Let f:[1,infty) [2,infty) be a differentiable function. If 101xf(t)dt = 5xf(x) - x5 - 9 for all x geq 1, then the value of f(3) is:

Differentiating both sides with respect to x eliminates the definite integral and leads to a first-order linear differential equation. Apply differentiation with respect to x using the Leibniz Rule on the left side, and product rule on the right side: Letting y = f(x), we rewrite it as: The integrating factor (I.F.) is: Multiply by the I.F. and integrate: Thus, the general solution is: Substitute x = 1 into the original integral equation: Now, substitute x = 1 and f(1) = 2 into our general solution to find C: Therefore, the complete function is: Evaluate the function at x = 3: Whenever a definite integral is defined from a constant to the variable x within an equation, differentiating immediately reduces it to a differential equation. Finding the value of f(a) at the lower bound is a standard method to get the constant of integration. Class 12 Mathematics: Differential Equations

Let f:[1,infty) [2,infty) be a differentiable function. If 101xf(t)dt = 5xf(x) - x5 - 9 for all x geq 1, then the value of f(3) is:

Differentiating both sides with respect to x eliminates the definite integral and leads to a first-order linear differential equation. Apply differentiation with respect to x using the Leibniz Rule on the left side, and product rule on the right side: Letting y = f(x), we rewrite it as: The integrating factor (I.F.) is: Multiply by the I.F. and integrate: Thus, the general solution is: Substitute x = 1 into the original integral equation: Now, substitute x = 1 and f(1) = 2 into our general solut

Let f:[1,infty) [2,infty) be a differentiable function. If 101xf(t)dt = 5xf(x) - x5 - 9 for all x geq 1, then the value of f(3) is:

Differentiating both sides with respect to x eliminates the definite integral and leads to a first-order linear differential equation. Apply differentiation with respect to x using the Leibniz Rule on the left side, and product rule on the right side: Letting y = f(x), we rewrite it as: The integrating factor (I.F.) is: Multiply by the I.F. and integrate: Thus, the general solution is: Substitute x = 1 into the original integral equation: Now, substitute x = 1 and f(1) = 2 into our general solution to find C: Therefore, the complete function is: Evaluate the function at x = 3: Whenever a definite integral is defined from a constant to the variable x within an equation, differentiating immediately reduces it to a differential equation. Finding the value of f(a) at the lower bound is a standard method to get the constant of integration. Class 12 Mathematics: Differential Equations

Let f:[1,infty) [2,infty) be a differentiable function. If 101xf(t)dt = 5xf(x) - x5 - 9 for all x geq 1, then the value of f(3) is:

Differentiating both sides with respect to x eliminates the definite integral and leads to a first-order linear differential equation. Apply differentiation with respect to x using the Leibniz Rule on the left side, and product rule on the right side: Letting y = f(x), we rewrite it as: The integrating factor (I.F.) is: Multiply by the I.F. and integrate: Thus, the general solution is: Substitute x = 1 into the original integral equation: Now, substitute x = 1 and f(1) = 2 into our general solut

Let f:[1,infty) [2,infty) be a differentiable function. If 101xf(t)dt = 5xf(x) - x5

More Differential Equations Previous-Year Questions

Q8 jee_main_2026_21_jan_morning Linear Differential Equations

Let

y = y(x)

be the solution curve of the differential equation

(1 + x^2)dy + (y - tan^-1x)dx = 0

y(0) = 1

. Then the value of

y(1)

is:

A. $frac2e^fracpi4 + fracpi4 - 1$
B. $frac2e^fracpi4 - fracpi4 - 1$
C. $frac4e^fracpi4 + fracpi2 - 1$
D. $frac4e^fracpi4 - fracpi2 - 1$

Solution

### Related Formula For a linear differential equation of the form

fracdydx + P(x)y = Q(x)

textIntegrating Factor (IF) = e^int P(x)dx

textSolution is y cdot textIF = int (Q(x) cdot textIF) dx + C

### Core Logic Rearrange the given differential equation to standard linear form:

(1 + x^2)dy = (tan^-1x - y)dx

fracdydx + fracyx^2 + 1 = fractan^-1xx^2 + 1

### Step 1: Find the Integrating Factor

P(x) = frac1x^2 + 1

textIF = e^int frac1x^2 + 1 dx = e^tan^-1x

### Step 2: Solve the Integral

y cdot e^tan^-1x = int e^tan^-1x cdot fractan^-1x1 + x^2 dx

Let

t = tan^-1x

, then

dt = frac11 + x^2 dx

. The integral becomes

int t e^t dt

. Using integration by parts:

int t e^t dt = t e^t - e^t + C

Substituting back

t = tan^-1x

y cdot e^tan^-1x = tan^-1x(e^tan^-1x) - e^tan^-1x + C

### Step 3: Apply Boundary Condition Given

y(0) = 1

1 cdot e^0 = 0 cdot e^0 - e^0 + C

1 = 0 - 1 + C Rightarrow C = 2

### Step 4: Evaluate at x = 1 Equation of curve:

y = tan^-1x - 1 + 2e^-tan^-1x

Evaluate at

x = 1

y(1) = tan^-1(1) - 1 + 2e^-tan^-1(1)

y(1) = fracpi4 - 1 + 2e^-pi/4

y(1) = frac2e^pi/4 + fracpi4 - 1

### Pattern Recognition A classic LDE integration trap:

int e^f(x) f(x) f'(x) dx

resolves trivially with the substitution

u = f(x)

turning it into

int u e^u du

, which always evaluates to

e^u(u - 1) + C

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Integrals

Q21 jee_main_2026_21_jan_morning Higher Order Differential Equations and AOD

Let

f: R to R

be a twice differentiable function such that the quadratic equation

f(x)m^2 - 2f'(x)m + f''(x) = 0

m

, has two equal roots for every

x in R

. If

f(0) = 1

f'(0) = 2

and

(alpha, beta)

is the largest interval in which the function

f(log_e x - x)

is increasing, then

alpha + beta

is equal to

Numerical Answer. Answer: 1 to 1

Solution

### Related Formula For a quadratic equation

Am^2 + Bm + C = 0

having equal roots, Discriminant

D = 0 Rightarrow B^2 - 4AC = 0

. ### Core Logic Given quadratic equation in

m

f(x)m^2 - 2f'(x)m + f''(x) = 0

has equal roots.

D = 0 Rightarrow (-2f'(x))^2 - 4(f(x))(f''(x)) = 0

4(f'(x))^2 = 4f(x)f''(x) Rightarrow (f'(x))^2 = f(x)f''(x)

### Step 1: Solve the Differential Equation Rewrite the DE:

fracf''(x)f'(x) = fracf'(x)f(x)

Integrate both sides:

int fracf''(x)f'(x) dx = int fracf'(x)f(x) dx

ln|f'(x)| = ln|f(x)| + ln|c| Rightarrow f'(x) = c cdot f(x)

Using given

f(0) = 1

and

f'(0) = 2

f'(0) = c cdot f(0) Rightarrow 2 = c(1) Rightarrow c = 2

Now we have

f'(x) = 2f(x) Rightarrow fracf'(x)f(x) = 2

. Integrate again:

ln|f(x)| = 2x + d

Use

f(0) = 1 Rightarrow ln(1) = 0 + d Rightarrow d = 0

. So,

ln f(x) = 2x Rightarrow f(x) = e^2x

. ### Step 2: Investigate increasing interval Let

g(x) = f(ln x - x) = e^2(ln x - x)

. For

g(x)

to be increasing,

g'(x) geq 0

g'(x) = 2e^2(ln x - x) cdot fracddx(ln x - x)

g'(x) = 2e^2(ln x - x) left(frac1x - 1right)

Since exponential is always positive,

g'(x) geq 0 Rightarrow frac1x - 1 geq 0

frac1 - xx geq 0

The critical points are

x=0, x=1

. Based on domain of

ln x

x > 0

. Sign scheme yields positive derivative in

x in (0, 1]

. Therefore,

(alpha, beta) = (0, 1) Rightarrow alpha = 0, beta = 1

. ### Step 3: Final Output

alpha + beta = 0 + 1 = 1

### Pattern Recognition The relation

(f')^2 = f cdot f''

is a classical indicator of exponential functions (

f = Ce^kx

). Solving via double logarithmic integration collapses the differential equation almost instantaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Applications of Derivatives

Q71 jee_main_2025_02_april_evening First Order Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

fracdydx + 2ysec^2 x = 2sec^2 x + 3tan x cdot sec^2 x

such that

y(0) = frac54

. Then

12left(yleft(fracpi4right) - e^-2right)

is equal to ____________.

Numerical Answer. Answer: 21 to 21

Solution

### Related Formula

textLinear Differential equation form: fracdydx + P(x) y = Q(x)

textIntegrating Factor: I.F. = e^int P(x) dx

textGeneral solution: y cdot I.F. = int Q(x) cdot I.F. \, dx + C

### Core Logic This is a first-order linear differential equation. We calculate the Integrating Factor first to write down the integral solution. ### Step 1: Find the Integrating Factor (I.F.) Here,

P(x) = 2sec^2 x

and

Q(x) = 2sec^2 x + 3tan x cdot sec^2 x

textI.F. = e^int 2sec^2 x dx = e^2tan x

### Step 2: Obtain the General Solution Multiply both sides by the integrating factor:

y cdot e^2tan x = int e^2tan x left( 2sec^2 x + 3tan x cdot sec^2 x right) dx

Let

t = tan x implies dt = sec^2 x dx

. The integral becomes:

int e^2t (2 + 3t) dt = int 2 e^2t dt + 3 int t e^2t dt

Applying integration by parts for the second term:

3 int t e^2t dt = 3 left[ fract e^2t2 - int frace^2t2 dt right] = frac3t e^2t2 - frac3e^2t4

Summing all parts:

y cdot e^2tan x = e^2t + frac3t e^2t2 - frac3e^2t4 + C = e^2tan x left[ 1 + frac3tan x2 - frac34 right] + C

y = frac3tan x2 + frac14 + C e^-2tan x

### Step 3: Apply the boundary conditions Using the initial boundary condition

y(0) = frac54

frac54 = frac3(0)2 + frac14 + C e^0 implies C = 1

Thus, the complete function is:

y(x) = frac3tan x2 + frac14 + e^-2tan x

### Step 4: Compute the final value Evaluate the function at

x = fracpi4

yleft(fracpi4right) = frac3(1)2 + frac14 + e^-2 = frac74 + e^-2

Now, calculate the requested value:

12 left( yleft(fracpi4right) - e^-2 right) = 12 left( frac74 right) = 21

### Pattern Recognition Integration by parts substitution: When integrating terms of the form

int e^at P(t) dt

(where

P(t)

is a polynomial), substituting the polynomial variable directly simplifies the exponential integration factors cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

Q72 jee_main_2025_02_april_morning Solving First Order Differential Equations

Let

f: mathbbR to mathbbR

be a thrice differentiable odd function satisfying

f'(x) geq 0

f'(x) = f(x)

f(0) = 0

f'(0) = 3

. Then

9f(log_e 3)

is equal to ________.

Numerical Answer. Answer: 36 to 36

Solution

### Related Formula Standard variable separable integration form:

int frac1sqrty^2 + a^2 \, mathrmdy = lnleft|y + sqrty^2 + a^2right| + C

### Core Logic The original paper solution states the structure equation setup as

f''(x) = f(x)

. Multiply by

f'(x)

on both sides to transform it into a integrable derivative form. ### Step 1: Integrate the derivative identity

f'(x) cdot f''(x) = f'(x) cdot f(x)

Integrate both sides with respect to

x

frac(f'(x))^22 = frac(f(x))^22 + C implies (f'(x))^2 = (f(x))^2 + C'

### Step 2: Find the constant of integration Use initial conditions

f(0) = 0

and

f'(0) = 3

3^2 = 0^2 + C' implies C' = 9

Thus,

(f'(x))^2 = (f(x))^2 + 9

. Given

f'(x) ge 0

f'(x) = sqrt(f(x))^2 + 9

### Step 3: Variable Separation and Solution Form Let

y = f(x) implies fracmathrmdymathrmdx = sqrty^2 + 9

int fracmathrmdysqrty^2 + 9 = int mathrmdx implies lnleft|y + sqrty^2 + 9right| = x + C_2

Substitute initial condition

x=0, y=0

ln|0 + sqrt9| = 0 + C_2 implies C_2 = ln 3

Therefore,

lnleft|y + sqrty^2 + 9right| = x + ln 3 implies y + sqrty^2 + 9 = 3e^x

. ### Step 4: Compute targeted value We need to evaluate at

x = ln 3

y + sqrty^2 + 9 = 3e^ln 3 = 3(3) = 9

sqrty^2 + 9 = 9 - y

Square both sides:

y^2 + 9 = 81 - 18y + y^2 implies 18y = 72 implies y = 4

Thus,

f(ln 3) = 4

. Multiply by

9

9 f(ln 3) = 9(4) = 36

### Pattern Recognition Multiplying a second derivative by the first derivative (

f'f''

) is a classic trick to convert a second-order linear differential equation into a first-order separable layout, opening a clear path to the solution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 12 Mathematics: Differential Calculus

Let $f:[1,infty) to [2,infty)$ $f:[1,\infty) \to [2,\infty)$ be a differentiable function. If $10int_1^xf(t)mathrmdt = 5xf(x) - x^5 - 9$ for all $x geq 1$ $x \geq 1$, then the value of $f(3)$ $f(3)$ is:

Solution & Explanation

Reference Study Guides

More Differential Equations Previous-Year Questions

Solution

Solution

Solution

Solution

More Differential Equations Questions — jee_main_2025_02_april_evening

We Map Every Repeating Question in Competitive Exams.

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

Let f:[1,infty) to [2,infty)$f:[1,\infty) \to [2,\infty)$ be a differentiable function. If 10int_1^xf(t)mathrmdt = 5xf(x) - x^5 - 9$10\int_{1}^{x}f(t)\mathrm{d}t = 5xf(x) - x^{5} - 9$ for all x geq 1$x \geq 1$, then the value of f(3)$f(3)$ is:

Solution & Explanation

Reference Study Guides

More Differential Equations Previous-Year Questions

Solution

Solution

Solution

Solution

More Differential Equations Questions — jee_main_2025_02_april_evening

We Map Every Repeating Question in Competitive Exams.

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

Welcome back

Let $f:[1,infty) to [2,infty)$ $f:[1,\infty) \to [2,\infty)$ be a differentiable function. If $10int_1^xf(t)mathrmdt = 5xf(x) - x^5 - 9$ for all $x geq 1$ $x \geq 1$, then the value of $f(3)$ $f(3)$ is: