RD Sharma Solutions Class 12 Mathematics Chapter 10 MCQ

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RD Sharma Solutions Class 12 Mathematics Chapter 10 MCQ

Updated on Jan 20, 2022 06:06 PM IST

It is pretty well-known among high school students that RD Sharma class 12th exercise MCQ is one of the best exercise solutions found in the market. The RD Sharma class 12 chapter 10 exercise MCQ solution is a fantastic study partner for students who practice diligently before exams. Experts in mathematics have joined hands to create some unique answers found in the RD Sharma solutions. Moreover, students can learn some new methods of calculations from these solutions, which would help them solve questions faster and more accurately.

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RD Sharma Class 12 Solutions Chapter 10 MCQ Differentiation - Other Exercise
Differentiation Excercise: MCQ
RD Sharma Chapter wise Solutions

Also Read - RD Sharma Solution for Class 9 to 12 Maths

The RD Sharma class 12 chapter 10 exercise MCQ is one of the best NCERT solutions trusted and recommended by hundreds of students in their exams. It can be used for both school and board exams. The 10th chapter of the maths book deals with Differentiations in trigonometric equations. It will teach students about Logarithmic differentiation, differentiation of inverse trigonometric functions, Differentiation of a function. Exercise MCQ includes 33 questions on all the concepts that students have learned in this chapter.

RD Sharma Class 12 Solutions Chapter 10 MCQ Differentiation - Other Exercise

Differentiation Excercise: MCQ

Differentiation exercise Multiple choice question 1

Answer:
$\frac{1}{2 e}$
Hint:
Use quotient rule to differentiate the function
Given:
$f (x) = \log_{x^{2}} (\log_{e}) x$
Solution:
$f (x) = \log_{x^{2}} (\log_{e}) x$

= \frac{\log (\log x)}{(\log (x))^{2}}

f (x) = \frac{\log (\log x)}{2 \log x}

Use quotient rule

f^{'} (x) = \frac{\frac{1}{2} [\log (\frac{1}{\log x}) (\frac{1}{x}) - \log (\log (x)) (\frac{1}{x})]}{(\log x)^{2}}

= \frac{1}{2} \frac{[\frac{(1 - \log (\log x))}{x}]}{(\log x)^{2}}

Put x=e

f^{'} (e) = (\frac{1}{2}) \frac{[\frac{(1 - \log (\log e))}{e}]}{(\log e)^{2}}

\begin{aligned} = (\frac{1}{2}) \frac{(1 - 0)}{e} [Since \log e = 1] \\ f^{'} (e) = \frac{1}{2 e} \end{aligned}

Differentiation exercise Multiple choice question 2

Answer:
$(x \log)^{- 1}$
Hint:
Use differential formula
Given:
$f (x) = \log x$
Solution:
We have

\begin{aligned} f (x) = \log x \\ f (\log x) = \log (\log x) \\ f^{'} (\log x) = \frac{1}{\log x} \frac{d}{d x} (\log x) \end{aligned}

= \frac{1}{x \log x}

f^{'} (\log x) = (x \log x)^{- 1}

Differentiation exercise Multiple choice question 3

Answer:
${(\frac{2}{3})}^{\frac{1}{2}}$
Hint:
Differentiate the given function and replace x by

\frac{π}{6}

and solve
Given:
$\cot^{- 1} {(\cos 2 x)^{\frac{1}{2}}} at x = \frac{π}{6}$
Solution:
$\begin{aligned} y = \cot^{- 1} (\sqrt{\cos 2 x}) \\ \frac{d y}{d x} = \frac{- 1}{1 + \cos 2 x} \frac{d}{d x} \sqrt{\cos 2 x} \end{aligned}$
$\frac{d y}{d x} = [\frac{- 1}{1 + \cos 2 x}] [\frac{1}{2} (\sqrt{\cos 2 x})] (- 2 \sin 2 x)$
$= \frac{\sin 2 x}{(1 + \cos 2 x) \sqrt{\cos 2 x}}$

At x = \frac{π}{6}

\frac{d y}{d x} = \frac{\sin \frac{π}{3}}{(1 + \cos \frac{π}{3}) \sqrt{\cos \frac{π}{3}}}

= \frac{(\frac{\sqrt{3}}{2})}{(1 + \frac{1}{2}) \sqrt{\frac{1}{2}}}

\begin{aligned} = \sqrt{\frac{2}{3}} \\ = {(\frac{2}{3})}^{\frac{1}{2}} \end{aligned}

Differentiation exercise Multiple choice question 4

Answer:
$\frac{x}{\sqrt{1 + x^{2}}}$
Hint:
Differentiate the function w.r.t x
Given:
$\sec (\tan^{- 1} x)$
Solution:

\begin{aligned} y = \sec (\tan^{- 1} x) \\ \frac{d y}{d x} = \sec (\tan^{- 1} x) \tan (\tan^{- 1} x) \times \frac{d}{d x} (\tan^{- 1} x) \end{aligned}

\begin{aligned} = \sec (\tan^{- 1} x) \tan (\tan^{- 1} x) \times \frac{1}{\sqrt{1 + x^{2}}} \\ = y \tan (\tan^{- 1} x) \times \frac{1}{\sqrt{1 + x^{2}}} \end{aligned}

= y (\frac{x}{\sqrt{1 + x^{2}}}) [\tan (\tan^{- 1} x) = x]

This is the equation of differential equation which have co-efficient=

(\frac{x}{\sqrt{1 + x^{2}}})

Differentiation exercise Multiple choice question 5

Answer:
$\frac{1}{2}$
Hint:
Differentiate the function and replace x by

\frac{π}{6}

solve
Given:
$f (x) = \tan^{- 1} \sqrt{\frac{1 + \sin x}{1 - \sin x}}, 0 \leq x \leq \frac{π}{2}, f^{'} (\frac{π}{6})$
Solution:

y = \tan^{- 1} (\sqrt{\frac{1 + \sin x}{1 - \sin x}})

= \tan^{- 1} (\sqrt{\frac{1 - \cos (\frac{π}{2} + x)}{1 + \cos (\frac{π}{2} + x)}})

= \tan^{- 1} (\sqrt{\frac{2 \sin^{2} (\frac{π}{4} + \frac{x}{2})}{2 \cos^{2} (\frac{π}{4} + \frac{x}{2})}})

y = \tan^{- 1} (\tan (\frac{π}{4} + \frac{x}{2})) = \frac{π}{4} + \frac{x}{2} [\tan^{- 1} (\tan x) = x]

\frac{d y}{d x} = \frac{1}{2}

Differentiation exercise Multiple choice question 6

Answer:
${(1 + \frac{1}{x})}^{x} (1 + \frac{1}{x}) - \frac{1}{x + 1}$
Hint:
Differentiate the function w.r.t x
Given:
$y = {(1 + \frac{1}{x})}^{x}$
Solution:
Let
$y = {(1 + \frac{1}{x})}^{x}$
Taking log on both sides

\begin{aligned} \log y = x \log (1 + \frac{1}{x}) \\ \frac{1}{y} \frac{d y}{d x} = x \frac{d}{d x} \log (1 + \frac{1}{x}) + \log (1 + \frac{1}{x}) \frac{d}{d x} (x) \end{aligned}

= x (\frac{1}{1 + \frac{1}{x}}) \frac{d}{d x} (1 + \frac{1}{x}) + \log (1 + \frac{1}{x})

\begin{aligned} = x \times \frac{x}{x + 1} (\frac{- 1}{x^{2}}) + \log (1 + \frac{1}{x}) \\ = \frac{x^{2}}{x + 1} \times \frac{- 1}{x^{2}} + \log (1 + \frac{1}{x}) \end{aligned}

\begin{aligned} \frac{1}{y} \frac{d y}{d x} = \frac{- 1}{x + 1} + \log (1 + \frac{1}{x}) \\ \frac{d y}{d x} = y [\frac{- 1}{x + 1} + \log (1 + \frac{1}{x})] \end{aligned}

= {(1 + \frac{1}{x})}^{x} [\log (1 + \frac{1}{x}) - \frac{1}{x + 1}]

Differentiation exercise Multiple choice question 7

Answer:
$\frac{\log x}{(1 + \log x)^{2}}$
Hint:
Differentiate the function w.r.t x
Given:
$x^{y} = e^{x - y}$
Solution:
$x^{y} = e^{x - y}$
Taking log on both sides

\log x^{y} = \log (e^{x - y})

y \log x = (x - y) \log e [∵ \log m^{n} = n \log m]

\begin{aligned} y \log x = (x - y) \cdot 1 [∵ \log e = 1] \\ y \log x = x - y \end{aligned}

\begin{aligned} x = y + y \log x \\ x = y (1 + \log x) \\ y = \frac{x}{1 + \log x} \end{aligned}

Differentiating y w.r.t x then

\frac{d y}{d x} = \frac{d}{d x} (\frac{x}{1 + \log x})

= \frac{(1 + \log x) \frac{d (x)}{d x} - x \frac{d}{d x} (1 + \log x)}{(1 + \log x)^{2}} [∵ \frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}]

= \frac{(1 + \log x) \cdot 1 - x {\frac{d (1)}{d x} + \frac{d (\log x)}{d x}}}{(1 + \log x)^{2}} [∵ \frac{d}{d x} {f (x) + g (x)} = \frac{d}{d x} f (x) + \frac{d}{d x} g (x), \frac{d (x)}{d x} = 1]

= \frac{1 + \log x - x (0 + \frac{1}{x})}{(1 + \log x)^{2}} [∵ \frac{d}{d x} \log x = \frac{1}{x}, \frac{d}{d x} (con s \tan t) = 0]

= \frac{1 + \log x - x \cdot \frac{1}{x}}{(1 + \log x)^{2}} = \frac{1 + \log x - 1}{(1 + \log x)^{2}}

= \frac{\log x}{(1 + \log x)^{2}}

Differentiation exercise Multiple choice question 9

Answer:
$| \sec θ |$
Hint:
Differentiate x and y w.r.t

θ

, then divide and solve
Given:
$x = a \cos^{3} θ, y = a \sin^{3} θ, \sqrt{1 + {(\frac{d y}{d x})}^{2}} =$
Solution:

\begin{aligned} x = a \cos^{3} θ \\ \frac{d x}{d θ} = a \frac{d}{d θ} (\cos^{3} θ) \\ \frac{d x}{d θ} = a (3) \cos^{2} \frac{d}{d θ} (\cos θ) \end{aligned}

\frac{d x}{d θ} = - 3 a \cos^{2} θ \sin θ

....................(1)

\begin{aligned} y = a \sin θ \\ \frac{d y}{d θ} = a \frac{d}{d θ} (\sin^{3} θ) \end{aligned}

= 3 a \sin^{2} θ \frac{d}{d θ} (\sin θ)

\frac{d y}{d θ} = 3 a \sin^{2} θ \cos θ

......................(2)
Dividing (2) by (1) we get

\frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{3 a \sin^{2} θ \cos θ}{- 3 a \cos^{2} θ \sin θ}

\begin{aligned} \frac{d y}{d x} = \frac{\sin θ}{- \cos θ} = - \tan θ \\ \sqrt{1 + {(\frac{d y}{d x})}^{2}} = \sqrt{1 + \tan^{2} θ} = \sqrt{\sec^{2} θ} = | \sec θ | \end{aligned}

Differentiation exercise Multiple choice question 10

Answer:
$\frac{- 2}{1 + x^{2}}$
Hint:
Differentiate y function w.r.t x
Given:
$y = \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$
Solution:
Let
$y = \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$
Differentiate y w.r.t x then

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

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

[∵ \frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}, \frac{d}{d x} (\frac{u}{v}) = v \frac{d u}{d x} - u \frac{d v}{d x}]

= \frac{1}{\sqrt{\frac{{(1 + x^{2})}^{2} - {(1 - x^{2})}^{2}}{{(1 + x^{2})}^{2}}}} [\frac{(1 + x^{2}) {\frac{d}{d x} (1) - \frac{d}{d x} (x^{2})} - (1 - x^{2}) {\frac{d}{d x} (1) + \frac{d}{d x} (x^{2})}}{{(1 + x^{2})}^{2}}]

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

= \frac{\sqrt{{(1 + x^{2})}^{2}}}{\sqrt{(1 + x^{4} + 2 x^{2}) - (1 + x^{4} - 2 x^{2})}} [\frac{(1 + x^{2}) (0 - 2 x) - (1 - x^{2}) (0 + 2 x)}{{(1 + x^{2})}^{2}}]

[\frac{d}{d x} (x^{n}) = n x^{n - 1}, (a^{2} + b^{2} + 2 a b) = (a + b)^{2} & (a - b)^{2} = a^{2} + b^{2} - 2 a b]

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

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

\begin{aligned} = \frac{1 + x^{2}}{\sqrt{(2 x)^{2}} [\frac{- 4 x}{{(1 + x^{2})}^{2}}}] = \frac{(1 + x^{2})}{2 x} \times \frac{- 4 x}{{(1 + x^{2})}^{2}} = \frac{- 2}{(1 + x^{2})} \\ ∴ \frac{d y}{d x} = \frac{- 2}{1 + x^{2}} \end{aligned}

Differentiation exercise Multiple choice question 11

Answer:
Does not exist
Hint:
$\sec^{- 1} α$ is not defined
Given:
$\sec^{- 1} (\frac{1}{2 x^{2} + 1}) W.r.t \sqrt{1 + 3 x} at x = - \frac{1}{3}$
Solution:
We know that $\sec^{- 1} α$ is not defined for

α \in (- 1, 1)

Here for

x = - \frac{1}{3}, \frac{1}{2 x^{2} + 1} = \frac{9}{11} \in (- 1, 1)

∴ \sec^{- 1} (\frac{1}{2 x^{2} + 1})

Is not defined at

x = - \frac{1}{3}

Derivative of

\sec^{- 1} (\frac{1}{2 x^{2} + 1})

does not exist at

x = - \frac{1}{3}

Differentiation exercise Multiple choice question 12

Answer:
$- 1$
Hint:
Differentiate the function w.r.t x
Given:
$\sqrt{x} + \sqrt{y} = 1, \frac{d y}{d x} a t (\frac{1}{4}, \frac{1}{4})$
Solution:
$\sqrt{x} + \sqrt{y} = 1$
Differentiating w.r.t x

\begin{aligned} \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0 \\ \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = \frac{- 1}{2 \sqrt{x}} \end{aligned}

\frac{d y}{d x} = - \frac{1}{2 \sqrt{x}} \times \frac{2 \sqrt{y}}{1} = \frac{- \sqrt{y}}{\sqrt{x}}

Now,

{[\frac{d y}{d x}]}_{(\frac{1}{4} \cdot \frac{1}{4})} = \frac{- \sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}} = - 1

Differentiation exercise Multiple choice question 13

Answer:
$- 1$
Hint:
Differentiate the function w.r.t x
Given:
$\sin (x + y) = \log (x + y)$
Solution:

\begin{aligned} \sin (x + y) = \log (x + y) \\ \cos (x + y) (1 + \frac{d y}{d x}) = \frac{1}{(x + y)} (1 + \frac{d y}{d x}) \end{aligned}

\begin{aligned} \cos (x + y) + \cos (x + y) \frac{d y}{d x} = \frac{1}{(x + y)} + \frac{1}{(x + y)} \frac{d y}{d x} \\ \cos (x + y) \frac{d y}{d x} - \frac{1}{(x + y)} \frac{d y}{d x} = \frac{1}{(x + y)} - \cos (x + y) \end{aligned}

\begin{aligned} [\cos (x + y) - \frac{1}{(x + y)}] \frac{d y}{d x} = \frac{1}{(x + y)} - \cos (x + y) \\ - [\frac{1}{(x + y)} - \cos (x + y)] \frac{d y}{d x} = \frac{1}{(x + y)} - \cos (x + y) \\ \frac{d y}{d x} = - 1 \end{aligned}

Differentiation exercise Multiple choice question 14

Answer:
1
Hint:
Differentiate the function w.r.t x
Given:
$u = \sin^{- 1} (\frac{2 x}{1 + x^{2}}), v = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$
Solution:
$u = \sin^{- 1} (\frac{2 x}{1 + x^{2}}), v = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$
Differentiating u w.r.t x then

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

= \frac{1}{\sqrt{1 - {(\frac{2 x}{1 + x^{2}})}^{2}}} [\frac{(1 + x^{2}) \frac{d}{d x} (2 x) - 2 x \frac{d}{d x} (1 + x^{2})}{{(1 + x^{2})}^{2}}]

[\begin{array}{l} \frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}} \\ \frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}} \end{array}]

= \frac{1}{\sqrt{\frac{{(1 + x^{2})}^{2} - (2 x)^{2}}{{(1 + x^{2})}^{2}}}} [\frac{(1 + x^{2}) 2 \cdot \frac{d (x)}{d x} - 2 x {\frac{d}{d x} (1) + \frac{d}{d x} (x^{2})}}{{(1 + x^{2})}^{2}}]

[\begin{array}{l} ∵ \frac{d}{d x} {f (x) + g (x)} = \frac{d}{d x} f (x) + \frac{d}{d x} g (x) \\ \frac{d}{d x} (a x^{n}) = a \frac{d}{d x} (x^{n}) \end{array}]

= \frac{\sqrt{{(1 + x^{2})}^{2}}}{\sqrt{1 + x^{4} + 2 x^{2} - 4 x^{2}}} \times [\frac{(1 + x^{2}) 2 \cdot 1 - 2 x (0 + 2 x)}{{(1 + x^{2})}^{2}}]

[∵ \frac{d}{d x} (cons \tan t) = 0, \frac{d}{d x} (x^{n}) = n x^{n - 1}]

= \frac{(1 + x^{2})}{\sqrt{1 + x^{2} - 2 x^{2}}} [\frac{(1 + x^{2}) 2 - 2 x (2 x)}{{(1 + x^{2})}^{2}}]

= \frac{(1 + x^{2})}{\sqrt{{(1 - x^{2})}^{2}}} [\frac{2 + 2 x^{2} - 4 x^{2}}{{(1 + x^{2})}^{2}}] [∵ a^{2} + b^{2} - 2 a b = (a - b)^{2}]

\begin{aligned} = \frac{(1 + x^{2})}{(1 - x^{2})} \times \frac{2 - 2 x^{2}}{{(1 + x^{2})}^{2}} = \frac{2 (1 - x^{2})}{(1 - x^{2}) (1 + x^{2})} = \frac{2}{1 + x^{2}} \\ ∴ \frac{d u}{d x} = \frac{2}{1 + x^{2}} \end{aligned}

................(1)
Differentiating v w.r.t x then,

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

= \frac{1}{1 + {(\frac{2 x}{1 - x^{2}})}^{2}} [\frac{(1 - x^{2}) \frac{d}{d x} (2 x) - \frac{d}{d x} (1 - x^{2}) \cdot 2 x}{{(1 - x^{2})}^{2}}]

[∵ \frac{d}{d x} (\tan^{- 1} x) = \frac{1}{1 + x^{2}}, \frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}]

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

[\begin{array}{l} \frac{d}{d x} {f (x) + g (x)} = \frac{d}{d x} f (x) + \frac{d}{d x} g (x) \\ \frac{d}{d x} (a x^{n}) = a \frac{d}{d x} (x^{n}) \end{array}]

= \frac{{(1 - x^{2})}^{2}}{1 + x^{4} - 2 x^{2} + 4 x^{2}} [\frac{(1 - x^{2}) \cdot 2 \cdot 1 - 2 x (0 - 2 x)}{{(1 - x^{2})}^{2}}]

[\begin{array}{l} ∵ \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (cons \tan t) = 0, \frac{d (x)}{d x} = 1 \\ (a - b)^{2} = a^{2} + b^{2} - 2 a b \end{array}]

= \frac{1}{1 + x^{4} + 2 x^{2}} [\frac{2 - 2 x^{2} + 4 x^{2}}{1}]

= \frac{2 + 2 x^{2}}{{(1 + x^{2})}^{2}} = \frac{2 (1 + x^{2})}{{(1 + x^{2})}^{2}} [∵ a^{2} + b^{2} + 2 a b = (a + b)^{2}]

∴ \frac{d v}{d x} = \frac{2}{1 + x^{2}}

..............................(2)
Thus

\frac{d y}{d x} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{d u}{d x} \times \frac{d x}{d v}

= \frac{2}{1 + x^{2}} \times \frac{1 + x^{2}}{2}

[Using (1) and (2)]

∴ \frac{d y}{d x} = 1

Differentiation exercise Multiple choice question 15

Answer:
$- \frac{1}{2}$
Hint:
Differentiate the function w.r.t x
Given:
$\frac{d}{d x} [\tan^{- 1} (\frac{\cos x}{1 + \sin x})]$
Solution:

u = \tan^{- 1} (\frac{\cos x}{1 + \sin x})

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

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

= \tan^{- 1} [\frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}}]

u = \tan^{- 1} [\frac{\tan \frac{π}{4} - \tan \frac{x}{2}}{1 + \tan \frac{π}{4} \times \tan \frac{x}{2}}]

= \tan^{- 1} [\tan (\frac{π}{4} - \frac{x}{2})]

\begin{aligned} u = \frac{π}{4} - \frac{x}{2} \\ \frac{d u}{d x} = 0 - \frac{1}{2} \\ \frac{d u}{d x} = - \frac{1}{2} \end{aligned}

Differentiation exercise Multiple choice question 16

Answer:
$\frac{x^{2} - 1}{x^{2} - 4}$
Hint:
Differentiate the function w.r.t x
Given:
$\frac{d}{d x} [\log {e^{x} {(\frac{x - 2}{x + 2})}^{\frac{3}{4}}}]$
Solution:

y = \frac{d}{d x} [\log (e^{x} {(\frac{x - 2}{x + 2})}^{2})]

= \frac{d}{d x} [x \log e + \frac{3}{4} \log (\frac{x - 2}{x + 2})]

y = \frac{d}{d x} [x + \frac{3}{4} \log (\frac{x - 2}{x + 2})]

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

\begin{aligned} = 1 + \frac{3}{4} \frac{(x + 2)}{(x - 2)} \times \frac{x + 2 - x + 2}{(x + 2)^{2}} \\ = 1 + \frac{3}{4} \frac{(x + 2)}{(x - 2)} \times \frac{4}{(x + 2)} \end{aligned}

\begin{aligned} = 1 + \frac{3}{(x^{2} - 4)} \\ \frac{d y}{d x} = \frac{x^{2} - 4 + 3}{x^{2} - 4} = \frac{x^{2} - 1}{x^{2} - 4} \end{aligned}

Differentiation exercise Multiple choice question 17

Answer:
$\frac{\cos 2 x}{2 y - 1}$
Hint:
Differentiate the function w.r.t x
Given:
$y = \sqrt{\sin x + y}$
Solution:
$y = \sqrt{\sin x + y}$
Squaring both side

\begin{aligned} y^{2} = \sin x + y \\ y^{2} - y = \sin x \\ 2 y \frac{d y}{d x} - \frac{d y}{d x} = \cos x \end{aligned}

Differentiation exercise Multiple choice question 18

Answer:
$- \frac{y}{x}$
Hint:
Differentiate the function w.r.t x
Given:
$3 \sin (x y) + 4 \cos (x y) = 5$
Solution:
$3 \sin (x y) + 4 \cos (x y) = 5$
$\begin{aligned} 3 \cos (x y) [x \frac{d y}{d x} + y] - 4 \sin (x y) [x \frac{d y}{d x} + y] = 0 \\ [x \frac{d y}{d x} + y] [3 \cos (x y) - 4 \sin (x y)] = 0 \end{aligned}$
$\begin{aligned} x \frac{d y}{d x} + y = 0 \\ x \frac{d y}{d x} = - y \\ \frac{d y}{d x} = - \frac{y}{x} \end{aligned}$

Differentiation exercise Multiple choice question 19

Answer:
$\frac{\sin^{2} (a + y)}{\sin a}$
Hint:
Differentiate the function w.r.t x
Given:
$\sin y = x \sin (a + y), then \frac{d y}{α x}$
Solution:
$\begin{aligned} \sin y = x \sin (a + y) \\ \frac{d}{d x} (\sin y) = \frac{d}{d x} [(a + y) \sin x] \end{aligned}$
$\cos y \frac{d y}{d x} = \sin (a + y) \frac{d}{d x} (x) + x \frac{d}{d x} [\sin (a + y)]$
$\begin{aligned} = \sin (a + y) \times 1 + x \cos (a + y) \frac{d y}{d x} \\ = \sin (a + y) + x \cos (a + y) \frac{d y}{d x} \end{aligned}$
$\cos y \frac{d y}{d x} = - x \cos (a + y) \frac{d y}{d x} = \sin (a + y)$
$(\cos y - \frac{\sin y}{\sin (a + y)} \times \cos (a + y)) \frac{d y}{d x} = \sin (a + y)$ $[\begin{array}{l} \sin y = 2 \sin x \cos x \\ x = \frac{\sin y}{\sin (a + y)} \end{array}]$
$(\frac{\sin (a + y) \cos y - \sin y \cos (a + y)}{\sin (a + y)}) \frac{d y}{d x} = \sin (a + y)$
$\begin{aligned} \frac{\sin (a + y - y)}{\sin (a + y)} \times \frac{d y}{d x} = \sin (a + y) \\ \frac{d y}{d x} = \frac{\sin^{2} (a + y)}{\sin a} \end{aligned}$

Differentiation exercise Multiple choice question 20

Answer:
2
Hint:
Differentiate the function w.r.t

\cos^{- 1} x

Given:

\cos^{- 1} (2 x^{2} - 1)

With respect to

\cos^{- 1} x

Solution:

\begin{aligned} u = \cos^{- 1} (2 x^{2} - 1) \\ x = \cos θ \\ θ = \cos^{- 1} x \\ \frac{d θ}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} \end{aligned}

\begin{aligned} u = \cos^{- 1} (\cos 2 θ) \\ u = 2 θ \\ \frac{d u}{d x} = 2 \frac{d θ}{d x} \\ \frac{d u}{d x} = \frac{- 2}{\sqrt{1 - x^{2}}} \end{aligned}

...............(1)

\begin{aligned} v = \cos^{- 1} x \\ v = \cos^{- 1} (\cos θ), v = θ \\ \frac{d v}{d x} = \frac{d θ}{d x} \\ \frac{d v}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} \end{aligned}

..................(2)
Divide (1) by (2)

\begin{aligned} \frac{d u}{d x} = \frac{- 2}{\frac{d v}{d x}} \sqrt{1 - x^{2}} \times \frac{\sqrt{1 - x^{2}}}{- 1} \\ \frac{d u}{d v} = 2 \end{aligned}

Differentiation exercise Multiple choice question 21

Answer:
$- 1 for x < - 3$
Hint:
Differentiate the function w.r.t x
Given:
$f (x) = \sqrt{x^{2} + 6 x + 9}, f^{'} (x)$
Solution:
$\begin{aligned} f (x) = \sqrt{x^{2} + 6 x + 9} \\ x^{2} + 6 x + 9 = (x + 3) (x + 3) = (x + 3)^{2} \end{aligned}$
$\begin{aligned} f (x) = \sqrt{(x + 3)^{2}} = | x + 3 | \\ f (x) = {\begin{array}{cc} x + 3 & x \geq - 3 \\ - x - 3 & x < - 3 \end{array} \end{aligned}$
$\begin{aligned} f^{'} (x) = {\begin{array}{cl} 1 & x \geq - 3 \\ - 1 & x < - 3 \end{array} \\ f^{'} (x) = - 1 For x < - 3 \end{aligned}$

Differentiation exercise Multiple choice question 22

Answer:
$- 2 x + 9, If 4 < x < 5$
Hint:
Differentiate the function w.r.t x
Given:
$f (x) = | x^{2} - 9 x + 20 | Then f^{'} (x) =$
Solution:

f (x) = | x^{2} - 9 x + 20 |

f (x) = {\begin{array}{cc} x^{2} - 9 x + 20 & - \infty < x \leq 4 \\ - (x^{2} - 9 x + 20) & 4 < x < 5 \\ x^{2} - 9 x + 20 & 5 \leq x < \infty \end{array}

f^{'} (x) = {\begin{cases} 2 x - 9 & \infty < x \leq 4 \\ - 2 x + 9 & 4 < x < 5 \\ 2 x - 9 & 5 \leq x < \infty \end{cases}

f^{'} (x) = - 2 x + 9 for 4 < x < 5

Differentiation exercise Multiple choice question 23

Answer:
None of these
Hint:
In case of function of one variable it’s a function that doesn’t have finite derivation
Given:

f (x) = \sqrt{x^{2} - 10 x + 25}, f^{'} (x) =

Solution:

f (x) = \sqrt{x^{2} - 10 x + 25}

\begin{aligned} = \sqrt{(x - 5)^{2}} \\ = | x - 5 | \end{aligned}

f (x) = {\begin{array}{cll} x - 5 & for & x > 5 \\ - (x - 5) & for & x < 5 \end{array}

L H D = lim_{x \to 5^{-}} \frac{f (x) - f (a)}{x - a}

= lim_{x \to 5^{-}} \frac{\sqrt{x^{2} - 10 x + 25} - \sqrt{5^{2} - 10 (5) + 25}}{x - 5}

= lim_{x \to 5^{-}} \frac{| x - 5 |}{(x - 5)} = lim_{x \to 5^{-}} \frac{- (x - 5)}{x - 5} = - 1

R H D = lim_{x \to 5^{+}} \frac{f (x) - f (a)}{x - a}

= lim_{x \to 5^{+}} \frac{\sqrt{x^{2} - 10 x + 25} - \sqrt{5^{2} - 10 (5) + 25}}{x - 5}

\begin{aligned} = lim_{x \to 5^{+}} \frac{| x - 5 |}{x - 5} \\ = lim_{x \to 5^{+}} \frac{x - 5}{x - 5} \\ = 1 \end{aligned}

L H D \neq R H D,

so function is not differentiable .

Differentiation exercise Multiple choice question 25

Answer:
0
Hint:
Differentiate the function w.r.t x
Given:
$f (x) = {(\frac{x^{l}}{x^{m}})}^{l + m} {(\frac{x^{m}}{x^{n}})}^{m + n} {(\frac{x^{n}}{x^{3}})}^{n + l}$
Solution:

f (x) = {(\frac{x^{l}}{x^{m}})}^{l + m} {(\frac{x^{m}}{x^{n}})}^{m + n} {(\frac{x^{n}}{x^{l}})}^{n + l}

\begin{aligned} = x^{(l - m) (l + m)} \times x^{(m - n) (m + n)} \times x^{(n - l) (n + l)} \\ = x^{l^{2} - m^{2}} \times x^{m^{2} - n^{2}} \times x^{n^{2} - l^{2}} \end{aligned}

\begin{aligned} f (x) = x^{(l^{2} - m^{2} + m^{2} - n^{2} + n^{2} - l^{2})} = x^{0} \\ f (x) = 1 \\ f^{'} (x) = 0 \end{aligned}

Differentiation exercise Multiple choice question 26

Answer:
0
Hint:
Differentiate the function w.r.t x
Given:
$y = \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} + \frac{1}{1 + x^{b - a} + x^{c - a}}$
Solution:
$y = \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} + \frac{1}{1 + x^{b - a} + x^{c - a}}$
$= \frac{1}{1 + \frac{x^{a}}{x^{b}} + \frac{x^{c}}{x^{b}}} + \frac{1}{1 + \frac{x^{b}}{x^{c}} + \frac{x^{a}}{x^{c}}} + \frac{1}{1 + \frac{x^{b}}{x^{a}} + \frac{x^{c}}{x^{a}}}$
$= \frac{x^{b}}{x^{a} + x^{b} + x^{c}} + \frac{x^{c}}{x^{a} + x^{b} + x^{c}} + \frac{x^{a}}{x^{a} + x^{b} + x^{c}}$
$\begin{aligned} = \frac{x^{b} + x^{c} + x^{a}}{x^{a} + x^{b} + x^{c}} \\ y = 1 \\ \frac{d y}{d x} = \frac{d (1)}{d x} = 0 \end{aligned}$

Differentiation exercise Multiple choice question 27

Answer:
$\frac{x^{2}}{y^{2}} \sqrt{\frac{1 - y^{6}}{1 - x^{6}}}$
Hint:
Differentiate the function w.r.t x
Given:
$\sqrt{1 - x^{6}} + \sqrt{1 - y^{6}} = a^{3} (x^{3} - y^{3})$
Solution:
We have, $\sqrt{1 - x^{6}} + \sqrt{1 - y^{6}} = a^{3} (x^{3} - y^{3})$
Putting

x^{3} = \sin A, y^{3} = \sin B

\begin{aligned} \sqrt{1 - \sin^{2} A} + \sqrt{1 - \sin^{2} B} = a (\sin A - \sin B) \\ \cos A + \cos B = a (\sin A - \sin B) \end{aligned}

\begin{aligned} 2 \cos (\frac{A + B}{2}) \cos (\frac{A - B}{2}) = 2 a \sin (\frac{A - B}{2}) \cos (\frac{A + B}{2}) \\ \cot (\frac{A - B}{2}) = a^{3} \end{aligned}

\begin{aligned} \frac{A - B}{2} = \cot^{- 1} (a^{3}) \\ A - B = 2 \cot^{- 1} (a^{3}) \\ \sin^{- 1} x^{3} - \sin^{- 1} y^{3} = 2 \cot^{- 1} (a^{3}) \end{aligned}

\begin{aligned} \frac{1}{\sqrt{1 - x^{6}}} \times \frac{d}{d x} (x^{3}) - \frac{1}{\sqrt{1 - y^{6}}} \times \frac{d}{d x} (y^{3}) = 0 \\ \frac{1}{\sqrt{1 - x^{6}}} \times 3 x^{2} - \frac{1}{\sqrt{1 - y^{6}}} \times 3 y^{2} \times \frac{d y}{d x} = 0 \end{aligned}

\frac{d y}{d x} = \frac{x^{2}}{y^{2}} \sqrt{\frac{1 - y^{6}}{1 - x^{6}}}

Differentiation exercise Multiple choice question 28

Answer:
1
Hint:
Differentiate the function w.r.t x
Given:
$y = \log \sqrt{\tan x}$
Solution:

\begin{aligned} y = \log \sqrt{\tan x} \\ \frac{d y}{d x} = \frac{1}{\sqrt{\tan x}} \times \frac{d}{d x} (\sqrt{\tan x}) \end{aligned}

= \frac{1}{\sqrt{\tan x}} \times \frac{1}{2 \sqrt{\tan x}} \times \frac{d}{d x} (\tan x)

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

Now

{(\frac{d y}{d x})}_{x = \frac{π}{4}} = \frac{{(\sec \frac{π}{4})}^{2}}{2 \tan (\frac{π}{4})} = \frac{2}{2 \times 1} = 1

Differentiation exercise Multiple choice question 29

Answer:
$\frac{y}{x}$
Hint:
Differentiate the function w.r.t x
Given:
$\sin^{- 1} (\frac{x^{2} - y^{2}}{x^{2} + y^{2}}) = \log a$
Solution:
$\sin^{- 1} (\frac{x^{2} - y^{2}}{x^{2} + y^{2}}) = \log a$
$\frac{x^{2} - y^{2}}{x^{2} + y^{2}} = \sin \log a$
$\frac{(x^{2} + y^{2}) (2 x - 2 y \frac{d y}{d x}) - (x^{2} - y^{2}) (2 x + 2 y \frac{d y}{d x})}{{(x^{2} + y^{2})}^{2}} = 0$
$\frac{2 x^{3} - 2 x^{2} y d y + 2 x y^{2} - 2 y^{3} \frac{d y}{d x} - 2 x^{3} - 2 x^{2} y \frac{d y}{d x} + 2 x y^{2} + 2 y^{3} \frac{d y}{d x}}{{(x^{2} + y^{2})}^{2}} = 0$
$\begin{aligned} - 4 x^{2} y \frac{d y}{d x} + 4 x y^{2} = 0 \\ - 4 x^{2} y \frac{d y}{d x} = - 4 x y^{2} \\ \frac{d y}{d x} = \frac{4 x y^{2}}{4 x^{2} y} = \frac{y}{x} \end{aligned}$

Differentiation exercise Multiple choice question 30

Answer:
$\frac{\cos^{2} (a + y)}{\cos a}$
Hint:
Differentiate the function w.r.t x
Given:
$\sin y = x \cos (a + y)$
Solution:

\begin{aligned} \sin y = x \cos (a + y) \\ \frac{d}{d x} (\sin y) = \frac{d}{d x} [x \cos (a + y)] \end{aligned}

\begin{aligned} \cos y \frac{d y}{d x} = 1 \times \cos (a + y) - x \sin (a + y) \frac{d}{d x} (a + y) \\ \cos y \frac{d y}{d x} = \cos (a + y) - x \sin (a + y) \frac{d y}{d x} \end{aligned}

\begin{aligned} \cos y \frac{d y}{d x} + x \sin (a + y) \frac{d y}{d x} = \cos (a + y) \\ [\cos y + x \sin (a + y)] \frac{d y}{d x} = \cos (a + y) \end{aligned}

[\cos y + \frac{\sin y}{\cos (a + y)} \times \sin (a + y)] \frac{d y}{d x} = \cos (a + y)

[\begin{array}{l} \sin y = x \cos (a + y) \\ x = \frac{\sin y}{\cos (a + y)} \end{array}]

[\frac{\cos (a + y) \cos y + \sin y \sin (a + y)}{\cos (a + y)}] \frac{d y}{d x} = \cos (a + y)

\begin{aligned} \frac{\cos (a + y - y)}{\cos (a + y)} \times \frac{d y}{d x} = \cos (a + y) \\ \frac{d y}{d x} = \frac{\cos^{2} (a + y)}{\cos a} \end{aligned}

Differentiation exercise Multiple choice question 31

Answer:
$- \frac{4 x}{1 - x^{4}}$
Hint:
Differentiate the function w.r.t x
Given:
$y = \log (\frac{1 - x^{2}}{1 + x^{2}})$
Solution:
$y = \log (\frac{1 - x^{2}}{1 + x^{2}})$
$\frac{d y}{d x} = \frac{1}{\frac{1 - x^{2}}{1 + x^{2}}} \frac{d}{d x} (\frac{1 - x^{2}}{1 + x^{2}})$
$= \frac{1 + x^{2}}{1 - x^{2}} [\frac{(1 + x^{2}) (- 2 x) - (1 - x^{2}) (2 x)}{{(1 + x^{2})}^{2}}]$
$\begin{aligned} = \frac{1}{1 - x^{2}} [\frac{- 2 x - 2 x^{3} - 2 x + 2 x^{3}}{1 + x^{4}}] \\ \frac{d y}{d x} = \frac{- 4 x}{1 - x^{4}} \end{aligned}$

Differentiation exercise Multiple choice question 32

Answer:
1
Hint:
Differentiate the function w.r.t x
Given:
$y = \tan^{- 1} (\frac{\sin x + \cos x}{\cos x - \sin x})$
Solution:
$y = \tan^{- 1} (\frac{\sin x + \cos x}{\cos x - \sin x})$
$\frac{d y}{d x} = \frac{1}{1 + {(\frac{\sin x + \cos x}{\cos x - \sin x})}^{2}} \frac{d}{d x} (\frac{\sin x + \cos x}{\cos x - \sin x})$
$\frac{d y}{d x} = \frac{(\cos x - \sin x)^{2}}{(\cos x - \sin x)^{2} + (\sin x + \cos x)^{2}} \times$ $[\frac{(\cos x - \sin x) \frac{d}{d x} (\sin x + \cos x) - (\sin x + \cos x) \frac{d}{d x} (\cos x - \sin x)}{(\cos x - \sin x)^{2}}]$
$= \frac{(\cos x - \sin x)^{2}}{(\cos x - \sin x)^{2} + (\sin x + \cos x)^{2}} [\frac{(\cos x - \sin x) (\cos x - \sin x) - (\sin x + \cos x) (- \sin x - \cos x)}{(\cos x - \sin x)^{2}}]$
$= \frac{(\cos x - \sin x)^{2}}{(\cos x - \sin x)^{2} + (\sin x + \cos x)^{2}} [\frac{(\cos x - \sin x) (\cos x - \sin x) + (\sin x + \cos x) (\sin x + \cos x)}{(\cos x - \sin x)^{2}}]$
$\begin{aligned} = \frac{(\cos x - \sin x)^{2}}{(\cos x - \sin x)^{2} (\sin x + \cos x)^{2}} \times \frac{(\cos x - \sin x)^{2} + (\sin x + \cos x)^{2}}{(\cos x - \sin x)^{2}} \\ \frac{d y}{d x} = 1 \end{aligned}$

Differentiation exercise Multiple choice question 33

Answer:
$(c) \frac{d y}{d x} = \frac{y - 1}{x + 1}$
Hint:
Differentiate the function w.r.t x
Given:
$\sec^{- 1} (\frac{1 + x}{1 - y})$
Solution:

\sec^{- 1} (\frac{1 + x}{1 - y}) = a

\begin{aligned} \sec a = \frac{1 + x}{1 - y} \\ (1 - y) \sec a = 1 + x \end{aligned}

Differentiate w.r.t x

- \sec a \frac{d y}{d x} = 1 [\frac{d (1)}{d x} = 0, \frac{d (x)}{d x} = 1]

\frac{d y}{d x} = \frac{- 1}{\sec a}

\begin{aligned} = \frac{- (1 - y)}{(1 + x)} \\ = \frac{y - 1}{1 + x} \end{aligned}

Hence, option (c) is correct.

The class 12 RD Sharma chapter 10 exercise MCQ solution can be practiced for school, board, and JEE exams. The syllabus covered in the book includes all the latest updates that NCERT Books receive every year. The RD Sharma class 12 solutions Differentiation ex MCQ book is also updated periodically to include updates so that students may not miss out on any question.

RD Sharma class 12 solutions chapter 10 ex MCQ is the ultimate guidebook for students who want to improve their knowledge on this particular chapter. The MCQ portion has numerous questions which might appear in board exams as it has in the past. Teachers also use RD Sharma class 12th exercise MCQ to give home assignments to students, so using the book will make it easier for them to get the answers correct.

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