NCERT Solutions for Miscellaneous Exercise Chapter 2 Class 12 - Inverse Trigonometric Functions

NCERT Solutions for Miscellaneous Exercise Chapter 2 Class 12 - Inverse Trigonometric Functions

Edited By Ramraj Saini | Updated on Dec 04, 2023 12:06 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Chapter 2 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 2 class 12 Inverse Trigonometric Functions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for Class 12 Maths chapter 2 miscellaneous exercise has a set of questions which are not introduced in earlier exercises. Class 12 Maths chapter 2 miscellaneous exercise basically deals with the questions related to proving of the inverse trigonometric functions equivalent. By using various methods and techniques, such equations are proved equal to each other. Students can easily find that many questions were asked from Class 12 Maths chapter 2 miscellaneous exercise in the board exams. Hence it is highly recommended to practice the NCERT solutions for Class 12 maths chapter 2 including miscellaneous exercise which is present in NCERT Class 12th book.

Miscellaneous exercise class 12 chapter 2 are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions: Miscellaneous Exercise

If $x \epsilon [0,\pi]$ then $\cos^{-1}(\cos x) = x$ , which is principal value of $\cos^{-1} x$ .

So, we have $\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )$

$where \:\frac{13\pi}{6} \notin \left [ 0, \pi \right ].$

$Hence\: we \:can\: write\: \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) \:as$

$=\cos^{-1}\left (\cos\left(2\pi+\frac{\pi}{6} \right ) \right )$

$=\cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right )$

$\frac{\pi}{6}\ \epsilon \left [ 0, \pi \right ]$

Therefore we have,

$\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) = \cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right ) = \frac{\pi}{6}$ .

We have given $\tan^{-1}\left(\tan\frac{7\pi}{6} \right )$ ;

so, as we know $\tan^{-1}\left(\tan x \right ) =x \:\:if\:\:x\epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

So, here we have $\frac{7\pi}{6} \notin \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$ .

Therefore we can write $\tan^{-1}\left(\tan\frac{7\pi}{6} \right )$ as:

$=\tan^{-1}\left(\tan \left (2\pi - \frac{5\pi}{6} \right ) \right )$ $\left [ \because \tan(2\pi - x) = -\tan x \right ]$

$=\tan^{-1}\left[-\tan \left ( \frac{5\pi}{6} \right ) \right ]$

$=\tan^{-1}\left[\tan \left (\pi- \frac{5\pi}{6} \right ) \right ]$

$=\tan^{-1}\left[\tan \left (\frac{\pi}{6} \right ) \right ]\:\:where\:\:\frac{\pi}{6} \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

$\therefore \tan^{-1} \left ( \tan \frac{7\pi}{6} \right ) = \tan^{-1} \left ( \tan \frac{\pi}{6} \right ) = \frac{\pi}{6}$ .

To prove: $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$ ;

$L.H.S=2\sin^{-1}\frac{3}{5}$

Assume that $\sin^{-1}\frac{3}{5} = x$

then we have $\sin x = \frac{3}{5}$ .

or $\cos x = \sqrt{1-\left (\frac{3}{5} \right )^2} = \frac{4}{5}$

Therefore we have

$\tan x = \frac{3}{4}\:\:or\:\:x = \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}$

Now,

We can write L.H.S as

$2\sin^{-1}\frac{3}{5} = 2\tan^{-1}\frac{3}{4}$

$=\tan^{-1} \left [\frac{2\times\frac{3}{4}}{1- \left ( \frac{3}{4} \right )^2} \right ]$ as we know $\left [2\tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \right ]$

$=\tan^{-1} \left [\frac{\frac{3}{2}}{\left ( \frac{16-9}{16} \right )} \right ] = \tan^{-1}\left ( \frac{3}{2}\times \frac{16}{7} \right )$

$=\tan^{-1} \frac{24}{7}=R.H.S$

L.H.S = R.H.S

Taking $\sin ^{-1} \frac{8}{17} = x$

then,

$\sin x = \frac{8}{17} \Rightarrow \cos x = \sqrt{1- \left ( \frac{8}{17} \right )^2} = \sqrt {\frac{225}{289}} = \frac{15}{17}.$

Therefore we have-

$\tan^{-1} x = \frac{8}{15} \Rightarrow x = \tan^{-1} \frac{8}{15}$

$\therefore \sin ^{-1} \frac{8}{17} = \tan ^{-1} \frac{8}{15}$ .............(1).

$Now, let\:\sin ^{-1} \frac{3}{5} = y$ ,

Then,

$\sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{3}{4}$ .............(2).

So, we have now,

L.H.S.

$\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5}$

using equations (1) and (2) we get,

$=\tan ^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}$

$=\tan^{-1} \frac{\frac{8}{15}+ \frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}}$ $[\because \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1 - xy}]$

$=\tan^{-1} (\frac{32+45}{60-24})$

$=\tan^{-1} (\frac{77}{36})$

= R.H.S.

Take $\cos^{-1}\frac{4}{5} = x$ and $\cos^{-1}\frac{12}{13} = y$ and $\cos^{-1}\frac{33}{65} = z$

then we have,

$\cos x = \frac{4}{5}$

$\sin x = \sqrt {1- \left ( \frac {4}{5} \right )^2} = \frac {3}{5}$

Then we can write it as:

$\tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$ or $x= \tan^{-1} \frac{3}{4}$

$\therefore \cos ^{-1} \frac{4}{5} = \tan^{-1} \frac{3}{4}$ ...............(1)

Now, $\cos^{-1}\frac{12}{13} = y$

$\cos y = \frac{12}{13} \Rightarrow$ $\sin y =\frac{5}{13}$

$\therefore \tan y = \frac{5}{12} \Rightarrow y = \tan^{-1} \frac{5}{12}$

So, $\cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12}$ ...................(2)

Also we have similarly;

$\cos^{-1}\frac{33}{65} = z$

Then,

$\cos^{-1}\frac{33}{65} = \tan^{-1} \frac{56}{33}$ ...........................(3)

Now, we have

L.H.S

$\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13}$ so, using (1) and (2) we get,

$=\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{5}{12}$

$=\tan^{-1}\left ( \frac{\frac{3}{4}+ \frac{5}{12}}{1-\left ( \frac{3}{4}\times \frac{5}{12} \right )} \right )$ $\because \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]$

$=\tan^{-1}\left ( \frac{36+20}{48-15} \right )$

$=\tan^{-1}\left ( \frac{56}{33} \right )$ or we can write it as;

$=\cos^{-1}\frac{33}{65}$

= R.H.S.

Hence proved.

Converting all terms in tan form;

Let $\cos^{-1}\frac{12}{13} = x$ , $\sin^{-1}\frac{3}{5} = y$ and $\sin^{-1}\frac{56}{65} = z$ .

now, converting all the terms:

$\cos^{-1}\frac{12}{13} = x$ or $\cos x = \frac{12}{13}$

We can write it in tan form as:

$\cos x = \frac{12}{13} \Rightarrow$ $\sin x = \frac{5}{13}$ .

$\therefore \tan x = \frac{5}{12} \Rightarrow x = \tan^{-1} \frac{5}{12}$

or $\cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12}$ ................(1)

$\sin^{-1}\frac{3}{5} = y$ or $\sin y = \frac{3}{5}$

We can write it in tan form as:

$\sin y = \frac{3}{5} \Rightarrow$ $\cos y = \frac{4}{5}$

$\therefore \tan y =\frac{3}{4} \Rightarrow y = \tan^{-1} \frac{3}{4}$

or $\sin^{-1}\frac{3}{5} = \tan^{-1} \frac{3}{4}$ ......................(2)

Similarly, for $\sin^{-1}\frac{56}{65} = z$ ;

we have $\sin^{-1}\frac{56}{65} = \tan^{-1} \frac{56}{33}$ .............(3)

Using (1) and (2) we have L.H.S

$\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5}$

$= \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{3}{4}$

On applying $\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy}$

We have,

$=\tan^{-1} \frac{\frac{5}{12}+\frac{3}{4}}{1-(\frac{5}{12}.\frac{3}{4})}$

$=\tan^{-1} (\frac{20+36}{48-15})$

$=\tan^{-1} (\frac{56}{33})$

$=\sin^{-1} (\frac{56}{65})$ ...........[Using (3)]

=R.H.S.

Hence proved.

Taking R.H.S;

We have $\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}$

Converting sin and cos terms in tan forms:

Let $\sin^{-1}\frac{5}{13} = x$ and $\cos^{-1}\frac{3}{5} = y$

now, we have $\sin^{-1}\frac{5}{13} = x$ or $\sin x = \frac{5}{13}$

$\sin x = \frac{5}{13} \:or\: \cos x =\frac{12}{13}\:or\:\tan x = \frac{5}{12}$

$\tan x = \frac{5}{12} \Rightarrow x =\tan^{-1} \frac{5}{12}$

$\therefore \sin^{-1} \frac{5}{13} = \tan^{-1} \frac{5}{12}$ ............(1)

Now, $\cos^{-1}\frac{3}{5} = y\Rightarrow \cos y = \frac{3}{5}$

$\cos y = \frac{3}{5} \:or\: \sin y = \frac{4}{5}\:or\:\tan y = \frac{4}{3}$

$\tan y = \frac{4}{3} \Rightarrow y = \tan^{-1} \frac{4}{3}$

$\therefore \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{4}{5}$ ................(2)

Now, Using (1) and (2) we get,

R.H.S.

$\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}$

$=\tan^{-1}\left ( \frac{\frac{5}{12}+\frac{4}{3}}{1- \frac{5}{12}\times \frac{4}{3}} \right )$ as we know $\left [ \tan^{-1} x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy} \right ]$

so,

$= \tan^{-1} \frac{63}{16}$

equal to L.H.S

Hence proved.

Applying the formlua:

$\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy}$ on two parts.

we will have,

$=\tan^{-1}\left (\frac{\frac{1}{5}+ \frac{1}{7}}{1- \frac{1}{5}\times \frac{1}{7}} \right ) + \tan^{-1}\left (\frac{\frac{1}{3}+ \frac{1}{8}}{1- \frac{1}{3}\times \frac{1}{8}} \right )$

$= \tan^{-1} \left ( \frac{7+5}{35-1} \right ) + \tan^{-1} \left ( \frac{8+3}{24-1} \right )$

$= \tan^{-1} \left ( \frac{12}{34} \right ) + \tan^{-1} \left ( \frac{11}{23} \right )$

$= \tan^{-1} \left ( \frac{6}{17} \right ) + \tan^{-1} \left ( \frac{11}{23} \right )$

$= \tan^{-1}\left [ \frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17}\times\frac{11}{23}} \right ]$

$= \tan^{-1}\left [ \frac{325}{325} \right ] = \tan^{-1} 1$

$=\frac{\pi}{4}$

Hence it s equal to R.H.S

Proved.

By observing the square root we will first put

$x= \tan^2 \theta$ .

Then,

we have $\tan^{-1} \sqrt{\tan^2 \theta} = \frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta}$

or, R.H.S.

$\frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \frac{1}{2}\cos^{-1}(cos2 \theta)$

$= \frac{1}{2}\times 2\theta = \theta$ .

L.H.S. $\tan^{-1} \sqrt{\tan^2 \theta} = \tan^{-1}(\tan \theta) = \theta$

hence L.H.S. = R.H.S proved.

Given that $\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right )$

By observing we can rationalize the fraction

$\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right )$

We get then,

$=\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right ) = \left(\frac{(\sqrt{1+\sin x} + \sqrt{1 - \sin x})^2}{{1+\sin x} - {1 + \sin x}} \right )$

$= \left(\frac{{1+\sin x} +{1 - \sin x} + 2\sqrt{(1+\sin x)(1-\sin x)} }{{1+\sin x} - {1 + \sin x}} \right )$

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

$= \cot \frac{x}{2}$

Therefore we can write it as;

$\cot^{-1}\left ( \cot \frac{x}{2} \right ) = \frac{x}{2}$

As L.H.S. = R.H.S.

Hence proved.

By using the Hint we will put $x = \cos 2\theta$ ;

we get then,

$=\tan^{-1}\left(\frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos2\theta}}{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}} \right )$

$=\tan^{-1}\left(\frac{\sqrt{2\cos^2 \theta} - \sqrt{2\sin^2\theta}}{\sqrt{2\cos^2 \theta} + \sqrt{2\sin^2\theta}} \right )$

$=\tan^{-1}\left(\frac{\sqrt2{\cos \theta} - \sqrt2{\sin\theta}}{\sqrt2{\cos \theta} + \sqrt2{\sin\theta}} \right )$

$=\tan^{-1}\left(\frac{{\cos \theta} - {\sin\theta}}{{\cos \theta} + {\sin\theta}} \right )$ dividing numerator and denominator by $\cos \theta$ ,

we get,

$= \tan^{-1}\left ( \frac{1-\tan \theta}{1+\tan \theta} \right )$

$= \tan^{-1} 1 - \tan^{-1} (\tan \theta)$ using the formula $\left [ \tan^{-1}x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy} \right ]$

$= \frac{\pi}{4} - \theta = \frac{\pi}{4}- \frac{1}{2}\cos^{-1}x$

As L.H.S = R.H.S

Hence proved

We have to solve the given equation:

$\frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\frac{1}{3} = \frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}$

Take $\frac{9}{4}$ as common in L.H.S,

$=\frac{9}{4}\left [ \frac{\pi}{2}- \sin^{-1}\frac{1}{3} \right ]$

or $=\frac{9}{4}\left [ \cos^{-1}\frac{1}{3} \right ]$ from $\left [ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \right ]$

Now, assume,

$\left [ \cos^{-1}\frac{1}{3} \right ] = y$

Then,

$\cos y = \frac{1}{3} \Rightarrow \sin y = \sqrt{1-(\frac{1}{3})^2} = \frac{2.\sqrt2}{3}$

Therefore we have now,

$y = \sin^{-1} \frac{2.\sqrt2}{3}$

So we have L.H.S then $= \frac{9}{4}\sin^{-1} \frac{2.\sqrt2}{3}$

That is equal to R.H.S.

Hence proved.

Given equation $2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x)$ ;

Using the formula:

$\left [ 2\tan^{-1}z = \tan^{-1} \frac{2z}{1-z^2} \right ]$

We can write

$2\tan^{-1}(\cos x) = \tan^{-1}\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ]$

$\tan^{-1}\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ] = \tan^{-1}\left [2cosec x \right ]$

So, we can equate;

$=\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ] = \left [2cosec x \right ]$

$=\left [ \frac{2\cos x}{\sin^2 x }\right ] = \left [ \frac{2}{sinx } \right ]$

that implies that $\cos x = \sin x$ .

or $\tan x =1$ or $x = \frac{\pi}{4}$

Hence we have solution $x = \frac{\pi}{4}$ .

Given equation is

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

L.H.S can be written as;

$\tan^{-1} \frac{1-x}{1+x} = \tan^{-1}1 - \tan^{-1}x$

Using the formula $\left [ \tan^{-1}x -\tan^{-1}y = \tan^{-1} \frac{x-y}{1+xy} \right ]$

So, we have $\tan^{-1}1 - \tan^{-1}x = \frac{1}{2} \tan^{-1}x$

$\Rightarrow \tan^{-1}1= \frac{3}{2} \tan^{-1}x$

$\Rightarrow\frac{\pi}{4}= \frac{3}{2} \tan^{-1}x$

$\Rightarrow \tan^{-1}x = \frac{\pi}{6}$

$\Rightarrow x= \tan \frac{\pi}{6} = \frac{1}{\sqrt3}$

Hence the value of $x= \frac{1}{\sqrt3}$ .

Let $\tan^{-1}x = y$ then we have;

$\tan y = x$ or

$y=\sin^{-1} \left ( \frac{x}{\sqrt{1+x^2}} \right ) \Rightarrow \tan^{-1} x = \sin^{-1} \left ( \frac{x}{\sqrt{1+x^2}} \right)$

$\Rightarrow \sin \left ( \tan^{-1} x \right ) = \sin\left ( \sin^{-1}\left ( \frac{x}{\sqrt{1+x^2}} \right ) \right ) = \frac{x}{\sqrt {1+x^2}}$

Hence the correct answer is D.

Given the equation: $\sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}$

we can migrate the $\sin^{-1}(1-x)$ term to the R.H.S.

then we have;

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

or $- 2\sin^{-1}x =\cos^{-1}(1-x)$ ............................(1)

from $\left [\because \cos^{-1}(1-x) + \sin^{-1}(1-x) = \frac{\pi}{2} \right ]$

Take $\sin^{-1}x = \Theta$ $\Rightarrow \sin \Theta = x$ or $\cos \Theta = \sqrt{1-x^2}$ .

So, we conclude that;

$\sin^{-1}x = \cos^{-1}\left ( \sqrt{1-x^2} \right )$

Therefore we can put the value of $\sin^{-1}x$ in equation (1) we get,

$- 2\cos^{-1}\left ( \sqrt{1-x^2} \right ) =\cos^{-1}(1-x)$

Putting x= sin y , in the above equation; we have then,

$\Rightarrow - 2\cos^{-1}\left ( \sqrt{1-(\sin y)^2} \right ) =\cos^{-1}(1-\sin y )$

$\Rightarrow - 2\cos^{-1}\left ( \sqrt{\cos^2 y} \right ) =\cos^{-1}(1-\sin y )$

$\Rightarrow - 2\cos^{-1}\left ( \cos y \right ) =\cos^{-1}(1-\sin y )$

$\Rightarrow \cos(-2y) = 1-\sin y$

$\Rightarrow - 2y=\cos^{-1}(1-\sin y )$

$\Rightarrow 1- 2\sin^2 y = 1-\sin y$

$\Rightarrow 2\sin^2 y - \sin y = 0$

$\Rightarrow \sin y(2 \sin y -1) = 0$

So, we have the solution;

$\sin y = 0\ or\ \frac{1}{2}$ Therefore we have $x = 0\ or\ x= \frac{1}{2}$ .

When we have $x= \frac{1}{2}$ , we can see that :

$L.H.S. = \sin ^{-1}\left ( 1 - \frac{1}{2} \right ) - 2\sin^{-1}\frac{1}{2} = - \sin^{-1}\frac{1}{2} = -\frac{\pi}{6}$

So, it is not equal to the R.H.S. $-\frac{\pi}{6} \neq \frac{\pi}{2}$

Thus we have only one solution which is x = 0

Hence the correct answer is (C).

Applying formula: $\left [ \tan^{-1} x - \tan^{-1}y = \tan^{-1} \left ( \frac{x-y}{1+xy} \right ) \right ]$ .

We get,

$\tan^{-1}\left (\frac{x}{y} \right )-\tan^{-1} \left ( \frac{x-y}{x+y} \right ) = \tan^{-1} \left [\frac{ \frac{x}{y} - \frac{x-y}{x+y}}{1+\left ( \frac{x}{y} \right ) \left ( \frac{x-y}{x+y} \right ) } \right ]$

$= \tan^{-1} \left [\frac{ \frac{x}{y} - \frac{x-y}{x+y}}{1+\left ( \frac{x}{y} \right ) \left ( \frac{x-y}{x+y} \right ) } \right ] = \tan^{-1} \left [ \frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y) + x(x-y)}{y(x+y)}} \right ]$

$= \tan^{-1}\left ( \frac{x^2+xy - xy + y^2}{xy + y^2 + x^2 - xy} \right )$

$= \tan^{-1}\left ( \frac{x^2 + y^2}{ y^2 + x^2 } \right ) = \tan^{-1} 1 = \frac{\pi}{4}$

Hence, the correct answer is C.

More About NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise

The NCERT Class 12 Maths chapter Inverse Trigonometric functions provided here is prepared by the experienced faculties. NCERT solutions for Class 12 Maths chapter 2 miscellaneous exercise covers the major syllabus of this chapter from exam perspective. As questions from this exercise are asked more than those of previous exercises. Therefore NCERT solutions for Class 12 Maths chapter 2 miscellaneous exercise becomes a must to do exercise for the examination.

Benefits of NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercises

• Miscellaneous exercise chapter 2 Class 12 NCERT syllabus has some of the questions which are very important from exam point of view.
• Questions mentioned in NCERT book Class 12 Maths chapter 2 miscellaneous solutions are of the level of JEE and NEET.
• It is a good source for revision also. Hence NCERT solutions for Class 12 Maths chapter 2 miscellaneous exercises can be referred directly to score well in the exam.
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Key Features Of NCERT Solutions For Class 12 Chapter 2 Miscellaneous Exercise

• Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 2, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 chapter 2 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this class 12 maths ch 2 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for class 12 chapter 2 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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NCERT Solutions Subject Wise

Subject wise NCERT Exemplar solutions

Happy learning!!!

1. What is the importance of miscellaneous exercise from exam perspective?

As direct questions are asked in the exam from this exercise, it is important to practice miscellaneous exercise before the examination. For more questions students can use NCERT exemplar.

2. List out the important topics of NCERT text book Class 12 Maths chapter 2?

The topics which are Important are among the following

• finding the inverse of sine, cos, tan etc. are important which are asked frequently in the exam

3. What is the overall importance of NCERT exercise in board examination ?

NCERT exercises are the favorite source of the Board examination. Hence it is advisable to go through the NCERT exercise.

4. Does memorization help in solving proof related questions?

Basic values of inverse trigonometric functions can be memorized, rest you will have to brainstorm in proof related questions.

5. What can be a basic approach to solve proof related questions ?

Process in step by step manner keeping in mind the final question can help in proving the desired direction.

6. How many total exercises are there in Class 12 Maths chapter 2 ?

In NCERT Class 12 Maths chapter 2, there are a total of 3 exercises.

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The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 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.

With increase of temperature, which of these changes?

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

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 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