NCERT Solutions for Exercise 2.2 Class 12 Maths Chapter 2 - Inverse Trigonometric Functions

# NCERT Solutions for Exercise 2.2 Class 12 Maths Chapter 2 - Inverse Trigonometric Functions

Edited By Ramraj Saini | Updated on Dec 03, 2023 01:48 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 2 Exercise 2.2

NCERT Solutions for Exercise 2.2 Class 12 Maths Chapter 2 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 exercise 2.1 talks about the principal values in a given range of various trigonometric functions. Exercise 2.2 Class 12 Maths basically deals with questions in which a certain range of the angle is provided. NCERT Solutions for Class 12 Maths chapter 2 exercise 2.2 has some questions which are quite tricky. Silly mistake chances are quite high. So practice them well before the exam with precision. Also below is the list of other NCERT exercises which can be referred to for more understanding.

12th class Maths exercise 2.1 answers 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 together using the link provided below.

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## Question:1 Prove the following: $3\sin^{-1}x = \sin^{-1}(3x - 4x^3),\;\;x\in\left[-\frac{1}{2},\frac{1}{2} \right ]$

Given to prove: $3\sin^{-1}x = \sin^{-1}(3x - 4x^3)$

where, $x\:\epsilon \left[-\frac{1}{2},\frac{1}{2} \right ]$ .

Take $\theta= \sin ^{-1}x$ or $x = \sin \theta$

Take R.H.S value

$\sin^{-1}(3x - 4x^3)$

= $\sin^{-1}(3\sin \theta - 4\sin^3 \theta)$

= $\sin^{-1}(\sin 3\theta)$

= $3\theta$

= $3\sin^{-1}x$ = L.H.S

Given to prove $3\cos^{-1} x = \cos^{-1}(4x^3 - 3x), \;\;x\in\left[\frac{1}{2},1 \right ]$ .

Take $\cos^{-1}x = \theta$ or $\cos \theta = x$ ;

Then we have;

R.H.S.

$\cos^{-1}(4x^3 - 3x)$

= $\cos^{-1}(4\cos^3 \theta - 3\cos\theta)$ $\left [ \because 4\cos^3 \theta - 3\cos\theta = \cos3 \theta \right ]$

= $\cos^{-1}(\cos3\theta)$

= $3\theta$

= $3\cos^{-1}x$ = L.H.S

Hence Proved.

Given to prove $\tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24} = \tan^{-1}\frac{1}{2}$

We have L.H.S

$\tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24}$

$=\tan^{-1}\frac{\frac{2}{11} + \frac{7}{24} }{1 - \left ( \frac{2}{11}\times\frac{7}{24} \right ) }$ $\left [ \because \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x +y}{1 - xy} \right ]$

$=\tan^{-1}\frac{11\times 24 }{\frac{11\times24 -14}{11\times 24} }$

$=\tan^{-1}\frac{48 + 77}{264 -14}$

$=\tan^{-1}\left ( \frac{125}{250}\right ) = \tan^{-1}\left ( \frac{1}{2} \right )$

= R.H.S

Hence proved.

Given to prove $2\tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{31}{17}$

Then taking L.H.S.

We have $2\tan^{-1} \frac{1}{2} + \tan^{-1}\frac{1}{7}$

$=\tan^{-1} \frac{2.\frac{1}{2}}{1 - \left ( \frac{1}{2} \right )^2} + \tan^{-1} \frac{1}{7}$ $\because 2\tan^{-1} x = \tan^{-1} \frac{2x}{1- x^2}$

$=\tan^{-1} \frac{1}{(\frac{3}{4})} + \tan^{-1} \frac{1}{7}$

$=\tan^{-1} \frac{4}{3} + \tan^{-1} \frac{1}{7}$

$=\tan^{-1} \frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3}.\frac{1}{7}}$ $\left [ \because \tan^{-1}x + \tan^{-1} y = \tan^{-1} \frac{x +y}{1- xy}\right ]$

$=\tan^{-1} \left ( \frac{\frac{28+3}{21}}{\frac{21-4}{21}} \right )$

$=\tan^{-1} \frac{31}{17}$

= R.H.S.

Hence proved.

We have $\tan^{-1}\frac{\sqrt{1 + x^2}- 1}{x}$

Take

$\therefore$ $\tan^{-1} \frac {\sqrt{1+x^2} - 1}{x} = \tan^{-1}\frac{\sqrt{1+\tan^2 \Theta - 1}}{\tan \Theta}$

$=\tan^{-1}(\frac{sec \Theta-1}{tan \Theta}) = \tan^{-1}\left ( \frac{1-cos \Theta}{sin \Theta} \right )$

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

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

$=\frac{1}{2}\tan^{-1}x$ is the simplified form.

Given that $\tan^{-1} \frac{1}{\sqrt{x^2 -1}},\;\; |x| > 1$

Take $x =cosec\ \Theta$ or $\Theta = cosec ^{-1}x$

$\therefore tan^{-1}\frac{1}{\sqrt{x^2-1}}$

$=tan^{-1} \frac{1}{\sqrt{cosec^2 \Theta -1}}$

$=tan^{-1}(\frac{1}{\cot \Theta})$

$=tan^{-1}(\tan \Theta)$ = $\Theta$

= $cosec^{-1}x$

$=\frac{\pi}{2}- \sec^{-1}x$ $[\because cosec^{-1}x + \sec^{-1}x = \frac{\pi}{2}]$

Given that $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1 + \cos x}} \right ),\;\; 0< x < \pi$

We have in inside the root the term : $\frac{1-\cos x}{1 + \cos x}$

Put $1-\cos x = 2\sin^2\frac{x}{2}$ and $1+\cos x = 2\cos^2\frac{x}{2}$ ,

Then we have,

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

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

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

Hence the simplest form is $\frac{x}{2}$

Given $\tan^{-1}\left(\frac{\cos x -\sin x }{\cos x + \sin x} \right )$ where $x\:\epsilon\:( \frac{-\pi}{4} < x < \frac{3\pi}{4})$

So,

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

Taking $\cos x$ common from numerator and denominator.

We get:

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

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

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

= $\frac{\pi}{4} - x$ is the simplest form.

Given that $\tan^{-1} \frac{x}{\sqrt{a^2 - x^2}}, \;\; |x| < a$

Take $x = a\sin \theta$ or

$\theta = \sin^{-1}\left ( \frac{x}{a} \right )$ and putting it in the equation above;

$\tan^{-1} \frac{a\sin \theta}{\sqrt{a^2 - (a\sin \theta)^2}}$

$=\tan^{-1} \frac{a\sin \theta}{a\sqrt{1 - \sin^2 \theta}}$

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

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

$=\theta = \sin^{-1}\left ( \frac{x}{a} \right )$ is the simplest form.

Given $\tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right )$

Here we can take $x = a\tan \theta \Rightarrow \frac{x}{a} = \tan \theta$

So, $\theta = \tan^{-1}\left ( \frac{x}{a} \right )$

$\tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right )$ will become;

$=\tan^{-1}\left(\frac{3a^2a\tan \theta -(a\tan \theta)^3}{a^3 - 3a(a\tan \theta)^2} \right ) = \tan^{-1}\left(\frac{3a^3\tan \theta -a^3\tan ^3 \theta}{a^3 - 3a^3\tan ^2 \theta} \right )$

and as $\left [ \because \left(\frac{3\tan \theta -\tan ^3 \theta}{ 1- 3\tan ^2 \theta} \right) =\tan 3\theta \right ]$ ;

$=3 \theta$

$=3 \tan^{-1}(\frac{x}{a})$

hence the simplest form is $3 \tan^{-1}(\frac{x}{a})$ .

Given equation:

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

So, solving the inner bracket first, we take the value of $\sin x^{-1} \frac{1}{2} = x.$

Then we have,

$\sin x = \frac{1}{2} = \sin \left ( \frac{\pi}{6} \right )$

Therefore, we can write $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$ .

$\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ] = \tan^{-1}\left[2\cos\left(2\times\frac{\pi}{6} \right ) \right ]$

$= \tan^{-1}\left[2\cos\left(\frac{\pi}{3} \right ) \right ] = \tan^{-1}\left[2\times\left(\frac{1}{2} \right ) \right ] = \tan^{-1}1 = \frac{\pi}{4}$ .

We have to find the value of $\cot(\tan^{-1}a + \cot^{-1}a)$

As we know $\left [\because \tan^{-1}x + \cot^{-1} x = \frac{\pi}{2} \right ]$ so,

Equation reduces to $\cot(\frac{\pi}{2}) = 0$ .

Taking the value $x = \tan \Theta$ or $\tan^{-1}x = \Theta$ and $y = \tan \Theta$ or $\tan^{-1} y = \Theta$ then we have,

= $\tan \frac{1}{2}\left[\sin^{-1}\frac{2\tan \Theta}{1+(\tan \Theta)^2} + cos^{-1}\frac{1-\tan^2 \Theta}{1+(\tan \Theta)^2} \right ]$ ,

= $\tan \frac{1}{2}\left[\sin^{-1}(\sin2\Theta) + cos^{-1} (\cos 2\Theta) \right ]$

$\because \left[\cos^{-1}(\frac{1-\tan^2 \Theta}{1+ \tan^2\Theta}) = \cos^{-1} (\cos2 \Theta) , \right ]$

$\because \left[\sin^{-1}(\frac{2\tan\Theta}{1+ \tan^2\Theta}) = \sin^{-1} (\sin2 \Theta) \right ]$

Then,

$=\tan \frac{1}{2}\left[2\tan^{-1}x + 2\tan^{-1}y \right ]$ $\because \left[\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right]$

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

$=\frac{x+y}{1-xy}$ Ans.

As we know the identity;

$sin^{-1} x + cos^{-1} x = \frac {\pi}{2},\ x\ \epsilon\ [-1,1]$ . it will just hit you by practice to apply this.

So, $\sin\left(\sin^{-1}\frac{1}{5} + \cos ^{-1}x \right ) =1$ or $\sin^{-1}\frac{1}{5} + \cos ^{-1}x =\sin^{-1}(1)$ ,

we can then write $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$ ,

putting in above equation we get;

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

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

Ans. $x = \frac{1}{5}$

Using the identity $\tan^{-1}x+\tan^{-1} y = \tan^{-1}{\frac{x+y}{1-xy}}$ ,

We can find the value of x;

So, $\tan^{-1}\frac{x-1}{x-2} + \tan^{-1}\frac{x+1}{x+2} =\frac{\pi}{4}$

on applying,

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

$=\tan^{-1}\frac{\frac{(x-1)(x+2)+(x-2)(x+1)}{x^2-4}}{1-\frac{x^2-1}{x^2-4}} = \tan^{-1} \left [ \frac{2x^2-4}{-3} \right ] = \frac{\pi}{4}$

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

= $2x^2=1$ or $x = \pm \frac{1}{\sqrt{2}}$ ,

Hence, the possible values of x are $\pm \frac{1}{\sqrt{2}}$ .

Given $\sin^{-1}\left (\sin\frac{2\pi}{3} \right )$ ;

We know that $\sin^{-1}(\sin x) = x$

If the value of x belongs to $\left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]$ then we get the principal values of $\sin^{-1}x$ .

Here, $\frac{2\pi}{3} \notin \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]$

We can write $\sin^{-1}\left (\sin\frac{2\pi}{3} \right )$ is as:

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

= $\sin^{-1}\left [ \sin \frac{\pi}{3} \right ]$ where $\frac{\pi}{3} \epsilon \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ]$

$\therefore \sin^{-1}\left (\sin\frac{2\pi}{3} \right )=\sin^{-1}\left [ \sin \frac{\pi}{3} \right ]=\frac{\pi}{3}$

As we know $\tan^{-1}\left ( \tan x \right ) =x$

If $x \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ).$ which is the principal value range of $\tan^{-1}x$ .

So, as in $\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$ ;

$\frac{3\pi}{4}\notin \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

Hence we can write $\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$ as :

$\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$ = $\tan^{-1}\left (\tan\frac{3\pi}{4} \right) = \tan^{-1}\left [ \tan(\pi - \frac{\pi}{4}) \right ] = \tan^{-1}\left [ \tan (\frac{-\pi}{4}) \right ]$

Where $-\frac{\pi}{4} \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

and $\therefore \tan^{-1}\left (\tan\frac{3\pi}{4} \right )=\tan^{-1}\left [ \tan (\frac{-\pi}{4}) \right ]=-\frac{\pi}{4}$

Given that $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )$

we can take $\sin^{-1}\frac{3}{5} = x$ ,

then $\sin x = \frac{3}{5}$

or $\cos x = \sqrt{1-\sin^{2}x}= \frac{4}{5}$

$\Rightarrow \tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

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

We have similarly;

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

Therefore we can write $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )$

$=\tan\left(\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{2}{3} \right )$

$=\tan\left[\tan^{-1}\left ( \frac{\frac{3}{4}+\frac{2}{3}}{1- \frac{3}{4}.\frac{2}{3}} \right ) \right ]$ from $As, \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]$

$=\tan \left (\tan^{-1} \frac{9+8}{12-6} \right ) = \tan \left (\tan^{-1} \frac{17}{6} \right )= \frac{17}{6}$

(A) $\frac{7\pi}{6}$

(B) $\frac{5\pi}{6}$

(C) $\frac{\pi}{3}$

(D) $\frac{\pi}{6}$

As we know that $\cos^{-1} (cos x ) = x$ if $x\epsilon [0,\pi]$ and is principal value range of $\cos^{-1}x$ .

In this case $\cos^{-1}\left(\cos\frac{7\pi}{6} \right )$ ,

$\frac{7\pi}{6} \notin [0,\pi]$

hence we have then,

$\cos^{-1}\left(\cos\frac{7\pi}{6} \right ) =$ $\cos^{-1} \left ( \cos \frac{-7\pi}{6} \right ) = \cos^{-1}\left [ \cos\left ( 2\pi - \frac{7\pi}{6} \right ) \right ]$

$\left [ \because \cos (2\pi + x) = \cos x \right ]$

$\therefore\ we\ have \cos^{-1}\left ( \cos \frac{7\pi}{6} \right ) = \cos^{-1}\left ( \cos \frac{5\pi}{6} \right ) = \frac{5\pi}{6}$

Hence the correct answer is $\frac{5\pi}{6}$ (B).

(A) $\frac{1}{2}$

(B)

(C) $\frac{1}{4}$

(D) $1$

Solving the inner bracket of $\sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right )$ ;

$\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right )$ or

Take $\sin^{-1}\left(-\frac{1}{2} \right ) = x$ then,

$\sin x =-\frac{1}{2}$ and we know the range of principal value of $\sin^{-1}x\ is\ \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ].$

Therefore we have $\sin^{-1}\left ( -\frac{1}{2} \right ) = -\frac{\pi}{6}$ .

Hence, $\sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right ) = \sin \left ( \frac{\pi}{3}+ \frac{\pi}{6} \right )= \sin \left ( \frac{3\pi}{6} \right ) = \sin\left ( \frac{\pi}{2} \right ) = 1$

Hence the correct answer is D.

(A) $\pi$

(B) $-\frac{\pi}{2}$

(C) 0

(D) $2\sqrt3$

We have $\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)$ ;

finding the value of $\cot^{-1}(-\sqrt3)$ :

Assume $\cot^{-1}(-\sqrt3) =y$ then,

$\cot y = -\sqrt 3$ and the range of the principal value of $\cot^{-1}$ is $(0,\pi)$ .

Hence, principal value is $\frac{5\pi}{6}$

Therefore $\cot^{-1} (-\sqrt3) = \frac {5\pi}{6}$

and $\tan^{-1} \sqrt3 = \frac{\pi}{3}$

so, we have now,

$\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)=\frac{\pi}{3} - \frac{5\pi}{6}$

$= \frac{2\pi - 5\pi}{6} = \frac{-3\pi}{6}$

or, $= \frac{ -\pi}{2}$

Hence the answer is option (B).

## More About NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2

The NCERT class 12 maths chapter Inverse Trigonometric Functions deals with questions taking from basic to advanced level. Exercise 2.2 Class 12 Maths has some moderate level of questions which are important for the examination. NCERT Solutions for class 12 maths chapter 2 exercise 2.2 along with NCERT exemplar questions is sufficient for a good understanding.

## Benefits of NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2

• The Class 12th maths chapter 2 exercise is described in a very easy manner. Students can comprehend by reading these notes.
• Exercise 2.2 Class 12 Maths is the extension of NCERT syllabus Exercise 2.1, as it has some difficult questions.
• NCERT book Class 12 Maths chapter 2 exercise 2.2 solutions has some questions which can take time, but one should make sure to complete them for better understanding.
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## Key Features Of NCERT Solutions for Exercise 2.2 Class 12 Maths Chapter 2

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 2.2 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 2.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 2.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 2.2 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 2.2 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 2.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
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### Also see-

NCERT Solutions Subject Wise

## Subject wise NCERT Exemplar solutions

Happy learning!!!

1. Is inverse trigonometric functions used in Physics also ?

Yes, just like integration, it is quite useful in Physics as well as Chemistry. These concepts are discussed in ex 2.2 class 12 comprehensively. To find the value of angles its quite useful. NCERT syllabus can be followed for the preparation of CBSE board exam.

2. Inverse functions of the trigonometric functions are known as ……..?

Inverse trigonometric functions

3. What is the use of Inverse trigonometric functions?

It is used to find angles from a given angle’s trigonometric ratios.

4. How to memorise principle values of basic inverse trigonometric functions ?

Make short notes and revise them multiple times. Practice questions so the brain retains it. For more questions use NCERT exemplar.

5. Is it required to memorise the principal value of basic inverse trigonometric functions ?

No, It is not mandatory but solving questions becomes easy if values are by heart.

6. How to solve proof related questions ?

Take step by step method to reach what is asked starting from the given conditions.

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• Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

• Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

• Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

Good Luck

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