NCERT Exemplar Class 11 Maths Solutions Chapter 13 Limits and Derivatives

Q: What are the important topics covered in NCERT Exemplar Class 11 Maths Chapter 13?

Chapter 13 of Class 11 Maths, titled "Limits and Derivatives," covers several important concepts that are foundational to understanding calculus. Key topics include limits of functions, which help in understanding the behavior of functions as they approach a specific point. The chapter also covers continuity and the conditions under which a function is continuous. Another critical concept is derivatives , where you learn how to find the rate of change of a function at a point, which is essential for understanding slopes of curves and real-world applications like speed and acceleration. Additionally, the chapter introduces you to derivatives from first principles , which is a fundamental method of finding derivatives. The chapter also includes problems on algebra of limits , limits involving infinity , and the chain rule for derivatives, which are useful tools in solving more complex calculus problems.

Q: How to solve limit problems in NCERT Exemplar Class 11 Maths Chapter 13?

To solve limit problems in Chapter 13, begin by analyzing the function involved. The first step is to substitute the given value into the function and check if the limit exists. If substituting the value results in an indeterminate form like 0/0?, then you must simplify the expression using algebraic techniques, such as factoring or rationalizing. Another approach is to use standard limit laws and theorems, such as the limit of a sum , product , or quotient of functions . If the expression is complex, you might need to apply trigonometric limits or limits involving infinity . In some cases, you can use L'Hopital's Rule , but this is more advanced and typically not covered in basic exercises. Remember to practice a variety of problems to understand the different strategies for solving limit problems effectively.

Q: What are the basic concepts of derivatives in Class 11 Maths?

In Class 11 Maths, the derivative of a function represents the rate of change of the function with respect to its variable. The basic concept involves understanding the slope of the tangent to the curve of a function at any given point. This is important for describing motion, growth rates, or changes in various contexts. Derivatives are typically found using two methods: algebraic differentiation and first principles . The derivative of a function can also represent how a function behaves as it increases or decreases, helping to identify maxima, minima, and points of inflection on a graph. The rules of differentiation, such as the power rule , product rule , quotient rule , and the chain rule , are essential for calculating derivatives efficiently and handling more complicated functions.

Q: How to find the derivative of a function using first principles?

f'(x) = lim(h → 0) [f(x + h) - f(x)] / h Here's the process: 1. Start by substituting f(x+h) and f(x) into the formula. 2. Simplify the expression in the numerator, which involves expanding and reducing terms. 3. The next step is to find the limit of the expression as h approaches zero. This step often requires factoring, expanding, or simplifying the terms in the numerator. 4. Finally, after taking the limit, you will obtain the derivative of the function. Using first principles is a more detailed method of finding derivatives and is especially useful for understanding the concept behind derivatives, even though more efficient rules like the power rule are often used in practice.

Q: What is L'Hôpital's Rule, and is it included in NCERT Class 11 Maths?

L'Hôpital's Rule is a technique used to evaluate limits that result in indeterminate forms such as 0/0? or ∞/∞. The rule states that for functions f(x) and g(x) with limits of the form 0/0 or ∞/∞?, the limit of f(x)/g(x) as x→a can be found by differentiating the numerator and denominator separately and then evaluating the limit of the resulting quotient: lim (x → a) [f(x) / g(x)] = lim (x → a) [f'(x) / g'(x)] This process is repeated if necessary, until a determinate form is obtained. However, L'Hôpital's Rule is generally not included in the NCERT Class 11 Maths curriculum, as the focus is primarily on basic limit calculations and derivative concepts. The rule is more commonly introduced in higher-level calculus, typically in Class 12 or in more advanced studies of calculus.

NCERT Exemplar Class 11 Maths Solutions Chapter 13 Limits and Derivatives

Edited By Komal Miglani | Updated on Mar 29, 2025 12:40 AM IST

Did you ever wonder how scientists determine the velocity of a car at a specific moment or how economists determine how rapidly financial patterns are changing? The concept of Limits and Derivatives is central to the understanding of such real-life applications. Limits assist us in determining what value a function is getting nearer to as we approach a particular number. Limits come in handy when we are unable to use the number in an equation directly. Derivatives indicate at what rate something is changing, for instance, the velocity of a traveling vehicle. Derivatives assist in determining the slope of a curve at any given point. In simple terms, limits indicate where a function is heading, and derivatives indicate at what rate it is changing. This chapter explains the fundamental concepts of calculus so that students can understand how things are changing extremely rapidly.

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To grasp this chapter well, students must concentrate on the fundamental concepts of limits, continuity, and differentiation. Solving the problems of Limits and Derivatives from the NCERT Exemplar exercises will create a strong concept. Solving more problems from NCERT Class 11 Maths Solutions Chapter Limits and Derivatives will make students proficient in solving problems. Regular practice, solving sample papers, and previous years' questions will make them precise and confident in solving calculus problems.

NCERT Exemplar Class 11 Maths Solutions Chapter 13

Class 11 Maths Chapter 13 exemplar solutions Exercise: 13.3
Page number: 239-245
Total questions: 80

Question:1

Evaluate $lim_{x \to 3} \frac{x^{2} - 9}{x - 3}$

Answer:

Given $lim_{x \to 3} \frac{x^{2} - 9}{x - 3} = lim_{x \to 3} \frac{(x - 3) (x + 3)}{x - 3} = lim_{x \to 3} x + 3 = 6$

Question:2

Evaluate $lim_{x \to 1 / 2} \frac{4 x^{2} - 1}{2 x - 1}$

Answer:

Given that $lim_{x \to 1 / 2} \frac{4 x^{2} - 1}{2 x - 1} = lim_{x \to 1 / 2} \frac{(2 x - 1) (2 x + 1)}{2 x - 1} = lim_{x \to 1 / 2} 2 x + 1 = 2$

Question:3

Evaluate $lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$

Answer:

$lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$

$= lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h [\sqrt{x + h} + \sqrt{x}]} \times (\sqrt{x + h} + \sqrt{x})$
[Rationalizing the denominator]
$lim_{h \to 0} \frac{x + h - x}{h [\sqrt{x + h} + \sqrt{x}]}$ $= lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}$
Put limit
$= \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2 \sqrt{x}}$

Question:4

Evaluate: $lim_{x \to 0} \frac{{(x + 2)}^{\frac{1}{3}} - 2^{\frac{1}{3}}}{x}$

Answer:

Given $lim_{x \to 0} \frac{{(x + 2)}^{\frac{1}{3}} - 2^{\frac{1}{3}}}{x}$
Now put x=x-2 limits change from 0 to 2
$= lim_{y \to 2} \frac{y^{\frac{1}{3}} - 2^{\frac{1}{3}}}{y - 2} = \frac{1}{3} {(2)}^{\frac{1}{3} - 1} = \frac{1}{3} 2^{- \frac{2}{3}}$ $[u \sin g lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n \cdot a^{n - 1}]$

Question:5

Evaluate : $lim_{x \to 1} \frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1}$

Answer:

$Given that lim_{x \to 1} \frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1} = lim_{x \to 1} \frac{{({(1 + x)}^{2})}^{3} - 1}{{(1 + x)}^{2} - 1}$
$= lim_{x \to 1} \frac{({(1 + x)}^{2} - 1) [{(1 + x)}^{4} + {(1 + x)}^{2} + 1]}{{(1 + x)}^{2} - 1}$

$= lim_{x \to 1} {(1 + x)}^{4} + {(1 + x)}^{2} + 1 = 2^{4} + 2^{2} + 1 = 21$

Question:6

Evaluate:
$lim_{x \to a} \frac{{(2 + x)}^{\frac{5}{2}} - {(2 + a)}^{\frac{5}{2}}}{x - a}$

Answer:
Given that $lim_{x \to a} \frac{{(2 + x)}^{\frac{5}{2}} - {(2 + a)}^{\frac{5}{2}}}{x - a}$

$= lim_{y \to a + 2} \frac{{(y)}^{\frac{5}{2}} - {(2 + a)}^{\frac{5}{2}}}{y - (a + 2)} = \frac{5}{2} {(a + 2)}^{\frac{5}{2} - 1} = \frac{5}{2} {(a + 2)}^{\frac{3}{2}}$ $[using lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n \cdot a^{n - 1}]$

Question:7

Evaluate:
$lim_{x \to 1} \frac{x^{4} - \sqrt{x}}{\sqrt{x} - 1}$
Answer:

$G i v e n lim_{x \to 1} \frac{x^{4} - \sqrt{x}}{\sqrt{x} - 1} = lim_{x \to 1} \frac{(x^{4} - \sqrt{x}) (\sqrt{x} + 1)}{(\sqrt{x} - 1) (\sqrt{x} + 1)}$

$= lim_{x \to 1} \frac{x^{4} \sqrt{x} + x^{4} - x - \sqrt{x}}{x - 1} = lim_{x \to 1} \frac{\sqrt{x} (x^{4} - 1) + x (x^{3} - 1)}{x - 1}$

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

$= lim_{x \to 1} \sqrt{x} (x^{3} + x^{2} + x + 1) + x (x^{2} + x + 1) = 1 * 4 + 1 * 3 = 7$

Question:8

Evaluate:
$lim_{x \to 2} \frac{x^{2} - 4}{\sqrt{3 x - 2} - \sqrt{x + 2}}$

Answer:

Given $lim_{x \to 2} \frac{x^{2} - 4}{\sqrt{3 x - 2} - \sqrt{x + 2}} = lim_{x \to 2} \frac{x^{2} - 4}{(\sqrt{3 x - 2} - \sqrt{x + 2})} \times \frac{\sqrt{3 x - 2} + \sqrt{x + 2}}{\sqrt{3 x - 2} + \sqrt{x + 2}}$

$= lim_{x \to 2} \frac{x^{2} - 4}{(3 x - 2 - x - 2)} \times (\sqrt{3 x - 2} + \sqrt{x + 2})$

$= lim_{x \to 2} \frac{(x - 2) (x + 2)}{2 (x - 2)} \times (\sqrt{3 x - 2} + \sqrt{x + 2})$

$= lim_{x \to 2} \frac{(x + 2)}{2} \times (\sqrt{3 x - 2} + \sqrt{x + 2}) = \frac{4}{2} \times (\sqrt{3 \times 2 - 2} + \sqrt{2 + 2}) = 8$

Question:9

Evaluate:
$lim_{x \to \sqrt{2}} \frac{x^{2} - 4}{x^{2} + 3 \sqrt{2} x - 8}$

Answer:

$\begin{aligned} = lim_{x \to \sqrt{2}} \frac{(x^{2} - 2) (x^{2} + 2)}{x^{2} + 4 \sqrt{2} x - \sqrt{2} x - 8} \\ = lim_{x \to \sqrt{2}} \frac{(x + \sqrt{2}) (x - \sqrt{2}) (x^{2} + 2)}{x (x + 4 \sqrt{2}) - \sqrt{2} (x + 4 \sqrt{2})} \\ = lim_{x \to \sqrt{2}} \frac{(x + \sqrt{2}) (x - \sqrt{2}) (x^{2} + 2)}{(x + 4 \sqrt{2}) (x - \sqrt{2})} \\ = lim_{x \to \sqrt{2}} \frac{(x + \sqrt{2}) (x^{2} + 2)}{x + 4 \sqrt{2}} \\ Put limit \\ = \frac{(\sqrt{2} + \sqrt{2}) (2 + 2)}{\sqrt{2} + 4 \sqrt{2}} \\ = \frac{2 \sqrt{2} \times 4}{5 \sqrt{2}} = \frac{8}{5} \end{aligned}$

Question:10

Evaluate:
$lim_{x \to 1} (\frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2})$

Answer:

Let us apply LH rule i.e. L.Hospita's rule to this question

$lim_{x \to 1} (\frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2}) lim_{x \to a} (\frac{f (x)}{g (x)}) = lim_{x \to a} (\frac{f^{^{'}} (x)}{g^{^{'}} (x)})$

$lim_{x \to 1} (\frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2}) = lim_{x \to 1} (\frac{7 x^{6} - 10 x^{4}}{3 x^{2} - 6 x}) = \frac{7 - 10}{3 - 6} = 1$

Question:11

Evaluate:
$lim_{x \to 0} (\frac{\sqrt{1 + x^{3}} - \sqrt{1 - x^{3}}}{x^{2}})$

Answer:

$Given that lim_{x \to 0} (\frac{\sqrt{1 + x^{3}} - \sqrt{1 - x^{3}}}{x^{2}})$
$= lim_{x \to 0} (\frac{\sqrt{1 + x^{3}} - \sqrt{1 - x^{3}}}{x^{2}} * \frac{\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}}}{\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}}})$

$= lim_{x \to 0} (\frac{1 + x^{3} - 1 + x^{3}}{x^{2}} * \frac{1}{\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}}}) = lim_{x \to 0} (\frac{2 x}{\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}}}) = 0$

Question:12

Evaluate:
$lim_{x \to - 3} (\frac{x^{3} + 27}{x^{5} + 243})$

Answer:

$\begin{aligned} = lim_{x \to - 3} \frac{\frac{x^{3} + (3)^{3}}{x + 3}}{\frac{x^{3} + (3)^{3}}{x + 3}} [Dividing the numerator and denominator by x + 3] \\ = \frac{lim_{x \to - 3} (\frac{x^{3} + (3)^{3}}{x + 3})}{lim_{x \to - 3} (\frac{x^{3} + (3)^{5}}{x + 3})} [lim_{x \to a} \frac{f (x)}{g (x)} = \frac{lim_{x \to a} f (x)}{lim_{x \to a} g (x)}] \\ = \frac{lim_{x \to - 3} (x^{2} - 3 x + 9)}{lim_{x \to - 3} (x^{4} - 3 x^{3} + 9 x^{2} - 27 x + 81)} \\ = \frac{9 + 9 + 9}{81 + 81 + 81 + 81 + 81} = \frac{3 \times 9}{5 \times 81} = \frac{1}{5 \times 3} = \frac{1}{15} \end{aligned}$

Question:13

Evaluate:
$lim_{x \to \frac{1}{2}} (\frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{4 x^{2} - 1})$

Answer:

Given $lim_{x \to \frac{1}{2}} (\frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{4 x^{2} - 1})$
$= lim_{x \to \frac{1}{2}} ((\frac{1}{2 x + 1}) \times (\frac{(8 x - 3) (2 x + 1) - 4 x^{2} - 1}{2 x - 1})) = lim_{x \to \frac{1}{2}} ((\frac{1}{2 x + 1}) \times (\frac{12 x^{2} + 2 x - 4}{2 x - 1}))$

$= lim_{x \to \frac{1}{2}} ((\frac{1}{2 x + 1}) \times (\frac{12 x^{2} + 8 x - 6 x - 4}{2 x - 1})) = lim_{x \to \frac{1}{2}} ((\frac{1}{2 x + 1}) \times (\frac{(3 x + 2) (4 x - 2)}{2 x - 1}))$

$= lim_{x \to \frac{1}{2}} ((\frac{1}{2 x + 1}) \times 2 (3 x + 2)) = \frac{1}{2 \times \frac{1}{2} + 1} \times 2 \times (3 \times \frac{1}{2} + 2) = \frac{7}{2}$

Question:14

Evaluate: Find ‘n’, if
$lim_{x \to 2} (\frac{x^{n} - 2^{n}}{x - 2}) = 80, n \in N$

Answer:

$We know that lim_{x \to 2} (\frac{x^{n} - 2^{n}}{x - 2}) = n {(2)}^{n - 1}$

$n {(2)}^{n - 1} = 80$

$n = 5 \times {(2)}^{5 - 1} = 5 \times 16 = 80$

Question:15

Evaluate:
$lim_{x \to a} (\frac{\sin 3 x}{\sin 7 x})$

Answer:

Given $lim_{x \to a} (\frac{\sin 3 x}{\sin 7 x}) = \frac{\sin 3 a}{\sin 7 a}$

Question:16

Evaluate:
$lim_{x \to 0} (\frac{\sin^{2} 2 x}{\sin^{2} 4 x})$

Answer:

$G i v e n lim_{x \to 0} (\frac{\sin^{2} 2 x}{\sin^{2} 4 x}) = lim_{x \to 0} (\frac{{(\frac{\sin 2 x}{2 x} * 2 x)}^{2}}{{(\frac{\sin 4 x}{4 x} * 4 x)}^{2}}) = lim_{x \to 0} (\frac{4}{16} * \frac{{(\frac{\sin 2 x}{2 x})}^{2}}{{(\frac{\sin 4 x}{4 x})}^{2}}) = \frac{4}{16} = \frac{1}{4}$

Question:17

Evaluate:
$lim_{x \to 0} (\frac{1 - \cos 2 x}{x^{2}})$

Answer:

Given that
$lim_{x \to 0} (\frac{1 - \cos 2 x}{x^{2}}) = lim_{x \to 0} (\frac{2 \sin^{2} x}{x^{2}}) = lim_{x \to 0} (2 * {(\frac{\sin x}{x})}^{2}) = 2$

Question:18

Evaluate:
$lim_{x \to 0} (\frac{2 \sin x - \sin 2 x}{x^{3}})$

Answer:

$G i v e n lim_{x \to 0} (\frac{2 \sin x - \sin 2 x}{x^{3}}) = lim_{x \to 0} (\frac{2 \sin x (1 - \cos x)}{x^{3}}) = lim_{x \to 0} (\frac{2 \sin x (2 \sin^{2} \frac{x}{2})}{x^{3}})$

$= lim_{x \to 0} (2 * \frac{\sin x}{x} * 2 * {(\frac{\sin \frac{x}{2}}{\frac{x}{2}})}^{2} * \frac{1}{4}) = 1$

Question:19

Evaluate:
$lim_{x \to 0} (\frac{1 - \cos m x}{1 - \cos n x})$

Answer:

$Given that lim_{x \to 0} (\frac{1 - \cos m x}{1 - \cos n x}) = lim_{x \to 0} (\frac{\sin^{2} \frac{m x}{2}}{\sin^{2} \frac{n x}{2}}) = lim_{x \to 0} (\frac{\frac{\sin^{2} \frac{m x}{2}}{{(\frac{m x}{2})}^{2}} * {(\frac{m x}{2})}^{2}}{\frac{\sin^{2} \frac{n x}{2}}{{(\frac{n x}{2})}^{2}} * {(\frac{n x}{2})}^{2}})$

$= lim_{x \to 0} (\frac{\frac{\sin^{2} \frac{m x}{2}}{{(\frac{m x}{2})}^{2}} * m^{2}}{\frac{\sin^{2} \frac{n x}{2}}{{(\frac{n x}{2})}^{2}} * n^{2}}) = \frac{m^{2}}{n^{2}}$

Question:20

Evaluate:
$lim_{x \to \frac{π}{3}} \frac{\sqrt{1 - \cos 6 x}}{\sqrt{2} (\frac{π}{3} - x)}$

Answer:

$lim_{x \to \frac{π}{3}} \frac{\sqrt{1 - \cos 6 x}}{\sqrt{2} (\frac{π}{3} - x)} H e r e \cos 6 x = 1 - 2 \sin^{2} 3 x$

$= lim_{x \to \frac{π}{3}} \frac{\sqrt{2 \sin^{2} 3 x}}{\sqrt{2} (\frac{π}{3} - x)} = lim_{x \to \frac{π}{3}} \frac{| \sin 3 x |}{(\frac{π - 3 x}{3})} = lim_{x \to \frac{π}{3}} \frac{| \sin (π - 3 x) |}{(\frac{π - 3 x}{3})} = lim_{x \to \frac{π}{3}} 3 * \frac{| \sin (π - 3 x) |}{π - 3 x} = 3 * 1 = 3$

Question:21

Evaluate:
$lim_{x \to \frac{π}{4}} \frac{\sin x - \cos x}{x - \frac{π}{4}}$

Answer:

$Given that lim_{x \to \frac{π}{4}} \frac{\sin x - \cos x}{x - \frac{π}{4}} = lim_{x \to \frac{π}{4}} \frac{\sqrt{2} (\sin x \cos \frac{π}{4} - \cos x \sin \frac{π}{4})}{x - \frac{π}{4}} = lim_{x \to \frac{π}{4}} \frac{\sqrt{2} \sin (x - \frac{π}{4})}{x - \frac{π}{4}} = \sqrt{2}$

Question:22

Evaluate:
$lim_{x \to \frac{π}{6}} \frac{\sqrt{3} \sin x - \cos x}{x - \frac{π}{6}}$

Answer:

$Given that lim_{x \to \frac{π}{6}} \frac{\sqrt{3} \sin x - \cos x}{x - \frac{π}{6}} = lim_{x \to \frac{π}{6}} \frac{2 (\frac{\sqrt{3}}{2} \sin x - \frac{1}{2} \cos x)}{x - \frac{π}{6}} = lim_{x \to \frac{π}{6}} \frac{2 (\cos \frac{π}{6} \sin x - \sin \frac{π}{6} \cos x)}{x - \frac{π}{6}}$

$= lim_{x \to \frac{π}{6}} \frac{2 \sin (x - \frac{π}{6})}{x - \frac{π}{6}} = 2$

Question:23

Evaluate:
$lim_{x \to 0} \frac{\sin 2 x + 3 x}{2 x + \tan 3 x}$

Answer:

$Given that lim_{x \to 0} \frac{\sin 2 x + 3 x}{2 x + \tan 3 x} = lim_{x \to 0} \frac{2 x * \frac{\sin 2 x}{2 x} + 3 x}{2 x + 3 x * \frac{\tan 3 x}{3 x}} = lim_{x \to 0} \frac{x (2 * \frac{\sin 2 x}{2 x} + 3)}{x (2 + 3 * \frac{\tan 3 x}{3 x})}$

$= lim_{x \to 0} \frac{(2 * \frac{\sin 2 x}{2 x} + 3)}{(2 + 3 * \frac{\tan 3 x}{3 x})} = \frac{2 + 3}{2 + 3} = \frac{5}{5} = 1$

Question:24

Evaluate:
$lim_{x \to a} \frac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}}$

Answer:

$Given that lim_{x \to a} \frac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}} = lim_{x \to a} \frac{(\sin x - \sin a) (\sqrt{x} + \sqrt{a})}{(\sqrt{x} - \sqrt{a}) (\sqrt{x} + \sqrt{a})} = lim_{x \to a} \frac{(\sin x - \sin a) (\sqrt{x} + \sqrt{a})}{(x - a)}$
$= lim_{x \to a} \frac{(2 \cos (\frac{x + a}{2}) \sin (\frac{x - a}{2})) (\sqrt{x} + \sqrt{a})}{(x - a)}$

$= lim_{x \to a} \cos (\frac{x + a}{2}) \frac{\sin (\frac{x - a}{2})}{\frac{x - a}{2}} (\sqrt{x} + \sqrt{a}) = \cos (a) * 1 * (2 \sqrt{a}) = 2 \sqrt{a} \cos a$

Question:25

Evaluate:
$lim_{x \to \frac{π}{6}} \frac{\cot^{2} x - 3}{c o s e c x - 2}$

Answer:

$Given that lim_{x \to \frac{π}{6}} \frac{\cot^{2} x - 3}{c o s e c x - 2} = lim_{x \to \frac{π}{6}} \frac{c o s e c^{2} x - 1 - 3}{c o s e c x - 2}$

$= lim_{x \to \frac{π}{6}} \frac{c o s e c^{2} x - 1 - 3}{c o s e c x - 2} = lim_{x \to \frac{π}{6}} \frac{(c o s e c x - 2) (c o s e c x + 2)}{c o s e c x - 2}$

$= lim_{x \to \frac{π}{6}} (c o s e c x + 2) = 2 + 2 = 4$

Question:26

Evaluate:
$lim_{x \to 0} \frac{(\sqrt{2} - \sqrt{1 + \cos x})}{\sin^{2} x}$

Answer:

$Given that lim_{x \to 0} \frac{(\sqrt{2} - \sqrt{1 + \cos x})}{\sin^{2} x} = lim_{x \to 0} \frac{(\sqrt{2} - \sqrt{1 + \cos x}) (\sqrt{2} + \sqrt{1 + \cos x})}{\sin^{2} x (\sqrt{2} + \sqrt{1 + \cos x})}$

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

$= lim_{x \to 0} \frac{1}{(1 + \cos x) (\sqrt{2} + \sqrt{1 + \cos x})} = \frac{1}{(1 + 1) * (2 \sqrt{2})} = \frac{1}{4 \sqrt{2}}$

Question:27

Evaluate:
$lim_{x \to 0} (\frac{\sin x - 2 \sin 3 x + \sin 5 x}{x})$

Answer:

$. Given that lim_{x \to 0} (\frac{\sin x - 2 \sin 3 x + \sin 5 x}{x})$

$= lim_{x \to 0} (\frac{\sin x}{x} - \frac{2 \sin 3 x}{3 x} * 3 + \frac{\sin 5 x}{5 x} * 5) = 1 - 2 * 3 + 5 = 0$

Question:28

Evaluate: If
$lim_{x \to 1} (\frac{x^{4} - 1}{x - 1}) = lim_{x \to k} (\frac{x^{3} - k^{3}}{x^{2} - k^{2}})$

Answer:

$Given that$

$lim_{x \to 1} (\frac{x^{4} - 1}{x - 1}) = 4 {(1)}^{4 - 1} = 4 lim_{x \to k} (\frac{x^{3} - k^{3}}{x^{2} - k^{2}})$

$= lim_{x \to k} (\frac{(x - k) (x^{2} + k^{2} + x k)}{(x - k) (x + k)}) = lim_{x \to k} (\frac{x^{2} + k^{2} + x k}{x + k}) = \frac{3 k}{2}$
$\frac{3 k}{2} = 4 k = 8 / 3$

Question:29

Differentiate each of the functions w.r.t x in
$\frac{x^{4} + x^{3} + x^{2} + 1}{x}$

Answer:

$Let y = \frac{x^{4} + x^{3} + x^{2} + 1}{x} \frac{d y}{d x} = \frac{d (x^{3})}{d x} + \frac{d (x^{2})}{d x} + \frac{d (x)}{d x} + \frac{d (x^{- 1})}{d x}$

$= 3 x^{2} + 2 x + 1 - \frac{1}{x^{2}} = \frac{3 x^{4} + 2 x^{3} + x^{2} - 1}{x^{2}}$

Question:30

Differentiate each of the functions w.r. to x in
${(x + \frac{1}{x})}^{3}$

Answer:

Let y= ${(x + \frac{1}{x})}^{3}$

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

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

$= 3 (x^{2} + 2 + \frac{1}{x^{2}} - 1 - \frac{2}{x^{2}} - \frac{1}{x^{4}})$

$= 3 x^{2} + 3 - \frac{3}{x^{2}} - \frac{3}{x^{4}}$

Question:31

Differentiate each of the functions w.r. to x in
$(3 x + 5) (1 + t a n x)$

Answer:

Given that $y = (3 x + 5) (1 + \tan x)$

Applying product rule of differentiation we get

$\frac{d y}{d x} = (1 + \tan x) \frac{d}{d x} (3 x + 5) + (3 x + 5) \frac{d}{d x} (1 + \tan x)$

$= 3 (1 + \tan x) + (3 x + 5) \sec^{2} x$

Question:32

Differentiate each of the functions w.r. to x in
$(s e c x - 1) (s e c x + 1)$

Answer:

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

Now applying the concept of chain rule

$\frac{d y}{d x} = \frac{d (\tan^{2} x)}{d x} = 2 \tan x \sec^{2} x$

Question:33

Differentiate each of the functions w.r. to x in
$\frac{3 x + 4}{5 x^{2} - 7 x + 9}$

Answer:

Given that $y = \frac{3 x + 4}{5 x^{2} - 7 x + 9}$

Applying division rule of differentiation that is

$\frac{d y}{d x} = \frac{d}{d x} (\frac{3 x + 4}{5 x^{2} - 7 x + 9})$

$= \frac{(5 x^{2} - 7 x + 9) \frac{d}{d x} (3 x + 4) - (3 x + 4) \frac{d}{d x} (5 x^{2} - 7 x + 9)}{{(5 x^{2} - 7 x + 9)}^{2}}$

$= \frac{(5 x^{2} - 7 x + 9) (3) - (3 x + 4) (10 x - 7)}{{(5 x^{2} - 7 x + 9)}^{2}}$

$= - \frac{5 (3 x^{2} + 8 x - 11)}{{(5 x^{2} - 7 x + 9)}^{2}}$
$= \frac{5 (3 x + 11) (1 - x)}{{(5 x^{2} - 7 x + 9)}^{2}}$

Question:34

Differentiate each of the functions w.r. to x in
$\frac{x^{5} - \cos x}{\sin x}$
Answer:

Given that $y = \frac{x^{5} - \cos x}{\sin x} = \frac{x^{5}}{\sin x} - \frac{\cos x}{\sin x}$

Applying division rule of differentiation that is

$\frac{d y}{d x} = \frac{\sin x \frac{d}{d x} (x^{5}) - x^{5} \frac{d}{d x} (s i n x)}{\sin^{2} x} - \frac{d}{d x} (\cot x) = \frac{5 x^{4} \sin x - x^{5} \cos x}{\sin^{2} x} + {cosec}^{2} x$

Question:35

Differentiate each of the functions w.r. to x in
$\frac{x^{2} \cos \frac{π}{4}}{\sin x}$

Answer:

$y = \frac{x^{2} \cos \frac{π}{4}}{\sin x}$

Applying division rule of differentiation that is

$\frac{d y}{d x} = \frac{\cos \frac{π}{4} (\sin x \frac{d}{d x} (x^{2}) - x^{2} \frac{d}{d x} (\sin x))}{\sin^{2} x} = \frac{1}{\sqrt{2}} (\frac{2 x \sin x - x^{2} \cos x}{\sin^{2} x})$

Question:36
Differentiate each of the functions w.r. to x in

$(ax\textsuperscript{2} + cotx) (p + q cosx)}} \\$

Answer:

Given that $y = (a x^{2} + \cot x) (p + q \cos x)$

Applying the division rule of differentiation that is

$\frac{d y}{d x} = (p + q \cos x) \frac{d}{d x} (a x^{2} + \cot x) + (a x^{2} + \cot x) \frac{d}{d x} (p + q \cos x)$
$= (p + q \cos x) (2 a x - c o s e c^{2} x) + (a x^{2} + \cot x) (- q \sin x)$

Question:37

Differentiate each of the functions w.r. to x in
$(\frac{a + b \sin x}{c + d \cos x})$

Answer:

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

Applying division rule of differentiation that is
$\frac{d y}{d t} = \frac{(c + d \cos x) \frac{d}{d x} (a + b \sin x) - (a + b \sin x) \frac{d}{d x} (c + d \cos x)}{{(c + d \cos x)}^{2}} = \frac{b \cos x (c + d \cos x) - (- d \sin x) (a + b \sin x)}{{(c + d \cos x)}^{2}}$

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

Question:38

Differentiate each of the functions w.r. to x in

(sin x + cosx)^2

Answer:

Given that
$y = {(\sin x + \cos x)}^{2} = \sin^{2} x + \cos^{2} x + 2 \sin x \cos x = 1 + 2 \sin x \cos x = 1 + \sin 2 x$
Applying the concept of chain rule
$\frac{d y}{d x} = \frac{d}{d x} (1 + \sin 2 x) = 0 + 2 * \cos 2 x = 2 \cos 2 x = 2 (\cos^{2} x - \sin^{2} x)$

Question:39

Differentiate each of the functions w.r. to x in

$(2x - 7)^{2} (3x + 5)^{3}$

Answer:

This question will involve the concept of both chain rule and product rule

Given that $y = {(2 x - 7)}^{2} {(3 x + 5)}^{3}$

Applying the product rule of differentiation
$\frac{d y}{d x} = {(3 x + 5)}^{3} \frac{d}{d x} {(2 x - 7)}^{2} + {(2 x - 7)}^{2} \frac{d}{d x} {(3 x + 5)}^{3}$

$= {(3 x + 5)}^{3} * 2 * (2 x - 7) * 2 + {(2 x - 7)}^{2} * 3 * {(3 x + 5)}^{2} * 3$

$= (2 x - 7) {(3 x + 5)}^{2} [4 (3 x + 5) + 9 (2 x - 7)] = (2 x - 7) {(3 x + 5)}^{2} (30 x - 43)$

Question:40

Differentiate each of the functions w.r. to x in
$x^{2} s i n x + c o s 2 x$

Answer:

This question will involve the concept of both chain rule and product rule

$G i v e n t h a t y = x^{2} \sin x + \cos 2 x$

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

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

Question:41

Differentiate each of the functions w.r.to x in
$s i n^{3} x c o s^{3} x$

Answer:

The question involves the concept of chain rule

$G i v e n t h a t y = \sin^{3} x \cos^{3} x y = \frac{1}{8} {(2 \sin x \cos x)}^{3}$

$= \frac{1}{8} \sin^{3} 2 x \frac{d y}{d x} = \frac{3}{8} \sin^{2} 2 x (2 \cos 2 x) = \frac{3}{4} \sin^{2} 2 x \cos 2 x$

Question:42

Differentiate each of the functions w.r. to x in
$\frac{1}{a x^{2} + b x + c}$

Answer:

The question involves the concept of chain rule

$G i v e n t h a t y = \frac{1}{a x^{2} + b x + c} = {(a x^{2} + b x + c)}^{- 1}$

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

Question:43

Differentiate each of the functions with respect to ‘x’Differentiate using first principle

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

Answer:

$Let f (x) = \cos (x^{2} + 1) \dots \dots . . (i)$

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

$Subtracting equation (i) f r o m e q u a t i o n (i i)$

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

$= lim_{Δ x \to 0} \frac{- 2 \sin [\frac{{(x + Δ x)}^{2} + 1 - x^{2} - 1}{2}] \sin [\frac{{(x + Δ x)}^{2} + 1 + x^{2} + 1}{2}]}{Δ x} = lim_{Δ x \to 0} \frac{- 2 \sin [\frac{2 x Δ x + {(Δ x)}^{2}}{2}] \sin [\frac{2 x^{2} + 2 x Δ x + 2}{2}]}{Δ x}$

$= lim_{Δ x \to 0} (\frac{2 x + Δ x}{2}) (\frac{- 2 \sin [\frac{2 x Δ x + {(Δ x)}^{2}}{2}] \sin [\frac{2 x^{2} + 2 x Δ x + 2}{2}]}{\frac{Δ x (2 x + Δ x)}{2}})$
$= lim_{Δ x \to 0} (\frac{2 x + Δ x}{2}) (- 2 \sin [\frac{2 x^{2} + 2 x Δ x + 2}{2}]) (\frac{\sin [\frac{2 x Δ x + {(Δ x)}^{2}}{2}]}{\frac{Δ x (2 x + Δ x)}{2}})$

$= - (\frac{2 x + 0}{2}) (2 \sin (x^{2} + 0 + 1)) (1) = - 2 x \sin (x^{2} + 1) is the required answer$

Question:44

Differentiate each of the functions with respect to ‘x’
Differentiate using first principle $\frac{a x + b}{c x + d}$

Answer:

$Let f (x) = \frac{a x + b}{c x + d} \dots . . (i)$

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

$Subtracting equation (i) from equation (i i)$

$\frac{f (x + Δ x) - f (x)}{Δ x} = \frac{\frac{a (x + Δ x) + b}{c (x + Δ x) + d} - \frac{a x + b}{c x + d}}{Δ x}$
$Taking the limit f^{^{'}} (x) = lim_{Δ x \to 0} \frac{\frac{a (x + Δ x) + b}{c (x + Δ x) + d} - \frac{a x + b}{c x + d}}{Δ x}$

$= lim_{Δ x \to 0} \frac{\frac{(c x + d) (a x + a Δ x + b) - (a x + b) (c x + c Δ x + d)}{(c x + c Δ x + d) (c x + d)}}{Δ x}$

$= lim_{Δ x \to 0} \frac{(a d - b c) Δ x}{(c x + c Δ x + d) (c x + d) Δ x}$

$= lim_{Δ x \to 0} \frac{(a d - b c)}{(c x + c Δ x + d) (c x + d)} = \frac{a d - b c}{{(c x + d)}^{2}} is the required answer$

Question:45

Differentiate each of the functions with respect to ‘x’
Differentiate using first principle $X^{\frac{2}{3}}$

Answer:

$f^{^{'}} (x) = lim_{Δ x \to 0} \frac{{(x + Δ x)}^{\frac{2}{3}} - x^{\frac{2}{3}}}{Δ x} = lim_{Δ x \to 0} \frac{x^{\frac{2}{3}} [{(1 + \frac{Δ x}{x})}^{\frac{2}{3}} - 1]}{Δ x} = lim_{Δ x \to 0} \frac{x^{\frac{2}{3}} [(1 + \frac{2}{3} \frac{Δ x}{x} + \dots) - 1]}{Δ x}$

Expanding by binomial theorem and rejecting the higher powers of $Δ x a s Δ x \to 0$

$= lim_{Δ x \to 0} \frac{x^{\frac{2}{3}} (\frac{2}{3} \frac{Δ x}{x})}{Δ x} = lim_{Δ x \to 0} (\frac{2}{3} \frac{x^{\frac{2}{3}}}{x}) = \frac{2}{3} x^{- \frac{1}{3}}$

Question:46

Differentiate each of the functions with respect to ‘x’
Differentiate using first principle x cos x.

Answer:

$G i v e n t h a t y = x \cos x \dots . (i)$

$y + Δ y = (x + Δ x) \cos (x + Δ x) \dots . (i i)$

$Subtracting equation (i) f r o m e q u a t i o n (i i)$

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

$= lim_{Δ x \to 0} (\frac{x \cos (x + Δ x) - x \cos x}{Δ x}) + lim_{Δ x \to 0} (\frac{Δ x \cos (x + Δ x)}{Δ x})$

$= lim_{Δ x \to 0} (\frac{x (- 2 \sin \frac{x + Δ x - x}{2} \sin \frac{x + Δ x + x}{2})}{Δ x}) + lim_{Δ x \to 0} \cos (x + Δ x)$

$= lim_{Δ x \to 0} \frac{- (2 \sin (x + \frac{Δ x}{2}) \sin (\frac{Δ x}{2}) x)}{Δ x} + lim_{Δ x \to 0} \cos (x + Δ x)$

$= lim_{Δ x \to 0} - x \sin (x + \frac{Δ x}{2}) \frac{(\sin (\frac{Δ x}{2}))}{\frac{Δ x}{2}} + lim_{Δ x \to 0} \cos (x + Δ x) = - x \sin x + \cos x$

Question:47

Evaluate each of the following limits

$lim_{y \to 0} \frac{(x + y) \sec (x + y) - x \sec x}{y}$

Answer:

$Given that$

$lim_{y \to 0} \frac{(x + y) \sec (x + y) - x \sec x}{y} = lim_{y \to 0} (\frac{x \sec (x + y) - x \sec x}{y} + \frac{y \sec (x + y)}{y})$

$= lim_{y \to 0} (\frac{x}{y} [\frac{\cos x - \cos (x + y)}{\cos x \cos (x + y)}] + \sec (x + y))$

$= lim_{y \to 0} (\frac{x}{y} [\frac{- 2 \sin (x + \frac{y}{2}) \sin (\frac{- y}{2})}{\cos x \cos (x + y)}] + \sec (x + y))$
$= lim_{y \to 0} (\frac{x \sin (x + \frac{y}{2})}{\cos x \cos (x + y)} [\frac{\sin (\frac{y}{2})}{\frac{y}{2}}] + \sec (x + y)) = x \sec x \tan x + \sec x = \sec x (x \tan x + 1)$

Question:48

Evaluate each of the following limits
$lim_{x \to 0} \frac{[\sin (α + β) x + \sin (α - β) x + \sin 2 α x]}{\cos 2 β x - \cos 2 α x} x$

Answer:

$lim_{x \to 0} \frac{[\sin (α + β) x + \sin (α - β) x + \sin 2 α x]}{\cos 2 β x - \cos 2 α x} x = lim_{x \to 0} \frac{[2 \sin α x \cos β x + 2 \sin α x \cos α x]}{[2 \sin (α + β) x \sin (α - β) x]} x$

$= lim_{x \to 0} \frac{[\sin α x (\cos β x + \cos α x)]}{[\sin (α + β) x \sin (α - β) x]} x = lim_{x \to 0} \frac{[\sin α x (2 \cos \frac{(α + β) x}{2} \cos \frac{(α - β) x}{2})]}{[\sin (α + β) x \sin (α - β) x]} x$

$= lim_{x \to 0} \frac{[\sin α x]}{[2 \sin \frac{(α + β) x}{2} \sin \frac{(α - β) x}{2}]} x$

$= lim_{x \to 0} \frac{[\frac{\sin α x}{α x}] * α x * x}{[2 (\frac{\sin \frac{(α + β) x}{2}}{\frac{(α + β) x}{2}}) (\frac{\sin \frac{(α - β) x}{2}}{\frac{(α - β) x}{2}}) * \frac{(α + β) x}{2} * \frac{(α - β) x}{2}]}$

$= \frac{α}{2 * \frac{(α + β)}{2} * \frac{(α - β)}{2}} = \frac{2 α}{α^{2} - β^{2}}$

Question:49

Evaluate each of the following limits
$lim_{x \to \frac{π}{4}} (\frac{\tan^{3} x - \tan x}{\cos (x + \frac{π}{4})})$
Answer:

$lim_{x \to \frac{π}{4}} (\frac{\tan^{3} x - \tan x}{\cos (x + \frac{π}{4})}) = lim_{x \to \frac{π}{4}} (\tan x) * lim_{x \to \frac{π}{4}} (\frac{\tan^{2} x - 1}{\cos (x + \frac{π}{4})})$

$= 1 * lim_{x \to \frac{π}{4}} (\frac{- (1 - \tan x) (1 + \tan x)}{\cos (x + \frac{π}{4})})$

$= lim_{x \to \frac{π}{4}} - (1 + \tan x) * lim_{x \to \frac{π}{4}} (\frac{(1 - \tan x)}{\cos (x + \frac{π}{4})})$

$= - 2 lim_{x \to \frac{π}{4}} (\frac{(\cos x - \sin x)}{\cos (x + \frac{π}{4}) \cos x})$
$= - 2 lim_{x \to \frac{π}{4}} (\frac{\sqrt{2} (\cos \frac{π}{4} \cos x - \sin \frac{π}{4} \sin x)}{\cos (x + \frac{π}{4}) \cos x})$

$= - 2 lim_{x \to \frac{π}{4}} (\frac{\sqrt{2} (\cos (x + \frac{π}{4}))}{\cos (x + \frac{π}{4}) \cos x}) = - 2 lim_{x \to \frac{π}{4}} (\frac{\sqrt{2}}{\cos x}) = - 4$

Question:50

Evaluate each of the following limits
$lim_{x \to π} (\frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})})$

Answer:

$lim_{x \to π} (\frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})}) = lim_{x \to π} (\frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})})$

$= lim_{x \to π} (\frac{\cos^{2} \frac{x}{4} + \sin^{2} \frac{x}{4} - 2 \sin \frac{x}{4} \cos \frac{x}{4}}{(\cos^{2} \frac{x}{4} - \sin^{2} \frac{x}{4}) (\cos \frac{x}{4} - \sin \frac{x}{4})})$
$= lim_{x \to π} (\frac{{(\cos \frac{x}{4} - \sin \frac{x}{4})}^{2}}{(\cos^{2} \frac{x}{4} - \sin^{2} \frac{x}{4}) (\cos \frac{x}{4} - \sin \frac{x}{4})}) = lim_{x \to π} (\frac{(\cos \frac{x}{4} - \sin \frac{x}{4})}{(\cos^{2} \frac{x}{4} - \sin^{2} \frac{x}{4})})$

$= lim_{x \to π} (\frac{1}{(\cos \frac{x}{4} + \sin \frac{x}{4})}) = \frac{1}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2}}$

Question:51

Evaluate each of the following limits
Show that $lim_{x \to 4} \frac{| x - 4 |}{x - 4}$ does not exist.

Answer:

$lim_{x \to 4} \frac{| x - 4 |}{x - 4}$

$L H L = lim_{x \to 4^{-}} \frac{- x + 4}{x - 4} = - 1$

$R H L = lim_{x \to 4^{+}} \frac{x - 4}{x - 4} = 1$

$L H L \neq R H L$

Hence, the limit doesnot exist

Question:52

Let $f (x) = \frac{k \cos x}{π - 2 x}$ when $x \neq \frac{π}{2}$ and
$f (x) = 3$ if $lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2})$ find the value of k.

Answer:

$L H L = lim_{x \to {\frac{π}{2}}^{-}} (\frac{k \cos (\frac{π}{2} - h)}{π - 2 (\frac{π}{2} - h)}) = lim_{h \to 0^{-}} (\frac{k \cos (\frac{π}{2} - h)}{2 h}) = lim_{h \to 0^{-}} (\frac{k \sin h}{2 h}) = \frac{k}{2}$

$R H L = lim_{x \to {\frac{π}{2}}^{+}} (\frac{k \cos (\frac{π}{2} + h)}{π - 2 (\frac{π}{2} + h)}) = lim_{h \to 0^{+}} (\frac{k \cos (\frac{π}{2} + h)}{- 2 h})$ $= lim_{h \to 0^{-}} (\frac{- k \sin h}{- 2 h}) = \frac{k}{2}$

$lim_{x \to \frac{π}{2}} f (x) = \frac{k}{2} = 3 k = 6$

Question:53

Evaluate each of the following limits
Let $f (x) = \begin{array}{ll} x + 2 & x \leq - 1 \\ c x^{2} & x > - 1 \end{array}$ find ‘c’ if $lim_{x \to - 1} f (x)$ exists.

Answer:

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

$R H L = lim_{x \to {- 1}^{+}} c x^{2} = c$

$L H L = R H L$

$c = 1$

Question:54

Choose the correct answer out of 4 options given against each Question
$lim_{x \to π} \frac{\sin x}{x - π}$ is
A. 1
B. 2
C. –1
D. –2

Answer:

$lim_{x \to π} \frac{\sin x}{x - π} = lim_{x \to π} \frac{\sin (π - x)}{- (π - x)} = - 1 lim_{x \to 0} \frac{\sin x}{x} = 1 π - x \to 0 x \to π$

Hence, the answer is option C

Question:55

Choose the correct answer out of 4 options given against each Question
$lim_{x \to 0} \frac{x^{2} \cos x}{1 - \cos x}$ is
A. 2
B. $\frac{3}{2}$
C. $\frac{- 3}{2}$
D. 1

Answer:

$lim_{x \to 0} \frac{x^{2} \cos x}{1 - \cos x} = lim_{x \to 0} \frac{x^{2} \cos x}{2 \sin^{2} \frac{x}{2}} = lim_{x \to 0} \frac{x^{2} \cos x}{\frac{2 x^{2}}{4} \frac{\sin^{2} \frac{x}{2}}{\frac{x^{2}}{4}}} = lim_{x \to 0} \frac{\cos x}{\frac{2}{4} \frac{\sin^{2} \frac{x}{2}}{\frac{x^{2}}{4}}} = 2$
Hence, the answer is option A

Question:56

Choose the correct answer out of 4 options given against each Question
$lim_{x \to 0} \frac{{(1 + x)}^{n} - 1}{x}$ is
A. n
B. 1
C. –n
D. 0
Answer:

$lim_{x \to 0} \frac{{(1 + x)}^{n} - 1}{x} = lim_{x \to 0} \frac{{(1 + x)}^{n} - 1^{n}}{(1 + x) - 1} = n {(1)}^{n - 1} = n$
Hence, the answer is option A

Question:57

Choose the correct answer out of 4 options given against each Question
$lim_{x \to 1} \frac{x^{m} - 1}{x^{n} - 1}$ is
A. 1

B. $\frac{m}{n}$
C. $- \frac{m}{n}$
D. $\frac{m^{2}}{n^{2}}$

Answer:

$lim_{x \to 1} \frac{x^{m} - 1}{x^{n} - 1} = lim_{x \to 1} \frac{\frac{x^{m} - 1}{x - 1}}{\frac{x^{n} - 1}{x - 1}} = \frac{m {(1)}^{m - 1}}{n {(1)}^{n - 1}} = \frac{m}{n}$
Hence, the answer is option B

Question:58

Choose the correct answer out of 4 options given against each Question

$lim_{θ \to 0} (\frac{1 - \cos 4 θ}{1 - \cos 6 θ})$ is

A. $\frac{4}{9}$
B. $\frac{1}{2}$
C. $- \frac{1}{2}$
D. $- 1$
Answer:

$lim_{θ \to 0} (\frac{1 - \cos 4 θ}{1 - \cos 6 θ}) = lim_{θ \to 0} (\frac{2 \sin^{2} 2 θ}{2 \sin^{2} 3 θ}) = lim_{θ \to 0} {(\frac{\sin 2 θ}{\sin 3 θ})}^{2} = lim_{θ \to 0} {(\frac{\frac{\sin 2 θ}{2 θ} * 2 θ}{(\frac{\sin 3 θ}{3 θ}) * 3 θ})}^{2} = \frac{4}{9}$
Hence, the answer is option A

Question:59

Choose the correct answer out of 4 options given against each Question
$lim_{x \to 0} (\frac{c o s e c x - \cot x}{x})$ is
A. $- \frac{1}{2}$

B. 1
C. $\frac{1}{2}$
D. –1

Answer:

$lim_{x \to 0} (\frac{c o s e c x - \cot x}{x}) = lim_{x \to 0} (\frac{1 - \cos x}{x \sin x}) = lim_{x \to 0} (\frac{2 \sin^{2} x / 2}{2 x \sin x / 2 \cos x / 2}) = lim_{x \to 0} (\frac{\sin x / 2}{x \cos x / 2}) = \frac{1}{2}$
Hence, the answer is option C

Question:60

Choose the correct answer out of 4 options given against each Question
$lim_{x \to 0} (\frac{\sin x}{\sqrt{x + 1} - \sqrt{1 - x}})$ is
A. 2
B. 0
C. 1
D. –1

Answer:

$lim_{x \to 0} (\frac{\sin x}{\sqrt{x + 1} - \sqrt{1 - x}}) = lim_{x \to 0} (\frac{\sin x (\sqrt{x + 1} + \sqrt{1 - x})}{(\sqrt{x + 1} - \sqrt{1 - x}) (\sqrt{x + 1} + \sqrt{1 - x})})$

$= lim_{x \to 0} (\frac{\sin x (\sqrt{x + 1} + \sqrt{1 - x})}{(x + 1 - 1 + x)}) = lim_{x \to 0} (\frac{\sin x (\sqrt{x + 1} + \sqrt{1 - x})}{2 x}) = 1$
Hence, the answer is option C

Question:61

Choose the correct answer out of 4 options given against each Question

$lim_{x \to \frac{π}{4}} (\frac{\sec^{2} x - 2}{\tan x - 1})$ is
A. 3
B. 1
C. 0
D. $\sqrt{2}$

Answer:

$lim_{x \to \frac{π}{4}} (\frac{\sec^{2} x - 2}{\tan x - 1}) = lim_{x \to \frac{π}{4}} (\frac{1 + \tan^{2} x - 2}{\tan x - 1}) = lim_{x \to \frac{π}{4}} (\frac{\tan^{2} x - 1}{\tan x - 1}) = lim_{x \to \frac{π}{4}} (\tan x + 1) = 2$
Hence, the answer is option D

Question:62

Choose the correct answer out of 4 options given against each Question
$lim_{x \to 1} (\frac{(\sqrt{x} - 1) (2 x - 3)}{2 x^{2} + x - 3})$ is

A. $\frac{1}{10}$
B. $\frac{- 1}{10}$
C. 1
D. None of these

Answer:

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

$= lim_{x \to 1} (\frac{(\sqrt{x} - 1) (2 x - 3)}{(2 x + 3) (\sqrt{x} - 1) (\sqrt{x} + 1)}) = lim_{x \to 1} (\frac{(2 x - 3)}{(2 x + 3) (\sqrt{x} + 1)}) = - \frac{1}{10}$
Hence, the answer is option B

Question:63

Choose the correct answer out of 4 options given against each Question
If $\begin{aligned} f (x) = \frac{\sin [x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{aligned}$ where [.] denotes the greatest integer function, then $lim_{x \to 0} f (x)$ is equal to
A. 1
B. 0
C. –1
D. None of these
Answer:

$L H L = lim_{x \to 0^{-}} (\frac{\sin [x]}{[x]}) = lim_{h \to 0^{-}} (\frac{\sin [0 - h]}{[0 - h]}) = lim_{h \to 0^{-}} (\frac{\sin [- h]}{[- h]}) = lim_{h \to 0^{-}} (\frac{\sin (- 1)}{- 1}) = \sin 1$
$R H L = lim_{x \to 0^{+}} (\frac{\sin [x]}{[x]}) = lim_{h \to 0^{+}} (\frac{\sin [0 + h]}{[0 + h]}) = lim_{h \to 0^{+}} (\frac{\sin [h]}{[h]}) = lim_{h \to 0^{+}} (\frac{\sin (0)}{0})$
Limit doesn’t exist
Hence, the answer is option D

Question:64

Choose the correct answer out of 4 options given against each Question

$lim_{x \to 0} (\frac{| \sin x |}{x})$ is
A. 1
B. –1
C. does not exist
D. None of these

Answer:

$L H L = lim_{x \to 0^{-}} (\frac{| \sin x |}{x}) = lim_{h \to 0^{-}} (\frac{| \sin (0 - h) |}{(0 - h)}) = lim_{h \to 0^{-}} (\frac{- \sin (- h)}{- h}) = - 1$

$R H L = lim_{x \to 0^{+}} (\frac{| \sin x |}{x}) = lim_{h \to 0^{+}} (\frac{| \sin (0 + h) |}{(0 + h)}) = lim_{h \to 0^{+}} (\frac{\sin (h)}{h}) = 1$
$So, limit does n^{^{'}} t exists Hence, the answer is option C$

Question:65

Choose the correct answer out of 4 options given against each Question
Let $f (x) = \begin{array}{l} x^{2} - 1, 0 < x < 2 \\ 2 x + 3, 2 \leq x < 3 \end{array}$ the quadratic equation whose roots are $lim_{x \to 2} f (x) and lim_{x \to 2} f (x)$ is
$A . x^{2} - 6 x + 9 = 0 B . x^{2} - 7 x + 8 = 0 C . x^{2} - 14 x + 49 = 0 D . x^{2} - 10 x + 21 = 0$

Answer:

$L H L = lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} (x^{2} - 1) = 3$

$R H L = lim_{x \to 2^{+}} f (x)$

$= lim_{x \to 2^{+}} (2 x + 3) = 7$

$The quadratic equation whose roots are 3 and 7 are (x - 3) (x - 7)$

$= x^{2} - 10 x + 21$
Hence, the answer is option D

Question:66

Choose the correct answer out of 4 options given against each Question
$lim_{x \to 0} \frac{(\tan 2 x - x)}{3 x - \sin x}$ is
A. 2
B. $\frac{1}{2}$
C. $\frac{- 1}{2}$
D. $\frac{1}{4}$

Answer:

$lim_{x \to 0} \frac{(\tan 2 x - x)}{3 x - \sin x} = lim_{x \to 0} \frac{(\frac{\tan 2 x}{x} - 1)}{(3 - \frac{\sin x}{x})} = lim_{x \to 0} \frac{(\frac{2 tan 2 x}{2 x} - 1)}{(3 - \frac{\sin x}{x})} = \frac{2 - 1}{3 - 1} = \frac{1}{2}$
Hence, the answer is option B

Question:67

Choose the correct answer out of 4 options given against each Question
Let f(x) = x – [x], $\in R, then f^{'} \frac{1}{2}$ is
A. 3/2
B. 1
C. 0
D. –1

Answer:

$L H D = lim_{h \to 0} \frac{(f (\frac{1}{2} - h) - f (\frac{1}{2}))}{- h} = lim_{h \to 0} \frac{((\frac{1}{2} - h) - [(\frac{1}{2} - h)] - (\frac{1}{2}) + [\frac{1}{2}])}{- h} = lim_{h \to 0} \frac{- h}{- h} = 1$

$R H D = lim_{h \to 0} \frac{(f (\frac{1}{2} + h) - f (\frac{1}{2}))}{h} = lim_{h \to 0} \frac{((\frac{1}{2} + h) - [(\frac{1}{2} + h)] - (\frac{1}{2}) + [\frac{1}{2}])}{h} = lim_{h \to 0} \frac{h}{h} = 1$
Hence, the answer is option B

Question:68

Choose the correct answer out of 4 options given against each Question
If $I f y = \sqrt{x} + \frac{1}{x}, t h e n \frac{d y}{d x}$ at x = 1 is
A. 1
B. $\frac{1}{2}$
C. $\frac{1}{\sqrt{2}}$
D. 0

Answer:

$y = \sqrt{x} + \frac{1}{x} \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} - \frac{1}{x^{2}}$

${(\frac{d y}{d x})}_{x = 1} = \frac{1}{2} - 1 = - \frac{1}{2}$

Hence, the answer is option D

Question:69

Choose the correct answer out of 4 options given against each Question
If $f (x) = \frac{x - 4}{2 \sqrt{x}}$ then f’(1) is
A. $\frac{5}{4}$
B. $\frac{4}{5}$
C. 1
D. 0

Answer:

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

$f^{^{'}} (x) = \frac{1}{2} [\frac{\sqrt{x} * 1 - (x - 4) * \frac{1}{2 \sqrt{x}}}{x}] = \frac{1}{2} [\frac{x + 4}{2 x^{\frac{3}{2}}}]$

$f^{^{'}} (1) = \frac{1}{2} [\frac{1 + 4}{2 (1)}] = \frac{5}{4}$
Hence, the answer is option A

Question:70

Choose the correct answer out of 4 options given against each Question

If $y = \frac{1 + \frac{1}{x^{2}}}{1 - \frac{1}{x^{2}}}$ then $\frac{d y}{d x}$ is
A. $\frac{- 4 x}{{(x^{2} - 1)}^{2}}$
B. $\frac{- 4 x}{x^{2} - 1}$
C. $\frac{1 - x^{2}}{4 x}$
D. $\frac{4 x}{x^{2} - 1}$

Answer:

$y = \frac{1 + \frac{1}{x^{2}}}{1 - \frac{1}{x^{2}}} = \frac{x^{2} + 1}{x^{2} - 1}$

$\frac{d y}{d x} = \frac{(x^{2} - 1) (2 x) - (x^{2} + 1) (2 x)}{{(x^{2} - 1)}^{2}} = \frac{(2 x) (x^{2} - 1 - x^{2} - 1)}{{(x^{2} - 1)}^{2}} = - \frac{4 x}{{(x^{2} - 1)}^{2}}$
Hence, the answer is option A

Question:71

Choose the correct answer out of 4 options given against each Question

If $y = \frac{\sin x + \cos x}{\sin x - \cos x}$ then ${\frac{d y}{d x}}_{a t x = 0}$ is

A. –2
B. 0
C. $\frac{1}{2}$
D. does not exist
Answer:

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

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

${(\frac{d y}{d x})}_{x = 0} = - \frac{2}{{(- 1)}^{2}} = - 2$
Hence, the answer is option A

Question:72

Choose the correct answer out of 4 options given against each Question

If $y = \frac{\sin (x + 9)}{\cos x}$ then $\frac{d y}{d x} at x = 0$ is
A. cos 9
B. sin 9
C. 0
D. 1

Answer:

$y = \frac{\sin (x + 9)}{\cos x} \frac{d y}{d x} = \frac{\cos x \cos (x + 9) - \sin (x + 9) (- \sin x)}{\cos^{2} x}$

$= \frac{\cos x \cos (x + 9) + \sin (x + 9) \sin x}{\cos^{2} x}$

$= \frac{\cos (x + 9 - x)}{\cos^{2} x} = \frac{\cos 9}{\cos^{2} x}$

${(\frac{d y}{d x})}_{x = 0} = \frac{\cos 9}{{(1)}^{2}} = \cos 9$

Hence, the answer is option A

Question:73

Choose the correct answer out of 4 options given against each Question

If $f (x) = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots + \frac{x^{100}}{100}$ then f’(1) is equal to
A. 1/100
B. 100
C. does not exist
D. 0

Answer:

$f (x) = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots + \frac{x^{100}}{100}$

$f^{^{'}} (x) = 0 + 1 + \frac{2 x}{2} + \frac{3 x^{2}}{3} + \dots + \frac{100 x^{99}}{100}$

$f^{^{'}} (x) = 0 + 1 + x + x^{2} + \dots x^{99}$

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

Hence, the answer is option B

Question:74

Choose the correct answer out of 4 options given against each Question

If $f (x) = \frac{x^{n} - a^{n}}{x - a}$ for some constant ‘a’, then f’(a) is
A. 1
B. 0
C. does not exist
D. 1/2

Answer:

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

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

$f^{^{'}} (a) = \frac{(a - a) (n a^{n - 1}) - (a^{n} - a^{n}) (1)}{{(a - a)}^{2}} = \frac{0}{0}$
Hence, the answer is option B

Question:75

Choose the correct answer out of 4 options given against each Question
If $f (x) = x^{100} + x^{99} +$ \ldots $x + 1,$ , then f’(1) is equal to
A. 5050
B. 5049
C. 5051
D. 50051

Answer:

$f (x) = x^{100} + x^{99} + \dots + x + 1$

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

$f^{^{'}} (1) = 100 + 99 + 98 + \dots + 2 + 1 = \frac{100 * 101}{2} = 5050$

Hence, the answer is option A

Question:76

Choose the correct answer out of 4 options given against each Question
If $f (x) = 1 - x + x^{2} - x^{3}$ \ldots $- x^{99} + x^{100}$ , then f’(1) is equal to
A. 150
B. –50
C. –150
D. 50

Answer:

$f (x) = 1 - x + x^{2} - x^{3} + \dots - x^{99} + x^{100}$

$f^{^{'}} (x) = 0 - 1 + 2 x - 3 x^{2} + \dots - 99 x^{98} + 100 x^{99}$

$f^{^{'}} (1)$ $= - 1 + 2 - 3 + 4 - \dots - 99 + 100$

$= (2 - 1) + (4 - 3) + (6 - 5) + \dots (100 - 99) = 1 + 1 + \dots + 1 (50 times) = 50$
Hence, the answer is option D

Question:77

Fill in the blanks
If $f (x) = \frac{\tan x}{x - π}, lim_{x \to π} f (x) =$

Answer:

$lim_{x \to π} \frac{\tan x}{x - π} = lim_{x \to π} \frac{- \tan (π - x)}{- (π - x)} = lim_{π - x \to 0} \frac{\tan (π - x)}{(π - x)} = 1$

Question:78

Fill in the blanks
$lim_{x \to 0} \sin m x \cot \frac{x}{\sqrt{3}} = 2$ then m = ________

Answer:

$lim_{x \to 0} \sin m x \cot \frac{x}{\sqrt{3}} = 2$

$lim_{x \to 0} m x * (\frac{\sin m x}{m x}) * (\frac{\frac{x}{\sqrt{3}}}{\tan \frac{x}{\sqrt{3}}}) * (\frac{\sqrt{3}}{x}) = 2$

$lim_{x \to 0} m x * (\frac{\sqrt{3}}{x}) = 2$

$\sqrt{3} m = 2 m = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$

Question:79

Fill in the blanks
If $y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots$ then $\frac{d y}{d x} =$ ........

Answer:

$y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots$

$\frac{d y}{d x} = 0 + 1 + \frac{2 x}{2!} + \frac{3 x^{2}}{3!} + \frac{4 x^{3}}{4!} + \dots$

$\frac{d y}{d x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots$

$\frac{d y}{d x} = y$

Question:80

Fill in the blanks
$lim_{x \to 3^{+}} \frac{x}{[x]} =$ ...............

Answer:

$lim_{x \to 3^{+}} \frac{x}{[x]} = lim_{h \to 0^{+}} \frac{3 + h}{[3 + h]} = lim_{h \to 0^{+}} \frac{3 + h}{3} = \frac{3}{3} = 1$

Important topics in Class 11 Maths NCERT Exemplar Solutions Chapter 13

13.1 Introduction
13.2 Intuitive Idea of Derivatives
13.3 Limits
13.3.1 Algebra of Limits
13.3.2 Limits of polynomials and rational functions
13.4 Limits of Trigonometric function
13.5 Derivatives
13.5.1 Algebra of Derivative of functions
13.5.2 Derivatives of polynomials and trigonometric functions
13.6 Miscellaneous Examples

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NCERT Exemplar Class 11 Mathematics Chapters

NCERT Exemplar Class 11 Mathematics Chapter 1: Sets

NCERT Exemplar Class 11 Mathematics Chapter 2: Relations and Functions

NCERT Exemplar Class 11 Mathematics Chapter 3: Trigonometric Functions

NCERT Exemplar Class 11 Mathematics Chapter 4: Principles of Mathematical Induction

NCERT Exemplar Class 11 Mathematics Chapter 5: Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Mathematics Chapter 6: Linear Inequalities

NCERT Exemplar Class 11 Mathematics Chapter 7: Permutations and Combinations

NCERT Exemplar Class 11 Mathematics Chapter 8: Binomial Theorem

NCERT Exemplar Class 11 Mathematics Chapter 9: Sequence and Series

NCERT Exemplar Mathematics Chapter 10: Straight Lines

NCERT Exemplar Class 11 Mathematics Chapter 11: Conic Sections

NCERT Exemplar Class 11 Mathematics Chapter 12: Introduction to Three Dimensional Geometry

NCERT Exemplar for Class 11 Mathematics Chapter 13: Limits and Derivatives

NCERT Exemplar Class 11 Mathematics Chapter 14: Mathematical Reasoning

NCERT Exemplar Class 11 Mathematics Chapter 15: Statistics

NCERT Exemplar Class 11 Mathematics Chapter 16: Probability

Importance of solving NCERT Exemplar Class 11 Maths Questions

NCERT Exemplar Class 11 Maths solutions chapter 13 covers the really important topic of limits and derivatives of any function which is a very important concept for mathematics as well as physics.

The students will learn about the limits of different functions from NCERT exemplar solutions for Class 11 Maths chapter 13
The students will be able to define the limits and derivatives of different trigonometric, polynomial and rational number functions.
The NCERT exemplar Class 11 Maths solutions chapter 13 covers various solved examples along with solutions for better understanding and learning of different concepts.
The students should practice the application of different formulas provided in the chapter along with solutions and solved examples, take help from Class 11 Maths NCERT exemplar solutions chapter 13.

NCERT Solutions for Class 11 Maths: 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

NCERT solutions of class 11 - Subject-wise

Here are the subject-wise links for the NCERT solutions of class 11:

NCERT Notes of class 11 - Subject Wise

Given below are the subject-wise NCERT Notes of class 11 :

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 11:

NCERT Exemplar Class 11 Solutions

Given below are the subject-wise exemplar solutions of class 11 NCERT:

NCERT Exemplar Class 11 Biology Solutions

NCERT Exemplar Class 11 Chemistry Solutions

NCERT Exemplar Class 11 Physics Solutions

Frequently Asked Questions (FAQs)

1. What are the important topics covered in NCERT Exemplar Class 11 Maths Chapter 13?

Chapter 13 of Class 11 Maths, titled "Limits and Derivatives," covers several important concepts that are foundational to understanding calculus. Key topics include limits of functions, which help in understanding the behavior of functions as they approach a specific point. The chapter also covers continuity and the conditions under which a function is continuous. Another critical concept is derivatives, where you learn how to find the rate of change of a function at a point, which is essential for understanding slopes of curves and real-world applications like speed and acceleration. Additionally, the chapter introduces you to derivatives from first principles, which is a fundamental method of finding derivatives. The chapter also includes problems on algebra of limits, limits involving infinity, and the chain rule for derivatives, which are useful tools in solving more complex calculus problems.

2. How to solve limit problems in NCERT Exemplar Class 11 Maths Chapter 13?

To solve limit problems in Chapter 13, begin by analyzing the function involved. The first step is to substitute the given value into the function and check if the limit exists. If substituting the value results in an indeterminate form like 0/0?, then you must simplify the expression using algebraic techniques, such as factoring or rationalizing. Another approach is to use standard limit laws and theorems, such as the limit of a sum, product, or quotient of functions. If the expression is complex, you might need to apply trigonometric limits or limits involving infinity. In some cases, you can use L'Hopital's Rule, but this is more advanced and typically not covered in basic exercises. Remember to practice a variety of problems to understand the different strategies for solving limit problems effectively.

3. What are the basic concepts of derivatives in Class 11 Maths?

In Class 11 Maths, the derivative of a function represents the rate of change of the function with respect to its variable. The basic concept involves understanding the slope of the tangent to the curve of a function at any given point. This is important for describing motion, growth rates, or changes in various contexts. Derivatives are typically found using two methods: algebraic differentiation and first principles. The derivative of a function can also represent how a function behaves as it increases or decreases, helping to identify maxima, minima, and points of inflection on a graph. The rules of differentiation, such as the power rule, product rule, quotient rule, and the chain rule, are essential for calculating derivatives efficiently and handling more complicated functions.

4. How to find the derivative of a function using first principles?

f'(x) = lim(h → 0) [f(x + h) - f(x)] / h
Here's the process:
1. Start by substituting f(x+h) and f(x) into the formula.
2. Simplify the expression in the numerator, which involves expanding and reducing terms.
3. The next step is to find the limit of the expression as h approaches zero. This step often requires factoring, expanding, or simplifying the terms in the numerator.
4. Finally, after taking the limit, you will obtain the derivative of the function. Using first principles is a more detailed method of finding derivatives and is especially useful for understanding the concept behind derivatives, even though more efficient rules like the power rule are often used in practice.

5. What is L'Hôpital's Rule, and is it included in NCERT Class 11 Maths?

L'Hôpital's Rule is a technique used to evaluate limits that result in indeterminate forms such as 0/0? or ∞/∞. The rule states that for functions f(x) and g(x) with limits of the form 0/0 or ∞/∞?, the limit of f(x)/g(x) as x→a can be found by differentiating the numerator and denominator separately and then evaluating the limit of the resulting quotient:

lim (x → a) [f(x) / g(x)] = lim (x → a) [f'(x) / g'(x)]

This process is repeated if necessary, until a determinate form is obtained. However, L'Hôpital's Rule is generally not included in the NCERT Class 11 Maths curriculum, as the focus is primarily on basic limit calculations and derivative concepts. The rule is more commonly introduced in higher-level calculus, typically in Class 12 or in more advanced studies of calculus.

Also Read

NCERT Exemplar Class 11 Maths Solutions Chapter 13 Limits and Derivatives

NCERT Exemplar Class 11 Biology Solutions Chapter 17 Breathing and Exchange of Gases

NCERT Exemplar Class 11 Biology Solutions Chapter 22 Chemical Coordination and Integration

NCERT Exemplar Class 11 Biology Solutions Chapter 12 Mineral Nutrition

NCERT Exemplar Class 11 Biology Solutions Chapter 16 Digestion and Absorption

NCERT Exemplar Class 11 Biology Solutions Chapter 15 Plant Growth and Development

NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development

NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

NCERT Solutions for Class 11 Biology Chapter 7 Structural Organisation in Animals

NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom

NCERT Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry Some Basic Principles and Techniques

NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

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

Environmental Chemistry Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 14 Notes - 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 Exemplar Class 11 Maths Solutions Chapter 13 Limits and Derivatives

NCERT Exemplar Class 11 Maths Solutions Chapter 13

Important topics in Class 11 Maths NCERT Exemplar Solutions Chapter 13

NCERT Exemplar Class 11 Mathematics Chapters

Importance of solving NCERT Exemplar Class 11 Maths Questions

NCERT solutions of class 11 - Subject-wise

NCERT Notes of class 11 - Subject Wise

NCERT Books and NCERT Syllabus

NCERT Exemplar Class 11 Solutions

Frequently Asked Questions (FAQs)

Also Read

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