Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be the function given by $f(x)=\cos \left(\tan ^{-1} x\ri...

Step 1: Express $ f(x) $ in simpler formGiven$ f(x) = \cos(\tan^{-1} x) $Let $\theta = \tan^{-1} x.$Then, by definition, $\tan \theta = x.$We can now use the identity:$ \cos \theta = \frac{1}{\sqrt{1+\tan^2 \theta}} $Therefore,$ f(x) = \frac{1}{\sqrt{1+x^2}} $Step 2: Differentiate $ f(x) $$ f'(x) = \frac{d}{dx}\left( (1+x^2)^{-1/2} \right) = -\frac{1}{2}(1+x^2)^{-3/2} \cdot (2x) = -\frac{x}{(1+x^2)^{3/2}} $Step 3: Determine where $ f'(x) $ is negative or positiveSince the denominator $(1+x^2)^{3/2} > 0$ for all $x$,the sign of $f'(x)$ depends only on the numerator $-x$:For $x > 0$: $f'(x) = -\frac{x}{(1+x^2)^{3/2}} 0.$For $x 0$⇒ $f$ isincreasingfor $x 0$ is true.✅Correct answer: Option AFinal Answer:$ \boxed{\text{Option A: } f \text{ is decreasing for } x > 0.} $

Inverse Trigonometric Functions - IAT-IISER Maths PYQs

Step 1: Express

f(x)

in simpler formGiven

f(x)%20%3D%20%5Ccos(%5Ctan%5E%7B-1%7D%20x)

Let

%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%20x.

Then, by definition,

%5Ctan%20%5Ctheta%20%3D%20x.

We can now use the identity:

%5Ccos%20%5Ctheta%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B1%2B%5Ctan%5E2%20%5Ctheta%7D%7D

Therefore,

f(x)%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B1%2Bx%5E2%7D%7D

Step 2: Differentiate

f(x)

f'(x)%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft(%20(1%2Bx%5E2)%5E%7B-1%2F2%7D%20%5Cright)%20%3D%20-%5Cfrac%7B1%7D%7B2%7D(1%2Bx%5E2)%5E%7B-3%2F2%7D%20%5Ccdot%20(2x)%20%3D%20-%5Cfrac%7Bx%7D%7B(1%2Bx%5E2)%5E%7B3%2F2%7D%7D

Step 3: Determine where

f'(x)

is negative or positiveSince the denominator

(1%2Bx%5E2)%5E%7B3%2F2%7D%20%3E%200

for all

x

,the sign of

f'(x)

depends only on the numerator

-x

:For

x%20%3E%200

f'(x)%20%3D%20-%5Cfrac%7Bx%7D%7B(1%2Bx%5E2)%5E%7B3%2F2%7D%7D%20%3C%200

⇒

f

isdecreasingfor

x%20%3E%200.

For

x%20%3C%200

f'(x)%20%3E%200

⇒

f

isincreasingfor

x%20%3C%200.

x%20%3D%200

, derivative is 0.Step 4: Choose the correct optionOnly the statement that

f

is decreasing for

x%20%3E%200

is true.✅Correct answer: Option AFinal Answer:

%5Cboxed%7B%5Ctext%7BOption%20A%3A%20%7D%20f%20%5Ctext%7B%20is%20decreasing%20for%20%7D%20x%20%3E%200.%7D

Inverse Trigonometric Functions

High-Yield Trend

Questions
1 MCQs

Official Solution

Inverse Trigonometric Functions

High-Yield Trend

Questions 1 MCQs

Official Solution

Questions
1 MCQs