JEE Advanced · Mathematics · 25. AOD

Let $f(x)$ be a non-constant twice differentiable function defined on $(-\infty, \infty)$, such that $f(x)=f(1-x)$ and $f^{\prime}\left(\frac{1}{4}\right)=0$. Then,

A
$f^{\prime}(x)$ vanishes atleast twice on $[0,1]$
B
$f^{\prime}\left(\frac{1}{2}\right)=0$
C
$\int_{\frac{1}{2}}^{\frac{1}{2}} f\left(x+\frac{1}{2}\right) \sin x d x=0$
D
$\int_0^{\frac{1}{2}} f(t) e^{\sin \pi t} d t=\int_{\frac{1}{2}}^1 f(1-t) e^{\sin \pi t} d t$

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Answer & Solution

Correct Answer

(A)
$f^{\prime}(x)$ vanishes atleast twice on $[0,1]$

Step-by-step Solution

Detailed explanation

Given that, $f(x)=f(1-x)$
On differentiating w.r.t. $x$, we get
\[
f^{\prime}(x)=-f^{\prime}(1-x)
\]
Let us put $\quad x=\frac{1}{2}$
\[
\Rightarrow \quad 2 f^{\prime}\left(\frac{1}{2}\right)=0 \Rightarrow f^{\prime}\left(\frac{1}{2}\right)=0
\]
Since, $f^{\prime}\left(\frac{1}{2}\right)=0$ and $f^{\prime}\left(\frac{1}{4}\right)=0$
$\Rightarrow f^{\prime \prime}(x)=0$ at two points in $[0,1]$.
Now, $\int_{-1 / 2}^{1 / 2} f\left(x+\frac{1}{2}\right) \sin x d x=0$
As, $f\left(x+\frac{1}{2}\right) \sin x$ is an odd function which is clear from the following explanation.
Let $g(x)=f\left(x+\frac{1}{2}\right) \sin x$,
\[
g(-x)=f\left(\frac{1}{2}-x\right) \sin (-x)=-\sin x f\left(1-\left(\frac{1}{2}-x\right)\right)=-\sin x f\left(\frac{1}{2}+x\right)=-g(x)
\]
Moreover, $\int_{1 / 2}^1 f(1-t) e^{\sin (\pi t)} d t=\int_0^{1 / 2} f(u) \cdot e^{\sin \pi u} d u$ where, $1-t=u$.

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.

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Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column 1	Column 2	Column 3
(I) $x^{2} + y^{2} = a^{2}$	(i) $m y = m^{2} x + a$	(P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$
(II) $x^{2} + a^{2} y^{2} = a^{2}$	(ii) $y = m x + a \sqrt{m^{2} + 1}$	(Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$
(III) $y^{2} = 4 a x$	(iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$	(R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$
(IV) $x^{2} - a^{2} y^{2} = a^{2}$	(iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$	(S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

For

a = \sqrt{2}

, if a tangent is drawn to a suitable conic (Column 1) at the point of contact

(- 1, 1)

, then which of the following options is the only Correct combination for obtaining its equation?

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A solid sphere of mass $1 kg$ and radius $1 m$ rolls without slipping on a fixed inclined plane with an angle of inclination $θ = 30 °$ from the horizontal. Two forces of magnitude $1 N$ each, parallel to the incline, act on the sphere, both at distance $r = 0.5 m$ from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is $m s^{- 2}$ . (Take $g = 10 m s^{- 2}$ )
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Let \(f(x)\) be a non-constant twice differentiable function defined on \((-\infty, \infty)\), such that \(f(x)=f(1-x)\) and \(f^{\prime}\left(\frac{1}{4}\right)=0\). Then,

Answer & Solution

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