Let for some function y = f(x), Then f(6) is equal to: (1) 1 (2) 2 (3) 6 (4) 3

Year	2026	2025	2024	Total
Questions	13	17	16	46

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Leibniz Rule for differentiating under the integral sign:

(d)/(dx) [ \intψ(x)φ(x) f(t) dt ] = f(φ(x))φ^ (x) - f(ψ(x))ψ^ (x)

Core Logic

Differentiate both sides of the integral equation with respect to $x$ $x$:

xf(x) = x² f^ (x) + 2xf(x) -xf(x) = x² f^ (x)

Step 1: Solving the Separable Differential Equation

Separating variables:

\int (f^ (x))/(f(x)) dx = \int -(1)/(x) dx ln |f(x)| = -ln x + ln c f(x) = (c)/(x)

Step 2: Applying Boundary Constraints

Given $f(2) = 3$ $f(2) = 3$:

3 = (c)/(2) c = 6 f(x) = (6)/(x)

Evaluating for $x = 6$ $x = 6$:

f(6) = (6)/(6) = 1

Pattern Recognition

Differentiating integral statements instantly converts complex integral equations into clean, separable differential equations.

Chapter Mix

Class 12 Maths: Differential Equations

More Differential Equations Previous-Year Questions — Page 10

Q11 jee_main_2024_31_jan_morning Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

(dy)/(dx) = (( x) + y)/( x( x - x x)), x in (0, (π)/(2))

satisfying the condition

y((π)/(4)) = 2

. Then,

y((π)/(3))

A. $\sqrt(3)(2 + ₑ\sqrt(3))$ $$\sqrt{3}(2 + \log_e\sqrt{3})$$
B. $\sqrt(3)2(2 + ₑ 3)$ $$\frac{\sqrt{3}}{2}(2 + \log_e 3)$$
C. $\sqrt(3)(1 + 2 ₑ 3)$ $$\sqrt{3}(1 + 2\log_e 3)$$
D. $\sqrt(3)(2 + ₑ 3)$ $$\sqrt{3}(2 + \log_e 3)$$

Solution

Core Logic

(dy)/(dx) = (( x)/( x) + y)/( x ((1)/( x) - ( ² x)/( x))) = ( x + y x)/( x (1 - ² x))

(dy)/(dx) = ( x + y x)/( x ² x) = ² x + (2y)/( 2x)

(dy)/(dx) - 2 (2x)y = ² x

Step 1: Integrating Factor

This is an LDE of form $(dy)/(dx) + Py = Q$ $\frac{dy}{dx} + Py = Q$.

I.F. = e\int -2 (2x) dx

Let $2x = t 2dx = dt$ $2x = t \implies 2dx = dt$.

I.F. = e-\int t dt = e-ln| (t/2)| = e-ln| x| = (1)/(| x|)

Step 2: Solution of LDE

y(I.F.) = \int Q(I.F.) dx + C

y(1)/( x) = \int ² x (1)/( x) dx + C

Let $x = t ² x dx = dt$ $\tan x = t \implies \sec^2 x dx = dt$.

y(1)/( x) = \int (dt)/(t) + C = ln| x| + C

y = x(ln| x| + C)

Step 3: Boundary Value

Given $y(π/4) = 2$ $y(\pi/4) = 2$:

2 = 1(ln 1 + C) C = 2

Thus, $y = x (ln| x| + 2)$ $y = \tan x (\ln|\tan x| + 2)$. At $x = π/3$ $x = \pi/3$:

y(π/3) = \sqrt(3)(ln\sqrt(3) + 2)

Chapter Mix

Class 12 Maths: Differential Equations

Let for some function $y = f(x)$ $y = f(x)$, $\int₀^x t f(t) dt = x² f(x), x > 0 and f(2) = 3.$ Then $f(6)$ $f(6)$ is equal to: (1) 1 (2) 2 (3) 6 (4) 3

Solution & Explanation

Related Formula

Core Logic

Step 1: Solving the Separable Differential Equation

Step 2: Applying Boundary Constraints

Pattern Recognition

Chapter Mix

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 10

Solution

Core Logic

Step 1: Integrating Factor

Step 2: Solution of LDE

Step 3: Boundary Value

Chapter Mix

More Differential Equations Questions — jee_main_2025_28_jan_morning

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Let for some function y = f(x)$y = f(x)$, ∫₀^x t f(t) dt = x² f(x), x > 0 and f(2) = 3.$$\int_0^x t f(t) dt = x^2 f(x), x > 0 \text{ and } f(2) = 3.$$ Then f(6)$f(6)$ is equal to: (1) 1 (2) 2 (3) 6 (4) 3

Solution & Explanation

Related Formula

Core Logic

Step 1: Solving the Separable Differential Equation

Step 2: Applying Boundary Constraints

Pattern Recognition

Chapter Mix

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 10

Solution

Core Logic

Step 1: Integrating Factor

Step 2: Solution of LDE

Step 3: Boundary Value

Chapter Mix

More Differential Equations Questions — jee_main_2025_28_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

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Let for some function $y = f(x)$ $y = f(x)$, $\int₀^x t f(t) dt = x² f(x), x > 0 and f(2) = 3.$ Then $f(6)$ $f(6)$ is equal to: (1) 1 (2) 2 (3) 6 (4) 3