Let f: R R be a thrice differentiable odd function satisfying f'(x) geq 0, f'(x) = f(x), f(0) = 0, f'(0) = 3. Then 9f(…

Q62 jee_main_2025_03_april_morning Linear Differential Equations

Let

g

be a differentiable function such that

int_0^xg(t)dt=x-int_0^xtg(t)dt

[cite: 568],

xge0

[cite: 569] and let

y=y(x)

satisfy the differential equation

fracdydx - ytan x = 2(x+1)sec x \, g(x)

[cite: 571, 575, 578, 581],

xin[0,fracpi2)

[cite: 581]. If

y(0)=0

[cite: 579] then

yleft(fracpi3right)

is equal to[cite: 582]:

A. $frac2pi3sqrt3$ $\frac{2\pi}{3\sqrt{3}}$
B. $frac4pi3$ $\frac{4\pi}{3}$
C. $frac2pi3$ $\frac{2\pi}{3}$
D. $frac4pi3sqrt3$ $\frac{4\pi}{3\sqrt{3}}$

Solution

### Related Formula Leibniz Integral Rule for differentiation:

fracmathrmdmathrmdxleft(int_0^x f(t)mathrmdtright) = f(x)

### Core Logic Differentiate the given integral relation using Leibniz rule [cite: 1344]:

fracmathrmdmathrmdxleft[int_0^xg(t)dtright] = fracmathrmdmathrmdxleft[x-int_0^xtg(t)dtright]

[cite: 1344]

g(x) = 1 - xg(x) implies g(x)(1+x) = 1 implies g(x) = frac11+x

[cite: 1345] Substitute

g(x)

into target differential equation configuration [cite: 1346]:

fracmathrmdymathrmdx - ytan x = 2(x+1)sec x cdot left(frac11+xright) = 2sec x

[cite: 1346] ### Step 1: Finding the Integrating Factor This matches a linear form

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

where

P(x) = -tan x

.

textI.F. = e^int -tan x \, mathrmdx = e^ln|cos x| = cos x

[cite: 1346] Write general functional solution template [cite: 1348]:

y cdot cos x = int (2sec x cdot cos x) \, mathrmdx = int 2 \, mathrmdx = 2x + C

[cite: 1348] Given boundary condition

y(0) = 0 implies 0 = 0 + C implies C = 0

[cite: 1349].

y(x) = frac2xcos x = 2xsec x

[cite: 1350] ### Step 2: Numeric substitution Substitute variable parameter values

x = fracpi3

[cite: 1352]:

yleft(fracpi3right) = 2left(fracpi3right)secleft(fracpi3right) = frac2pi3 cdot 2 = frac4pi3

[cite: 1351] ### Pattern Recognition Integral functional definitions are codes for simpler underlying derivatives. Applying Leibniz rule immediately extracts the true variable functions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

Q66 jee_main_2025_04_april_evening Linear Differential Equations

If a curve

y = y(x)

passes through the point

left(1, fracpi2right)

and satisfies the differential equation

(7x^4 cot y - e^x mathrmcosec\,y) fracdxdy = x^5

,

x geq 1

, then at

x = 2

, the value of cosine is:

A. $frac2mathrme^2 - mathrme64$ $\frac{2\mathrm{e}^2 - \mathrm{e}}{64}$
B. $frac2mathrme^2 + mathrme64$ $\frac{2\mathrm{e}^2 + \mathrm{e}}{64}$
C. $frac2mathrme^2 - mathrme128$
D. $frac2mathrme^2 + mathrme128$

Solution

### Core Logic Let's rearrange the given differential equation by expressing

fracdydx

:

x^5 fracdydx = 7x^4 cot y - e^x csc y

Dividing both sides by

x^5

:

fracdydx = frac7x cot y - frace^xx^5 csc y

Multiply the entire equation by

sin y

to clear denominators:

sin y fracdydx - frac7x cos y = -frace^xx^5

This can be transformed into a linear form by substituting

t = -cos y

. Then

fracdtdx = sin y fracdydx

. ### Step 1: Solving the Linear ODE Substituting

t

leads to:

fracdtdx + frac7x t = -frace^xx^5

This is a standard linear first-order ODE with

P(x) = frac7x

. The Integrating Factor (I.F.) is:

textI.F. = e^int frac7x dx = e^7 ln x = x^7

The general solution is:

t cdot x^7 = int left(-frace^xx^5right) cdot x^7 dx = -int x^2 e^x dx

### Step 2: Evaluating the Integration and Constant Using integration by parts for

int x^2 e^x dx

:

int x^2 e^x dx = x^2 e^x - 2xe^x + 2e^x

Substituting this back:

-cos y cdot x^7 = -e^x(x^2 - 2x + 2) + C

cos y cdot x^7 = e^x(x^2 - 2x + 2) - C

Since the curve passes through

left(1, fracpi2right)

:

cosleft(fracpi2right) cdot (1)^7 = e^1(1^2 - 2(1) + 2) - C implies 0 = e(1) - C implies C = e

### Step 3: Calculating cos y at x = 2 Now substitute

x = 2

and

C = e

into our equation block:

cos y cdot (2^7) = e^2(2^2 - 2(2) + 2) - e

cos y cdot 128 = e^2(4 - 4 + 2) - e = 2e^2 - e

cos y = frac2e^2 - e128

### Pattern Recognition When trigonometric terms are mixed inside an ODE containing derivative blocks like

fracdxdy

or

fracdydx

, check if clearing denominators using

sin y

or

cos y

reveals a standard substitution path for a Linear ODE. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

Q58 jee_main_2025_04_april_morning Area Bounded by Curves

Let

f colon [0, infty) to mathbbR

be a differentiable function such that

f(x) = 1 - 2x + int_0^x e^x - t f(t) mathrmdt

for all

x in [0, infty)

. Then the area of the region bounded by

y = f(x)

and the coordinate axes is

A. $sqrt5$ $\sqrt{5}$
B. $frac12$ $\frac{1}{2}$
C. $sqrt2$ $\sqrt{2}$
D. 2

Solution

### Related Formula Leibniz Integral Rule for differentiation under integral sign:

fracmathrmdmathrmdxleft(int_phi(x)^psi(x) f(t)mathrmdtright) = f(psi(x))psi'(x) - f(phi(x))phi'(x)

### Core Logic Rewrite equation to isolate the integral kernel:

y = 1 - 2x + e^x int_0^x e^-t f(t)mathrmdt

Differentiating with respect to

x

using product rule and Leibniz rule:

fracmathrmdymathrmdx = -2 + e^x int_0^x e^-t f(t)mathrmdt + e^x cdot left(e^-x f(x)right)

Notice that

e^x int_0^x e^-t f(t)mathrmdt = y - (1 - 2x)

. Substitute this back:

fracmathrmdymathrmdx = -2 + [y - 1 + 2x] + y implies fracmathrmdymathrmdx - 2y = 2x - 3

### Step 1: Solve Differential Equation Integrating factor

textI.F. = e^int -2 mathrmdx = e^-2x

.

y e^-2x = int (2x - 3)e^-2xmathrmdx = frac-(2x - 3)2e^-2x - frac12e^-2x + c

From original equation, at

x=0

,

f(0) = 1

. Evaluating

c

:

1 = frac32 - frac12 + c implies c = 0

Thus,

y = -x + 1 implies x + y = 1

. ### Step 2: Calculate Area The boundary line is

x + y = 1

. The area bounded by this line and coordinate axes is a right triangle with intercepts

(1,0)

and

(0,1)

:

textArea = frac12 times 1 times 1 = frac12

### Pattern Recognition Integral equations of convolution type (

e^x-t

) always simplify directly into standard linear ordinary differential equations of first or second order when tracking Leibniz rules properly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 12 Mathematics: Area Under Curves

Q63 jee_main_2025_07_april_evening Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

(x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)cos x

,

y(0) = 1

. Then

int_-3^3 y(x) dx

is:

A. $24$ $24$
B. $36$ $36$
C. $30$ $30$
D. $18$ $18$

Solution

### Related Formula For a linear differential equation

fracdydx + Py = Q

, the Integrating Factor (IF) is defined as:

textIF = e^int P \, dx

### Core Logic Divide the full differential equation by

(x^2+1)

:

fracdydx - left(frac2xx^2+1right)y = frac(x^2+1)^2 cos xx^2+1 = (x^2+1)cos x

This is a standard Linear Differential Equation with:

P = -frac2xx^2+1, quad Q = (x^2+1)cos x

textIF = e^int -frac2xx^2+1\,dx = e^-ln(x^2+1) = frac1x^2+1

### Step 1: Solve for General Solution The solution format is

y cdot textIF = int Q cdot textIF \, dx

:

y cdot frac1x^2+1 = int (x^2+1)cos x cdot frac1x^2+1 \, dx

fracyx^2+1 = sin x + c

Using the boundary condition

y(0) = 1

:

frac10+1 = sin(0) + c implies c = 1

y = (x^2+1)(sin x + 1)

### Step 2: Definite Integration Evaluation We need to evaluate

int_-3^3 y \, dx

:

int_-3^3 (x^2+1)(sin x + 1) \, dx = int_-3^3 (x^2sin x + x^2 + sin x + 1) \, dx

By symmetry of odd/even functions over symmetric intervals

[-a, a]

:

int_-3^3 x^2sin x \, dx = 0

(since it is an odd function)

int_-3^3 sin x \, dx = 0

(since it is an odd function) Thus, we are left with the even components:

int_-3^3 (x^2 + 1) \, dx = 2 int_0^3 (x^2 + 1) \, dx = 2 left[ fracx^33 + x right]_0^3 = 2(9 + 3) = 24

### Pattern Recognition Splitting a symmetric interval integral into odd and even parts immediately simplifies calculations by dropping all odd functions down to zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 12 Mathematics: Integral Calculus

Q54 jee_main_2025_24_jan_evening Linear Differential Equations

Let

f:(0,infty)rightarrow R

be a function which is differentiable at all points of its domain and satisfies the condition

x^2f^prime(x)=2xf(x)+3,

with

f(1)=4

Then

2f(2)

is equal to: [cite: 3263, 3264]

A. $29$ $29$
B. $19$ $19$
C. $39$ $39$
D. $23$ $23$

Solution

### Related Formula The quotient rule derivative identity is given by:

fracddxleft(fracf(x)x^2right) = fracx^2 f'(x) - 2x f(x)x^4

### Core Logic Rearrange the given differential condition:

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

### Step 1: Divide by

x^4

To convert the left-hand side into an exact derivative form, divide the full relation by

x^4

:

fracx^2 f'(x) - 2x f(x)x^4 = frac3x^4

fracddxleft(fracf(x)x^2right) = 3x^-4

### Step 2: Integration and Evaluating Constant Integrating both sides with respect to

x

[cite: 3893, 3894]:

fracf(x)x^2 = int 3x^-4 dx = -x^-3 + C = -frac1x^3 + C

f(x) = -frac1x + Cx^2

Using the given value

f(1) = 4

[cite: 3264, 3896]:

4 = -frac11 + C(1)^2 Rightarrow 4 = -1 + C Rightarrow C = 5

Thus, the function is

f(x) = -frac1x + 5x^2

. ### Step 3: Calculating

2f(2) Substitute

x = 2

to compute

2f(2)

:

2 times f(2) = 2 times left[ -frac12 + 5(2)^2 right]

2 times f(2) = 2 times left[ -frac12 + 20 right] = -1 + 40 = 39

[cite: 3900, 3901] ### Pattern Recognition Recognizing the structure

x^2 f'(x) - 2x f(x)

as a partial quotient rule is faster than formatting it into standard linear order

\frac{dy}{dx} + P(x)y = Q(x)$ format, though both methods lead to the identical integration parameters safely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

Let $f: mathbbR to mathbbR$ $f: \mathbb{R} \to \mathbb{R}$ be a thrice differentiable odd function satisfying $f'(x) geq 0$ $f'(x) \geq 0$, $f'(x) = f(x)$ $f'(x) = f(x)$, $f(0) = 0$ $f(0) = 0$, $f'(0) = 3$ $f'(0) = 3$. Then $9f(log_e 3)$ $9f(\log_e 3)$ is equal to ________.

Solution & Explanation

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 3

Solution

Solution

Solution

Solution

Solution

More Differential Equations Questions — jee_main_2025_02_april_morning

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: mathbbR to mathbbR$f: \mathbb{R} \to \mathbb{R}$ be a thrice differentiable odd function satisfying f'(x) geq 0$f'(x) \geq 0$, f'(x) = f(x)$f'(x) = f(x)$, f(0) = 0$f(0) = 0$, f'(0) = 3$f'(0) = 3$. Then 9f(log_e 3)$9f(\log_e 3)$ is equal to ________.

Solution & Explanation

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 3

Solution

Solution

Solution

Solution

Solution

More Differential Equations Questions — jee_main_2025_02_april_morning

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 $f: mathbbR to mathbbR$ $f: \mathbb{R} \to \mathbb{R}$ be a thrice differentiable odd function satisfying $f'(x) geq 0$ $f'(x) \geq 0$, $f'(x) = f(x)$ $f'(x) = f(x)$, $f(0) = 0$ $f(0) = 0$, $f'(0) = 3$ $f'(0) = 3$. Then $9f(log_e 3)$ $9f(\log_e 3)$ is equal to ________.