JEE Advanced · Mathematics · 25. AOD

Paragraph:

Let $f:[0,1] \rightarrow \mathbb{R}$ (the set of all real numbers) be a function. Suppose the function $f$ is twice differentiable, $f(0)=f(1)=0$ and satisfies $f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^{x}, x \in[0,1] .$

Question:

Which of the following is true for $f(x)$ ?

A $0 < f (x) < \infty$
B $- \frac{1}{2} < f (x) < \frac{1}{2}$
C $- \frac{1}{4} < f (x) < 1$
D $- \infty < f (x) < 0$

Verified Solution

Answer & Solution

Correct Answer

(D) $- \infty < f (x) < 0$

Step-by-step Solution

Detailed explanation

f^{″} (x) - 2 f^{'} (x) + f (x) \geq e^{x}

f^{″} (x) \cdot e^{- x} - f^{'} (x) e^{- x} - f^{'} (x) e^{- x} + f (x) e^{- x} \geq 1

\frac{d}{dx} (f^{'} (x) e^{- x}) - \frac{d}{dx} (f (x) \cdot e^{- x}) \geq 1

\frac{d}{dx} (f^{'} (x) e^{- x} - f (x) e^{- x}) \geq 1

\Rightarrow \frac{d^{2}}{{dx}^{2}} (e^{- x} f (x)) \geq 1

\forall x \in [0, 1]

Let

ϕ (x) = e^{- x} f (x)

\Rightarrow ϕ (x)

is concave upward

f (0) = f (1) = 0

\Rightarrow ϕ (0) = 0 = ϕ (1)

\Rightarrow f (x) < 0

\Rightarrow ϕ (x) < 0

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 (x) = x + \log_{e} x - x \log_{e} x, x \in (0, \infty) .

•

Column 1 contains information about zeros of

f (x), f^{'} (x)

and

f^{''} (x) .

•

Column 2 contains information about the limiting behaviour of

f (x), f^{'} (x)

and

f^{''} (x)

at infinity.

•

Column 3 contains information about increasing-decreasing nature of

f (x)

and

f^{'} (x) .

Column 1	Column 2	Column 3
(I) $f (x) = 0$ for some $x \in (1, e^{2})$	(i) $\lim_{x \to \infty} f (x) = 0$	(P) $f$ is increasing in (0, 1)
(II) $f^{'} (x) = 0$ for some $x i n (1, e)$	(ii) $\lim_{x \to \infty} f (x) = - \infty$	(Q) $f$ is decreasing in $(e, e^{2})$
(III) $f^{'} (x) = 0$ for some $x \in (0, 1)$	(iii) $\lim_{x \to \infty} f^{'} (x) = - \infty$	(R) $f^{'}$ is increasing in (0, 1)
(IV) $f^{''} (x) = 0$ for some $x \in (1, e)$	(iv) $\lim_{x \to \infty} f^{''} (x) = 0$	(S) $f^{'}$ is decreasing in $(e, e^{2})$

Which of the following options is the only Correct combination?

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