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JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Definite Integrals

Let

f

be a twice differentiable function such that

f (x) = \int_{0}^{x} tan (t - x) dt - \int_{0}^{x} f (t) tan tdt, x \in (- \frac{π}{2}, \frac{π}{2})

Then

f^{''} (\frac{π}{6}) + 12 f^{'} (- \frac{π}{6}) + f (\frac{π}{6})

is equal to ____ .

Answer: 5

Solution

Evaluate the first integral by

u = t - x

:

\int_{0}^{x} tan (t - x) d t = \int_{- x}^{0} tan u d u = [- ln ∣ cos u ∣]_{- x}^{0} = ln ∣ cos x ∣

. Thus

f (x) = ln cos x - \int_{0}^{x} f (t) tan t d t

, with

f (0) = 0

. Differentiating,

f^{'} (x) = - tan x - f (x) tan x = - tan x (1 + f (x))

. Separating,

\frac{df}{1 + f} = - tan x d x

, so

ln ∣1 + f ∣ = ln ∣ cos x ∣ + C

, giving

1 + f = A cos x

; and

f (0) = 0

forces

A = 1

, hence

f (x) = cos x - 1

. Then

f^{'} (x) = - sin x

and

f^{''} (x) = - cos x

, so

f^{''} (\frac{π}{6}) + 12 f^{'} (- \frac{π}{6}) + f (\frac{π}{6}) = - \frac{3}{2} + 12 \cdot \frac{1}{2} + (\frac{3}{2} - 1) = 6 - 1 = 5

.

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Integer answer

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