JEE Advanced · Mathematics · 22. Functions

Let $f(x)=(1-x)^{2} \sin ^{2} x+x^{2}$ for all $x \in I R$ and let $g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t$ for all $x \in(1, \infty)$.

Question: Consider the statements:

$P$ : There exists some $x \in \mathrm{R}$ such that $f(x)+2 x$

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

$Q$ : There exists some $x \in \mathrm{R}$ such that $2 f(x)+1$ $=2 x(1+x)$

Then

A both $P$ and $Q$ are true
B $P$ is true and $Q$ is false
C $P$ is false and $Q$ is true
D both $P$ and $Q$ are false

Verified Solution

Answer & Solution

Correct Answer

(C) $P$ is false and $Q$ is true

Step-by-step Solution

Detailed explanation

For the statement $P, f(x)+2 x=2\left(1+x^{2}\right)$

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

$\Rightarrow(1-x)^{2} \sin ^{2} x=x^{2}-2 x+1+1$

$\Rightarrow(1-x)^{2} \sin ^{2} x=(1-x)^{2}+1$

$\Rightarrow(1-x)^{2} \cos ^{2} x=-1$, which is not possible for any real value of $x$.

Hence $P$ is not true.

Let $H(x)=2 f(x)+1-2 x(1+x)$

$H(0)=2 f(0)+1-0=1$

and $H(1)=2 f(1)+1-4=-3$

Hence, $H(x)$ has a solution in $(0,1)$

Therefore, $Q$ is true.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Let

f_{1} : R \to R, f_{2} (- \frac{π}{2}, \frac{π}{2}) \to R, f_{3} : (- 1, e^{\frac{π}{2}} - 2) \to R

and

f_{4} : ℝ \to ℝ

be functions defined by
(i)

f_{1} (x) = \sin (\sqrt{1 - e^{- x^{2}}})

(ii)

f_{2} (x) = \{\begin{cases} \frac{|\sin x|}{\tan^{- 1} (x)} & , x \neq 0 \\ 1 & , x = 0 \end{cases}

, where the inverse trigonometric function

\tan^{- 1} x

assumes values in

(- \frac{π}{2}, \frac{π}{2})

(iii)

f_{3} (x) = [\sin (\log_{e} (x + 2))]

, where, for

t \in R, [t]

denotes the greatest integer less than or equal to

t

,
(iv)

f_{4} (x) = \{\begin{matrix} x^{2} \sin (\frac{1}{x}) & , x \neq 0 \\ 0 & , x = 0 \end{matrix}

LIST-I	LIST-II
A. the function $f_{1}$ is	P. NOT continuous at $x = 0$
B. The function $f_{2}$ is	Q. continuous at $x = 0$ and NOT differentiable at $x = 0$
C. The function $f_{3}$ is	R. differentiable at $x = 0$ and its derivative is NOT continuous at $x = 0$
D. The function $f_{4}$ is	S. differentiable at $x = 0$ and its derivative is continuous at $x = 0$

The correct option is :

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