Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2026 · 28 January · Evening Shift · Mathematics · Application of Derivatives

Let f be a differentiable function satisfying f(x)=1-2 x+∫_ 0 ^ x...

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

Correct answer

0

Step-by-step solution

Given f(x)=1−2x+∫0xe(x−t)f(t) dtf(x) = 1 - 2x + \int_0^x e^{(x-t)} f(t) \, dt.
Differentiating and simplifying:
e−xf′(x)−e−xf(x)=−2e−x+(1−2x)e−x(−1)+e−xf(x)e^{-x}f'(x) - e^{-x}f(x) = -2e^{-x} + (1-2x)e^{-x}(-1) + e^{-x}f(x)
This reduces to:
f′(x)−2f(x)=2x−3f'(x) - 2f(x) = 2x - 3
Solving the linear ODE dydx−2y=2x−3\frac{dy}{dx} - 2y = 2x - 3:
y⋅e−2x=∫e−2x(2x−3) dxy \cdot e^{-2x} = \int e^{-2x}(2x-3)\,dx
On solving, we get f(x)=1−xf(x) = 1 - x.
Now, g(x)=∫0x(3−t)15(t−4)6(t+12)17 dtg(x) = \int_0^x (3-t)^{15}(t-4)^6(t+12)^{17} \, dt
g′(x)=(3−x)15(x−4)6(x+12)17g'(x) = (3-x)^{15}(x-4)^6(x+12)^{17}
=−(x−3)15(x−4)6(x+12)17= -(x-3)^{15}(x-4)^6(x+12)^{17}
Sign analysis of g′(x)g'(x):
g′(x)g'(x)changes from +→−+ \to -at x=3x = 3→ local maxima (q=3)(q = 3)
g′(x)g'(x)changes from −→+- \to +at x=−12x = -12→ local minima (p=−12)(p = -12)
∣p+q∣=∣−12+3∣=9|p + q| = |-12 + 3| = 9
Practice Application of Derivatives on Quantrex Academy →

More from Application of Derivatives

Consider the function f: (0, ∞) arrow (-∞, ∞) given by f(x) = x \, _e(x) - x + 1 . Then… JEE Advanced 2026Let P be the point on the parabola y = x^2 such that the slope of the tangent to the… JEE Advanced 2026Let f: ℝ arrow ℝ be a differentiable function such that f((x+y)/(3)) = (f(x) + f(y))/(3)… JEE Main 2026_ 0 ≤ x ≤ π (16 ((x)/(2)) ^3((x)/(2))) is equal to: JEE Main 2026Let f(x) be a polynomial of degree 5 , and have extrema at x = 1 and x = -1 . If _ x → 0… JEE Main 2026The number of critical points of the function f(x) = cases |( x)/(x)|, & x ≠ 0 1, & x = 0… JEE Main 2026Consider the following three statements for the function f:(0, ∞) arrow ℝ defined by… JEE Main 2026Let (2 α, α) be the largest interval in which the function f(t)= |t+1| t^ 2 , t 2 , is… JEE Main 2026Let f(x)=x^ 2025 -x^ 2000 , x ∈[0,1] and the minimum value of the function f(x) in the… JEE Main 2026Let f: ℝ → ℝ be a twice differentiable function such that f''(x) > 0 for all x ∈ ℝ and… JEE Main 2026Let f: R arrow R be a twice differentiable function such that the quadratic equation f(x)… JEE Main 2026Let ℝ denote the set of all real numbers. Define the function f: R arrow R by f(x)=\… JEE Advanced 2025Let ℝ denote the set of all real numbers. For a real number x , let [x] denote the… JEE Advanced 2025Let ℝ denote the set of all real numbers. Let f: ℝ arrow ℝ be defined by f(x)= cases (6… JEE Advanced 2025Let the function f(x)=(x)/(3)+(3)/(x)+3, x ≠ 0 be strictly increasing in (-∞, α_1) U(α_2… JEE Main 2025Let f: R arrow R be a polynomial function of degree four having extreme values at x=4 and… JEE Main 2025 Full Application of Derivatives list Application of Derivatives JEE Maths PYQsJEE Mathematics PYQs All JEE Main PYQs