, we get

$f^{'} (x) = 3 a c^{2} x^{2} + 4 a c d x + a d^{2} + 2 b c^{2} x + 2 b c d + 0$

$f^{'} (x) = 3 a c^{2} x^{2} + 4 a c d x + a d^{2} + 2 b c^{2} x + 2 b c d$

$f^{'} (x) = 3 a c^{2} x^{2} + (4 a c d + 2 b c^{2}) x + a d^{2} + 2 b c d$

Question:5 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $a x + b / c x + d$

Answer:

Given,

$f (x) = \frac{a x + b}{c x + d}$

Now, as we know, the derivative of any function

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} d (\frac{d y_{1}}{d x}) - y_{1} (\frac{d y_{2}}{d x})}{y_{2}^{2}}$

Hence, the derivative of f(x) is

$\frac{d (\frac{a x + b}{c x + d})}{d x} = \frac{(c x + d) d (\frac{d (a x + b)}{d x}) - (a x + b) (\frac{d (c x + d)}{d x})}{(c x + d)^{2}}$

$\frac{d (\frac{a x + b}{c x + d})}{d x} = \frac{(c x + d) a - (a x + b) c}{(c x + d)^{2}}$

$\frac{d (\frac{a x + b}{c x + d})}{d x} = \frac{a c x + a d - a c x - b c}{(c x + d)^{2}}$

$\frac{d (\frac{a x + b}{c x + d})}{d x} = \frac{a d - b c}{(c x + d)^{2}}$

Hence, the Derivative of the function is

$\frac{a d - b c}{(c x + d)^{2}}$ .

Question:6 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}$

Answer:

Given,

$f (x) = \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}$

can also be written as

$f (x) = \frac{x + 1}{x - 1}$

Now, as we know, the derivative of any function

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} d (\frac{d y_{1}}{d x}) - y_{1} (\frac{d y_{2}}{d x})}{y_{2}^{2}}$

Hence, the derivative of f(x) is

$\frac{d (\frac{x + 1}{x - 1})}{d x} = \frac{(x - 1) d (\frac{d (x + 1)}{d x}) - (x + 1) (\frac{d (x - 1)}{d x})}{(x - 1)^{2}}$

$\frac{d (\frac{x + 1}{x - 1})}{d x} = \frac{(x - 1) 1 - (x + 1) 1}{(x - 1)^{2}}$

$\frac{d (\frac{x + 1}{x - 1})}{d x} = \frac{x - 1 - x - 1}{(x - 1)^{2}}$

$\frac{d (\frac{x + 1}{x - 1})}{d x} = \frac{- 2}{(x - 1)^{2}}$

Hence, the Derivative of the function is

$\frac{- 2}{(x - 1)^{2}}$

Question:7 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{1}{a x^{2} + b x + c}$

Answer:

Given,

$f (x) = \frac{1}{a x^{2} + b x + c}$

Now, as we know, the derivative of any such function is given by

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} d (\frac{d y_{1}}{d x}) - y_{1} (\frac{d y_{2}}{d x})}{y_{2}^{2}}$

Hence, the derivative of f(x) is

$\frac{d (\frac{1}{a x^{2} + b x + c})}{d x} = \frac{(a x^{2} + b x + c) d (\frac{d (1)}{d x}) - 1 (\frac{d (a x^{2} + b x + c)}{d x})}{(a x^{2} + b x + c)^{2}}$

$\frac{d (\frac{1}{a x^{2} + b x + c})}{d x} = \frac{0 - (2 a x + b)}{(a x^{2} + b x + c)^{2}}$

$\frac{d (\frac{1}{a x^{2} + b x + c})}{d x} = - \frac{(2 a x + b)}{(a x^{2} + b x + c)^{2}}$

Question:8 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{a x + b}{p x^{2} + q x + r}$

Answer:

Given,

$f (x) = \frac{a x + b}{p x^{2} + q x + r}$

Now, as we know, the derivative of any function

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} d (\frac{d y_{1}}{d x}) - y_{1} (\frac{d y_{2}}{d x})}{y_{2}^{2}}$

Hence, the derivative of f(x) is

$\frac{d (\frac{a x + b}{p x^{2} + q x + r})}{d x} = \frac{(p x^{2} + q x + r) d (\frac{d (a x + b)}{d x}) - (a x + b) (\frac{d (p x^{2} + q x + r)}{d x})}{(p x^{2} + q x + r)^{2}}$

$\frac{d (\frac{a x + b}{p x^{2} + q x + r})}{d x} = \frac{(p x^{2} + q x + r) a - (a x + b) (2 p x + q)}{(p x^{2} + q x + r)^{2}}$

$\frac{d (\frac{a x + b}{p x^{2} + q x + r})}{d x} = \frac{a p x^{2} + a q x + a r - 2 a p x^{2} - a q x - 2 b p x - b q}{(p x^{2} + q x + r)^{2}}$

$\frac{d (\frac{a x + b}{p x^{2} + q x + r})}{d x} = \frac{- a p x^{2} + a r - 2 b p x - b q}{(p x^{2} + q x + r)^{2}}$

Question:9 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{p x^{2} + q x + r}{a x + b}$

Answer:

Given,

$f (x) = \frac{p x^{2} + q x + r}{a x + b}$

Now, as we know, the derivative of any function

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} d (\frac{d y_{1}}{d x}) - y_{1} (\frac{d y_{2}}{d x})}{y_{2}^{2}}$

Hence, the derivative of f(x) is

$\frac{d (\frac{p x^{2} + q x + r}{a x + b})}{d x} = \frac{(a x + b) d (\frac{d (p x^{2} + q x + r)}{d x}) - (p x^{2} + q x + r) (\frac{d (a x + b)}{d x})}{(a x + b)^{2}}$

$\frac{d (\frac{p x^{2} + q x + r}{a x + b})}{d x} = \frac{(a x + b) (2 p x + q) - (p x^{2} + q x + r) (a)}{(a x + b)^{2}}$

$\frac{d (\frac{p x^{2} + q x + r}{a x + b})}{d x} = \frac{2 a p x^{2} + a q x + 2 b p x + b q - a p x^{2} - a q x - a r}{(a x + b)^{2}}$

$\frac{d (\frac{p x^{2} + q x + r}{a x + b})}{d x} = \frac{a p x^{2} + 2 b p x + b q - a r}{(a x + b)^{2}}$

Question 10 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{a}{x^{4}} - \frac{b}{x^{2}} + \cos x$

Answer:

Given

$f (x) = \frac{a}{x^{4}} - \frac{b}{x^{2}} + \cos x$

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

And the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying that property, we get

$f^{'} (x) = \frac{- 4 a}{x^{5}} - (\frac{- 2 b}{x^{3}}) + (- \sin x)$

$f^{'} (x) = \frac{- 4 a}{x^{5}} + \frac{2 b}{x^{3}} - \sin x$

Question 11 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers): $4 \sqrt{x} - 2$

Answer:

Given

$f (x) = 4 \sqrt{x} - 2$

It can also be written as

$f (x) = 4 x^{\frac{1}{2}} - 2$

Now,

As we know, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

And the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying that property, we get

$f^{'} (x) = 4 (\frac{1}{2}) x^{- \frac{1}{2}} - 0$

$f^{'} (x) = 2 x^{- \frac{1}{2}}$

$f^{'} (x) = \frac{2}{\sqrt{x}}$

Question 12 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$(a x + b)^{n}$

Answer:

Given

$f (x) = (a x + b)^{n}$

Now, ass we know the chain rule of derivative,

$[f (g (x))]^{'} = f^{'} (g (x)) \times g^{'} (x)$

And, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

Also the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying those properties, we get,

$f^{'} (x) = n (a x + b)^{n - 1} \times a$

$f^{'} (x) = a n (a x + b)^{n - 1}$

Question:13 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $(a x + b)^{n} (c x + d)^{m}$

Answer:

Given

$f (x) = (a x + b)^{n} (c x + d)^{m}$

Now, ass we know the chain rule of derivative,

$[f (g (x))]^{'} = f^{'} (g (x)) \times g^{'} (x)$

And the Multiplication property of the derivative,

$\frac{d (y_{1} y_{2})}{d x} = y_{1} \frac{d y_{2}}{d x} + y_{2} \frac{d y_{1}}{d x}$

And, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

Also the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying those properties, we get,

$f^{'} (x) = (a x + b)^{n} (m (c x + d)^{m - 1}) + (c x + d) (n (a x + b)^{n - 1})$

$f^{'} (x) = m (a x + b)^{n} (c x + d)^{m - 1} + n (c x + d) (a x + b)^{n - 1}$

Question 14 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers): $\sin (x + a)$

Answer:

Given,

$f (x) = \sin (x + a)$

Now, ass we know the chain rule of derivative,

$[f (g (x))]^{'} = f^{'} (g (x)) \times g^{'} (x)$

Applying this property, we get,

$f^{'} (x) = \cos (x + a) \times 1$

$f^{'} (x) = \cos (x + a)$

Question:15 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\csc x \cot x$

Answer:

Given,

$f (x) = \csc x \cot x$

The Multiplication Property of the derivative,

$\frac{d (y_{1} y_{2})}{d x} = y_{1} \frac{d y_{2}}{d x} + y_{2} \frac{d y_{1}}{d x}$

Applying the property

$\frac{d (\csc x) (\cot x))}{d x} = \csc x \frac{d \cot x}{d x} + \cot x \frac{d \csc x}{d x}$

$\frac{d (\csc x) (\cot x))}{d x} = \csc x (- \csc^{2} x) + \cot x (- \csc x \cot x)$

$\frac{d (\csc x) (\cot x))}{d x} = - \csc^{3} x - \cot^{2} x \csc x$

Hence, thee, the derivative of the function is $- \csc^{3} x - \cot^{2} x \csc x$ .

Question:16 Find the derivative of the following functions (it is to be understood that a, b, c, d,p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{\cos x}{1 + \sin x}$

Answer:

Given,

$f (x) = \frac{\cos x}{1 + \sin x}$

Now, as we know, the derivative of any function

$\frac{d (\frac{\cos x}{1 + \sin x})}{d x} = \frac{(1 + \sin x) d (\frac{d \cos x}{d x}) - \cos x (\frac{d (1 + \sin x)}{d x})}{(1 + \sin x)^{2}}$

Hence, thee derivative of f(x) is

$\frac{d (\frac{\cos x}{1 + \sin x})}{d x} = \frac{(1 + \sin x) (- \sin x) - \cos x (\cos x)}{(1 + \sin x)^{2}}$

$\frac{d (\frac{\cos x}{1 + \sin x})}{d x} = \frac{- \sin x - \sin^{2} x - \cos^{2} x}{(1 + \sin x)^{2}}$

$\frac{d (\frac{\cos x}{1 + \sin x})}{d x} = \frac{- \sin x - 1}{(1 + \sin x)^{2}}$

$\frac{d (\frac{\cos x}{1 + \sin x})}{d x} = - \frac{1}{(1 + \sin x)}$

Question:17 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\frac{\sin x + \cos x}{\sin x - \cos x}$

Answer:

Given

$f (x) = \frac{\sin x + \cos x}{\sin x - \cos x}$

can also be written as

$f (x) = \frac{\tan x + 1}{\tan x - 1}$

which further can be written as

$f (x) = - \frac{\tan x + t a n (π / 4)}{1 - \tan (π / 4) \tan x}$

$f (x) = - \tan (x - π / 4)$

Now,

$f^{'} (x) = - \sec^{2} (x - π / 4)$

$f^{'} (x) = - \frac{1}{\cos^{2} (x - π / 4)}$

Question:18 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{\sec x - 1}{\sec x + 1}$

Answer:

Given,

$f (x) = \frac{\sec x - 1}{\sec x + 1}$

which also can be written as

$f (x) = \frac{1 - \cos x}{1 + \cos x}$

Now,

As we know, the derivative of such a function

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} d (\frac{d y_{1}}{d x}) - y_{1} (\frac{d y_{2}}{d x})}{y_{2}^{2}}$

So, the derivative of the function is,

$\frac{d (\frac{1 - \cos x}{1 + \cos x})}{d x} = \frac{(1 + \cos x) d (\frac{d (1 - \cos x)}{d x}) - (1 - \cos x) (\frac{d (1 + \cos x)}{d x})}{(1 + \cos x)^{2}}$

$\frac{d (\frac{1 - \cos x}{1 + \cos x})}{d x} = \frac{(1 + \cos x) (- (- \sin x)) - (1 - \cos x) (- \sin x)}{(1 + \cos x)^{2}}$

$\frac{d (\frac{1 - \cos x}{1 + \cos x})}{d x} = \frac{\sin x + \sin x \cos x + \sin x - \cos x \sin x}{(1 + \cos x)^{2}}$

$\frac{d (\frac{1 - \cos x}{1 + \cos x})}{d x} = \frac{2 \sin x}{(1 + \cos x)^{2}}$

This can also be written as

$\frac{d (\frac{1 - \cos x}{1 + \cos x})}{d x} = \frac{2 \sec x \tan x}{(1 + \sec x)^{2}}$ .

Question:19 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\sin^{n} x$

Answer:

Given,

$f (x) = \sin^{n} x$

Now, as we know the chain rule of derivative,

$[f (g (x))]^{'} = f^{'} (g (x)) \times g^{'} (x)$

And, the property,

$f^{'} (x^{n}) = n x^{n - 1}$

Applying those properties, we get

$f^{'} (x) = n \sin^{n - 1} x \cos x$

Hence Derivative of the given function is $n \sin^{n - 1} x \cos x$

Question 20 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{a + b \sin x}{c + d \cos x}$

Answer:

Given Function

$f (x) = \frac{a + b \sin x}{c + d \cos x}$

Now, as we know, the derivative of any function of this type is:

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} d (\frac{d y_{1}}{d x}) - y_{1} (\frac{d y_{2}}{d x})}{y_{2}^{2}}$

Hence, the derivative of the given function will be:

$\frac{d (\frac{a + b \sin x}{c + d \cos x})}{d x} = \frac{(c + d \cos x) (\frac{d (a + b \sin x)}{d x}) - (a + b \sin x) (\frac{d (c + d \cos x x)}{d x})}{(c + d \cos x)^{2}}$

$\frac{d (\frac{a + b \sin x}{c + d \cos x})}{d x} = \frac{(c + d \cos x) (b \cos x) - (a + b \sin x) (d (- \sin x))}{(c + d \cos x)^{2}}$

$\frac{d (\frac{a + b \sin x}{c + d \cos x})}{d x} = \frac{c b \cos x + b d \cos^{2} x + a d \sin x + b d \sin^{2} x}{(c + d \cos x)^{2}}$

$\frac{d (\frac{a + b \sin x}{c + d \cos x})}{d x} = \frac{c b \cos x + a d \sin x + b d (\sin^{2} x + \cos^{2} x)}{(c + d \cos x)^{2}}$

$\frac{d (\frac{a + b \sin x}{c + d \cos x})}{d x} = \frac{c b \cos x + a d \sin x + b d}{(c + d \cos x)^{2}}$

Question:21 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{\sin (x + a)}{\cos x}$

Answer:

Given,

$f (x) = \frac{\sin (x + a)}{\cos x}$

Now, as we know, the derivative of any function

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} d (\frac{d y_{1}}{d x}) - y_{1} (\frac{d y_{2}}{d x})}{y_{2}^{2}}$

Hence, the derivative of the given function is:

$\frac{d (\frac{\sin (x + a)}{\cos x})}{d x} = \frac{(\cos x) (\frac{d (\sin (x + a))}{d x}) - \sin (x + a) (\frac{d (\cos x)}{d x})}{(\cos x)^{2}}$

$\frac{d (\frac{\sin (x + a)}{\cos x})}{d x} = \frac{(\cos x) (\cos (x + a)) - \sin (x + a) (- \sin (x))}{(\cos x)^{2}}$

$\frac{d (\frac{\sin (x + a)}{\cos x})}{d x} = \frac{(\cos x) (\cos (x + a)) + \sin (x + a) (\sin (x))}{(\cos x)^{2}}$

$\frac{d (\frac{\sin (x + a)}{\cos x})}{d x} = \frac{\cos (x + a - x)}{(\cos x)^{2}}$

$\frac{d (\frac{\sin (x + a)}{\cos x})}{d x} = \frac{\cos (a)}{(\cos x)^{2}}$

Question:22 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers): $x^{4} (5 \sin x - 3 \cos x)$

Answer:

Given

$f (x) = x^{4} (5 \sin x - 3 \cos x)$

Now, as we know, the Multiplication property of the derivative,

$\frac{d (y_{1} y_{2})}{d x} = y_{1} \frac{d y_{2}}{d x} + y_{2} \frac{d y_{1}}{d x}$

Hence, the derivative of the given function is:

$\frac{d (x^{4} (5 \sin x - 3 \cos x))}{d x} = x^{4} \frac{d (5 \sin x - 3 \cos x)}{d x} + (5 \sin x - 3 \cos x) \frac{d x^{4}}{d x}$

$\frac{d (x^{4} (5 \sin x - 3 \cos x))}{d x} = x^{4} (5 \cos x + 3 \sin x) + (5 \sin x - 3 \cos x) 4 x^{3}$

$\frac{d (x^{4} (5 \sin x - 3 \cos x))}{d x} = 5 x^{4} \cos x + 3 x^{4} \sin x + 20 x^{3} \sin x - 12 x^{3} \cos x$

Question 23 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $(x^{2} + 1) \cos x$

Answer:

Given

$f (x) = (x^{2} + 1) \cos x^{}$

Now, as we know the product rule of derivative,

$\frac{d (y_{1} y_{2})}{d x} = y_{1} \frac{d y_{2}}{d x} + y_{2} \frac{d y_{1}}{d x}$

The derivative of the given function is

$\frac{d ((x^{2} + 1) \cos x)}{d x} = (x^{2} + 1) \frac{d \cos x}{d x} + \cos x \frac{d (x^{2} + 1)}{d x}$

$\frac{d ((x^{2} + 1) \cos x)}{d x} = (x^{2} + 1) (- \sin x) + \cos x (2 x)$

$\frac{d ((x^{2} + 1) \cos x)}{d x} = - x^{2} \sin x - \sin x + 2 x \cos x$

Question:24 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $(a x^{2} + \sin x) (p + q \cos x)$

Answer:

Given,

$f (x) = (a x^{2} + \sin x) (p + q \cos x)$

Now, as we know, the Multiplication property of the derivative (the product rule)

$\frac{d (y_{1} y_{2})}{d x} = y_{1} \frac{d y_{2}}{d x} + y_{2} \frac{d y_{1}}{d x}$

And also the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying those properties, we get,

$\frac{d ((a x^{2} + \sin x) (p + q \cos x))}{d x} = (a x^{2} + \sin x) \frac{d (p + q \cos x)}{d x} + (p + q x) \frac{d (a x^{2} + s i n x)}{d x}$

$\frac{d ((a x^{2} + \sin x) (p + q \cos x))}{d x} = (a x^{2} + \sin x) (- q \sin x) + (p + q x) (2 a x + \cos x)$

$f^{'} (x) = - a q x^{2} \sin x - q \sin^{2} x + 2 a p x + p \cos x + 2 a q x^{2} + q x \cos x$

$f^{'} (x) = x^{2} (- a q \sin x + 2 a q) + x (2 a p + q \cos x) + p \cos x - q \sin^{2} x$

Question:25 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $(x + \cos x) (x - \tan x)$

Answer:

Given,

$f (x) = (x + \cos x) (x - \tan x)$

And the Multiplication property of the derivative,

$\frac{d (y_{1} y_{2})}{d x} = y_{1} \frac{d y_{2}}{d x} + y_{2} \frac{d y_{1}}{d x}$

Also the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying those properties, we get,

$\frac{d ((x + \cos x) (x - \tan x))}{d x} = (x + \cos x) \frac{d (x - \tan x)}{d x} + (x - \tan x) \frac{d (x + \cos x)}{d x}$

$= (x + \cos x) (1 - \sec^{2} x) + (x - \tan x) (1 - \sin x)$

$= (x + \cos x) (- \tan^{2} x) + (x - \tan x) (1 - \sin x)$

$= (- \tan^{2} x) (x + \cos x) + (x - \tan x) (1 - \sin x)$

Question:26 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r, and s are fixed non-zero constants and m and n are integers):

$\frac{4 x + 5 \sin x}{3 x + 7 \cos x}$

Answer:

Given,

$f (x) = \frac{4 x + 5 \sin x}{3 x + 7 \cos x}$

Now, as we know, the derivative of any function

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} d (\frac{d y_{1}}{d x}) - y_{1} (\frac{d y_{2}}{d x})}{y_{2}^{2}}$

Also the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying those properties, we get

$\frac{d (\frac{4 x + 5 \sin x}{3 x + 7 \cos x})}{d x} = \frac{(3 x + 7 \cos x) d (\frac{d (4 x + 5 \sin x)}{d x}) - (4 x + 5 \sin x) (\frac{d (3 x + 7 \cos x)}{d x})}{(3 x + 7 \cos x)^{2}}$

$\frac{d (\frac{4 x + 5 \sin x}{3 x + 7 \cos x})}{d x} = \frac{(3 x + 7 \cos x) (4 + 5 \cos x) - (4 x + 5 \sin x) (3 - 7 \sin x)}{(3 x + 7 \cos x)^{2}}$

$= \frac{12 x + 28 \cos x + 15 x \cos x + 35 \cos^{2} x - 12 x - 15 \sin x + 28 x \sin x + 35 \sin^{2} x}{(3 x + 7 \cos x)^{2}}$

$= \frac{12 x + 28 \cos x + 15 x \cos x - 12 x - 15 \sin x + 28 x \sin x + 35 (\sin^{2} x + \cos^{2} x)}{(3 x + 7 \cos x)^{2}}$

$= \frac{28 \cos x + 15 x \cos x - 15 \sin x + 28 x \sin x + 35}{(3 x + 7 \cos x)^{2}}$

Question:27 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q,,r, and s are fixed non-zero constants and m and n are integers):

$\frac{x^{2} \cos (π / 4)}{\sin x}$

Answer:

Given,

$f (x) = \frac{x^{2} \cos (π / 4)}{\sin x}$

As we know, the derivative of any function

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} d (\frac{d y_{1}}{d x}) - y_{1} (\frac{d y_{2}}{d x})}{y_{2}^{2}}$

Hence, the derivative of the given function is

$\frac{d (\frac{x^{2} \cos (π / 4)}{\sin x})}{d x} = \frac{(\sin x) d (\frac{d (x^{2} \cos (π / 4))}{d x}) - (x^{2} \cos (π / 4)) (\frac{d \sin x}{d x})}{\sin^{2} x}$

$\frac{d (\frac{x^{2} \cos (π / 4)}{\sin x})}{d x} = \frac{(\sin x) (2 x \cos (π / 4)) - (x^{2} \cos (π / 4)) (\cos x)}{\sin^{2} x}$

$\frac{d (\frac{x^{2} \cos (π / 4)}{\sin x})}{d x} = \frac{2 x \sin x \cos (π / 4) - x^{2} \cos x \cos (π / 4)}{\sin^{2} x}$

Question:28 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q,, r, and s are fixed non-zero constants and m and n are integers):

$\frac{x}{1 + \tan x}$

Answer:

Given

$f (x) = \frac{x}{1 + \tan x}$

Noas As we know, the derivative of any function

$\frac{d (\frac{x}{1 + \tan x})}{d x} = \frac{(1 + \tan x) d (\frac{d x}{d x}) - x (\frac{d (1 + \tan x)}{d x})}{(1 + \tan x)^{2}}$

$\frac{d (\frac{x}{1 + \tan x})}{d x} = \frac{(1 + \tan x) (1) - x (\sec^{2} x)}{(1 + \tan x)^{2}}$

$\frac{d (\frac{x}{1 + \tan x})}{d x} = \frac{1 + \tan x - x \sec^{2} x}{(1 + \tan x)^{2}}$

Question:29 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $(x + \sec x) (x - \tan x)$

Answer:

Given

$f (x) = (x + \sec x) (x - \tan x)$

Now, ass we know the Multiplication property of the derivative,

$\frac{d (y_{1} y_{2})}{d x} = y_{1} \frac{d y_{2}}{d x} + y_{2} \frac{d y_{1}}{d x}$

Also the property

$\frac{d (y_{1} + y_{2})}{d x} = \frac{d y_{1}}{d x} + \frac{d y_{2}}{d x}$

Applying those properties, we get,

The derivative of the given function is,

$\frac{d ((x + \sec x) (x - \tan x)}{d x} = (x + \sec x) \frac{d (x - \tan x)}{d x} + (x - \tan x) \frac{d (x + \sec x)}{d x}$

$\frac{d ((x + \sec x) (x - \tan x)}{d x} = (x + \sec x) (1 - \sec^{2} x) + (x - \tan x) (1 + \sec x \tan x)$

Question:30 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q,,r, and s are fixed non-zero constants and m and n are integers):

$\frac{x}{\sin^{n} x}$

Answer:

Given,

$f (x) = \frac{x}{\sin^{n} x}$

As we know, the derivative of any function

$\frac{d (\frac{y_{1}}{y_{2}})}{d x} = \frac{y_{2} d (\frac{d y_{1}}{d x}) - y_{1} (\frac{d y_{2}}{d x})}{y_{2}^{2}}$

As the chain rule of derivatives,

$[f (g (x))]^{'} = f^{'} (g (x)) \times g^{'} (x)$

Hence, the derivative of the given function is

$\frac{d (\frac{x}{\sin^{n} x})}{d x} = \frac{\sin^{n} x d (\frac{d x}{d x}) - x (\frac{d (\sin^{n} x)}{d x})}{s i n^{2 n} x}$

$\frac{d (\frac{x}{\sin^{n} x})}{d x} = \frac{\sin^{n} x (1) - x (n \sin^{n - 1} x \times \cos x)}{s i n^{2 n} x}$

$\frac{d (\frac{x}{\sin^{n} x})}{d x} = \frac{\sin^{n} x - x \cos x n \sin^{n - 1} x}{s i n^{2 n} x}$

$\frac{d (\frac{x}{\sin^{n} x})}{d x} = \frac{\sin^{n - 1} x (\sin x - n x \cos x)}{s i n^{2 n} x}$

$\frac{d (\frac{x}{\sin^{n} x})}{d x} = \frac{(\sin x - n x \cos x)}{s i n^{n + 1} x}$

Students can use the below links to analyze all the exercises separately as the difficulty level of the exercises is easy to moderate level. So after analyzing each exercise, they can attempt the next on their own before checking the answers.

NCERT Solutions for Class 11 Mathematics - Chapter Wise

NCERT Solutions for Class 11 Mathematics Chapter 1: Sets
NCERT Solutions for Class 11 Mathematics Chapter 2: Relations and Functions
NCERT Solutions for Class 11 Mathematics Chapter 3: Trigonometric Functions
NCERT Solutions for Class 11 Mathematics Chapter 4: Complex Numbers and Quadratic Equations
NCERT Solutions for Class 11 Mathematics Chapter 5: Linear Inequalities
NCERT Solutions for Class 11 Mathematics Chapter 6: Permutations and Combinations
NCERT Solutions for Class 11 Mathematics Chapter 7: Binomial Theorem
NCERT Solutions for Class 11 Mathematics Chapter 8: Sequences and Series
NCERT Solutions for Class 11 Mathematics Chapter 9: Straight Lines
NCERT Solutions for Class 11 Mathematics Chapter 10: Conic Sections
NCERT Solutions for Class 11 Mathematics Chapter 11: Introduction to Three-Dimensional Geometry
NCERT Solutions for Class 11 Mathematics Chapter 12: Limits and Derivatives
NCERT Solutions for Class 11 Mathematics Chapter 13: Statistics
NCERT Solutions for Class 11 Mathematics Chapter 14: Probability

Importance of Solving NCERT Questions for Class 11 Chapter 12

Limits and Derivatives is not only a useful chapter in class 11 but also for higher studies and competitive exams. Strengthening basic concepts is a necessity for students so that later they do not face any difficulties solving Limits and derivatives questions in higher studies or competitive examinations.
Some important facts about solving statistics in class 11 are listed below.

Students can study strategically at their own pace after accessing Class 10 Maths NCERT Solutions chapter 12. This will boost their confidence to attempt other questions from this chapter.
Class 11 Maths Chapter 12 NCERT solutions are solved by subject-matter experts and are very reliable at the same time. The solutions provide shortcuts as well as detailed explanations with necessary formulae that will help students to understand the answers better.
These solutions will help students manage their time efficiently in this chapter and understand which questions are easier to approach and which are time-consuming. This will be helpful during the exam.
NCERT solutions for class 11 Maths chapter 12 Limits and Derivatives is designed to give the students step-by-step solutions for a particular question.

NCERT Solutions for Class 11- Subject Wise

Students can check the following links for more in-depth learning.

NCERT Books and NCERT Syllabus

Here is the latest NCERT syllabus, which is useful for students before strategizing their study plan. Also, it contains links to some reference books which are important for further studies.

Frequently Asked Questions (FAQs)

1. What are the important topics covered in NCERT Solutions for Class 11 Maths Chapter 12?

Important Topics in NCERT Solutions for Class 11 Maths Chapter 12 (Limits and Derivatives)

This chapter introduces limits and derivatives, forming the foundation of calculus.

1. Limits: Understanding approaching values, left-hand and right-hand limits, and standard limits like:

$lim_{x \to 0} \frac{\sin x}{x} = 1$

$lim_{x \to 0} (1 + x)^{n} = e^{x}$

2. Limit Evaluation Methods:

Direct substitution, factorization, rationalization.

3. Derivatives: First principle definition:

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

Power rule, sum, product, and quotient rules.

2. How to solve limit problems in Class 11 Maths Chapter 12?

To solve limit problems in Class 11 Maths Chapter 12 (Limits and Derivatives), follow these steps:
1. Substitution: Directly put the given value of $x$ .
2. Factorization: Factor and cancel common terms if you get $\frac{0}{0}$ .
3. Rationalization: Multiply by the conjugate for roots.
4. Trigonometric Limits: Use identities like $lim_{x \to 0} \frac{\sin x}{x} = 1$ .
5. L'Hôpital's Rule (for indeterminate forms): Differentiate numerator and denominator separately.

3. What are the basic formulas of limits and derivatives in Class 11?

Basic Formulas of Limits and Derivatives in Class 11

Limits

1. $lim_{x \to a} c = c$

2. $lim_{x \to a} x = a$

3. $lim_{x \to a} x^{n} = a^{n}$

4. $lim_{x \to 0} \frac{\sin x}{x} = 1$

5. $lim_{x \to 0} \frac{1 - \cos x}{x} = 0$

Derivatives

1. $\frac{d}{d x} (c) = 0$

2. $\frac{d}{d x} (x^{n}) = n x^{n - 1}$

3. $\frac{d}{d x} (\sin x) = \cos x$

4. $\frac{d}{d x} (\cos x) = - \sin x$

5. $\frac{d}{d x} (e^{x}) = e^{x}$

4. How do you find derivatives using the first principle in NCERT Class 11 Maths?

To find derivatives using the first principle, we use the definition:

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$
Steps to Find the Derivative Using the First Principle:
1. Substitute $f (x + h)$ into the formula.
2. Find $f (x + h) - f (x)$ .
3. Simplify the expression and divide by $h$ .
4. Apply the limit $h \to 0$ .

Example: $f (x) = x^{2}$

$f^{'} (x) = lim_{h \to 0} \frac{(x + h)^{2} - x^{2}}{h} = lim_{h \to 0} \frac{x^{2} + 2 x h + h^{2} - x^{2}}{h} = lim_{h \to 0} (2 x + h) = 2 x$

5. What is the easiest way to understand limits and derivatives?

Easiest Way to Understand Limits and Derivatives
1. Limits (Approaching a Value)
Think of limits as "getting closer to a value" without necessarily reaching it.
Example: $lim_{x \to 2} (x^{2}) = 4$ , as $x^{2}$ gets closer to 4 when $x$ approaches 2 .
Use substitution first; if it gives $\frac{0}{0}$ , try factoring or rationalizing.
2. Derivatives (Rate of Change)
The derivative measures how a function changes at a point (slope of a curve).
Formula: $f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ .
Start with simple functions like $x^{2}$ to see patterns in differentiation.

Also Read

NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

NCERT Solutions for Class 11 Maths Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium

NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Biology Solutions Chapter 5 Morphology of Flowering Plants

NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions

NCERT Exemplar Class 11 Chemistry solutions Chapter 11 P-Block Elements

NCERT Class 11 Biology Chapter 21 Notes Neural Control And Coordination- Download PDF Notes

NCERT Class 11 Chemistry Chapter 4 Notes Chemical Bonding and Molecular Structure - Download PDF

Hydrogen Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 9 Notes - Download PDF

The S-Block Elements Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 10 Notes - Download PDF

The P-Block Elements Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 11 Notes - Download PDF

NCERT Class 11 Chemistry Chapter 13 Notes Hydrocarbons - Download PDF

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

Limits and Derivatives Class 11 Questions And Answers PDF Free Download

Limits and Derivatives Class 11 Solutions - Important Formulae

Limits

Properties of Limits:

Standard Limits:

Derivatives:

Standard Derivatives:

Limits and Derivatives Class 11 NCERT Solutions

NCERT Solutions for Class 11 Mathematics - Chapter Wise

Importance of Solving NCERT Questions for Class 11 Chapter 12

NCERT Solutions for Class 11- Subject Wise

NCERT Books and NCERT Syllabus

Frequently Asked Questions (FAQs)

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