Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2013-Paper-2 SECTION - 2 : (Paragraph Type) This section …
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2013-Paper-2 Single Correct MCQ
Published on: August 13, 2026

SECTION - 2 : (Paragraph Type)

This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph.

Each question has four choices

A
, 0 < f(x) <
B
,
C
and
D
out of which ONLY ONE is correct. Paragraph for Questions (i) and (ii) Let f : [0, 1] 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 ” (x) - 2f ’ (x) + f(x) e x , x [0, 1]. (i) Which of the following is true for 0 < x < 1 ? (ii) If the function e -x f(x) assumes its minimum in the interval [0, 1] at x = , which of the following is true ?

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Text Solution

Verified by Experts
The correct answer is:
CHECK THE SOLUTION.

(i)

Let g (x) = e – x f (x)

and g ” (x) > 1 > 0

So, g (x) is concave upward and g (0) = g (1) = 0

Hence, g (x) < 0 x (0, 1)

e – x f (x) < 0

f (x) < 0 x (0, 1)

Alternate Solution

f"(x) – 2f ’ (x) + f (x) e x

Let g (x) = f (x) e – x –

g (0) = 0, g (1) =

Since g is concave up so it will always lie below the chord joining the extremities which is y =

(ii)

Let, g (x) = e – x f (x)

As g ” (x) > 0 so g ’ (x) is increasing.

So, for x < 1/4, g ’ (x) < g ’ (1/4) = 0

(f ’ (x) – f (x))e – x < 0

f ’ (x) < f (x) in (0, 1/4).

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