Let f be a differentiable function satisfying \(\mathrm{f}(\mathrm{x})=1-2 \mathrm{x}+\int_{0}^{\mathrm{x}} \mathrm{e}^{(\mathrm{x}-\mathrm{t})} \mathrm{f}(\mathrm{t}) \mathrm{dt}, \mathrm{x} \in \mathbf{R}\) and let \(g(x)=\int_{0}^{x}(f(t)+2)^{15}(t-4)^{6}(t+12)^{17} d t, x \in \mathbf{R}\). If p and q are respectively the points of local minima and local maxima of g , then the value of \(|\mathrm{p}+\mathrm{q}|\) is equal to ______。

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Application of Derivatives

Let f be a differentiable function satisfying

f (x) = 1 - 2 x + \int_{0}^{x} e^{(x - t)} f (t) dt, x \in R

and let

g (x) = \int_{0}^{x} (f (t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} d t, x \in R

.
If p and q are respectively the points of local minima and local maxima of g , then the value of

∣ p + q ∣

is equal to ______。

Answer: 9

Write the relation as

f (x) = 1 - 2 x + e^{x} \int_{0}^{x} e^{- t} f (t) d t

and differentiate:

f^{'} (x) = - 2 + e^{x} \int_{0}^{x} e^{- t} f (t) d t + f (x)

. Since

e^{x} \int_{0}^{x} e^{- t} f (t) d t = f (x) - 1 + 2 x

, this reduces to

f^{'} (x) - 2 f (x) = 2 x - 3

. The solution with

f (0) = 1

f (x) = 1 - x

. Then

g^{'} (x) = (f (x) + 2)^{15} (x - 4)^{6} (x + 12)^{17} = (3 - x)^{15} (x - 4)^{6} (x + 12)^{17} .

The zeros are

x = - 12, 3, 4

. The factor

(x - 4)^{6}

has even exponent, so

g^{'}

does not change sign at

x = 4

. Across

x = - 12

the sign of

g^{'}

changes from negative to positive — a local minimum at

p = - 12

; across

x = 3

it changes from positive to negative — a local maximum at

q = 3

. Therefore

∣ p + q ∣ = ∣ - 12 + 3∣ = 9

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Integer answer