JEE Mathematics Differential Equations MCQs Set 18

Practice MCQs for JEE Mathematics Differential Equations

Review structured MCQ sets for JEE Mathematics Differential Equations. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Access Differential Equations Questions and Solutions

Access the complete set of multiple-choice questions for Differential Equations below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: If \( f \) is twice differentiable and \( \lambda^2f(x)-2\lambda f'(x)+f''(x)=0 \) has two equal values of \( \lambda \) for all \( x \), with \( f(0)=1,\ f'(0)=2 \), then \( f(\log_e2) \) is equal to
(a) 3
(b) 1
(c) 4
(d) 2
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. Equal roots need \( f'^2=ff'' \), so \( \frac{f'}{f} \) is constant: \( f=e^{kx} \).
2. With \( f(0)=1,\ f'(0)=2 \): \( f=e^{2x} \).
3. So \( f(\ln2)=e^{2\ln2}=4 \).

Teacher's Note:
1. Work from the given condition to find the function or quantity, then substitute.
2. Verify the answer on a simple case.

Question 2: A tangent to the curve \( y=f(x) \) at \( P(x,y) \) cuts the x-axis and y-axis at \( A \) and \( B \) respectively with \( BP:AP=3:1 \). If \( f(1)=1 \), which of the following is/are correct? (i) the equation of the curve is \( x\frac{dy}{dx}+3y=0 \) (ii) the curve passes through \( \left(2,\frac18\right) \) (iii) the equation of the curve is \( x\frac{dy}{dx}-3y=0 \) (iv) the normal at \( (1,1) \) is \( x+3y=4 \)
(a) (i) and (ii) only
(b) (ii) and (iii) only
(c) (iii) and (iv) only
(d) (i) and (iv) only
Show Answer & Explanation

Answer: (a) (i) and (ii) only

Explanation:
1. The tangent has intercepts \( A\left(x-\frac{y}{y'},0\right) \) and \( B(0,\,y-xy') \). Since \( P \) divides \( BA \) with \( BP:PA=3:1 \), \( x=\frac34\left(x-\frac y{y'}\right) \).
2. This gives \( xy'+3y=0 \), so \( y=\frac{C}{x^3} \); \( f(1)=1 \) gives \( y=\frac1{x^3} \), which passes through \( \left(2,\frac18\right) \).
3. The normal at \( (1,1) \) (tangent slope \( -3 \)) is \( x-3y+2=0 \), not \( x+3y=4 \). So (i) and (ii).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Question 3: The order and degree of the differential equation of the family of circles with centres on the y-axis are respectively
(a) 2, 2
(b) 1, 1
(c) 2, 1
(d) 1, 2
Show Answer & Explanation

Answer: (c) 2, 1

Explanation:
1. Such circles: \( x^2+y^2+2gy+c=0 \) with two arbitrary constants, so the order is 2.
2. Eliminating both constants gives a relation linear in the highest derivative \( y'' \), so the degree is 1.

Teacher's Note:
1. Order equals the number of independent arbitrary constants.
2. Degree is the power of the highest-order derivative after clearing radicals.

Question 4: Let \( f \) be differentiable on \( \mathbb R \) with \( f(1)=1 \). If the y-intercept of the tangent at any point \( P(x,y) \) of \( y=f(x) \) equals the cube of the abscissa of \( P \), the value of \( f(-3) \) is
(a) 9
(b) 27
(c) 3
(d) 1
Show Answer & Explanation

Answer: (a) 9

Explanation:
1. The tangent's intercept is \( y-xy'=x^3 \), i.e. \( \frac d{dx}\left(\frac yx\right)=-x \).
2. So \( y=-\frac{x^3}2+Cx \); \( f(1)=1 \) gives \( C=\frac32 \).
3. Then \( f(-3)=\frac{27}2-\frac92=9 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Question 5: The general solution of \( \dfrac{d^3y}{dx^3}-7\dfrac{d^2y}{dx^2}+16\dfrac{dy}{dx}-12y=0 \) is
(a) \( (A\cos x+B\sin x)e^{2x}+c_3e^{3x} \)
(b) \( (c_1x+c_2)e^{-3x}+c_3e^{2x} \)
(c) \( c_1e^{2x}+c_2e^{-2x}+c_3e^{-3x} \)
(d) \( (c_1x+c_2)e^{2x}+c_3e^{3x} \)
Show Answer & Explanation

Answer: (d) \( (c_1x+c_2)e^{2x}+c_3e^{3x} \)

Explanation:
1. Auxiliary equation: \( D^3-7D^2+16D-12=0 \), i.e. \( (D-2)^2(D-3)=0 \).
2. Roots 2, 2, 3 give \( y=(c_1x+c_2)e^{2x}+c_3e^{3x} \).

Teacher's Note:
1. A repeated root contributes a factor (c1 x + c2).
2. Check D = 2: 8 - 28 + 32 - 12 = 0.

Question 6: Let \( g(x)=\log f(x) \), where \( f \) is a twice differentiable positive function on \( (0,\infty) \) with \( f(x+1)=xf(x) \). Then for \( N=1,2,3,\dots \), \( g''\left(N+\frac12\right)-g''\left(\frac12\right) \) equals
(a) \( 4\left\{1+\frac19+\frac1{25}+\dots+\frac1{(2N+1)^2}\right\} \)
(b) \( -4\left\{1+\frac19+\frac1{25}+\dots+\frac1{(2N-1)^2}\right\} \)
(c) \( 1+\frac19+\frac1{25}+\dots+\frac1{(2N-1)^2} \)
(d) \( -4\left\{1+\frac19+\frac1{25}+\dots+\frac1{(2N+1)^2}\right\} \)
Show Answer & Explanation

Answer: (b) \( -4\left\{1+\frac19+\frac1{25}+\dots+\frac1{(2N-1)^2}\right\} \)

Explanation:
1. \( g(x+1)=\ln x+g(x) \); differentiating twice: \( g''(x+1)-g''(x)=-\frac1{x^2} \).
2. Put \( x=k-\frac12 \) for \( k=1,\dots,N \) and add: \( g''(N+\tfrac12)-g''(\tfrac12)=-\sum_{k=1}^N\frac{1}{(k-\frac12)^2}=-4\sum_{k=1}^N\frac1{(2k-1)^2} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Question 7: If the family of curves is \( y=a\cos(x+b) \), where \( a,b \) are arbitrary constants, and \( y'=\frac{dy}{dx},\ y''=\frac{d^2y}{dx^2} \), then the differential equation is
(a) \( y''=yy' \)
(b) \( y''=y+y' \)
(c) \( y''=y \)
(d) \( y''+y=0 \)
Show Answer & Explanation

Answer: (d) \( y''+y=0 \)

Explanation:
1. Differentiating twice: \( y'=-a\sin(x+b) \) and \( y''=-a\cos(x+b)=-y \).
2. Hence \( y''+y=0 \) (the two constants a and b are eliminated by one second-order equation).

Teacher's Note:
1. Two arbitrary constants give a second-order equation.
2. The second derivative of a cosine is minus the function.

Question 8: \( y=(c_1\cos x+c_2\sin x)e^{-x}+c_3e^{x} \) is the general solution of the differential equation
(a) \( \frac{d^3y}{dx^3}-y=0 \)
(b) \( \frac{d^2y}{dx^2}-y=0 \)
(c) \( \frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}-2y=0 \)
(d) \( \frac{d^3y}{dx^3}-\frac{d^2y}{dx^2}+2y=0 \)
Show Answer & Explanation

Answer: (c) \( \frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}-2y=0 \)

Explanation:
1. The roots are \( -1\pm i \) and \( 1 \).
2. The auxiliary equation is \( (D-1)(D^2+2D+2)=D^3+D^2-2=0 \).
3. So the equation is \( y'''+y''-2y=0 \).

Teacher's Note:
1. Complex roots a +- ib appear as e^(ax)(cos bx, sin bx).
2. Multiply the factors to rebuild the auxiliary equation.

Question 9: Consider the equation \( y^3-3y+x=0 \) which defines functions implicitly. For \( x\in(-\infty,-2)\cup(2,\infty) \) it defines a unique real-valued differentiable function \( y=f(x) \); for \( x\in(-2,2) \) it defines a unique real-valued differentiable function \( y=g(x) \) with \( g(0)=0 \). If \( f(-10\sqrt2)=2\sqrt2 \), then \( f''(-10\sqrt2) \) equals
(a) \( \frac{4\sqrt2}{7^3\cdot3^2} \)
(b) \( \frac{4\sqrt2}{7^3\cdot3} \)
(c) \( -\frac{4\sqrt2}{7^3\cdot3} \)
(d) \( -\frac{4\sqrt2}{7^3\cdot3^2} \)
Show Answer & Explanation

Answer: (d) \( -\frac{4\sqrt2}{7^3\cdot3^2} \)

Explanation:
1. \( x=3y-y^3 \) gives \( y'=\frac1{3(1-y^2)} \) and \( y''=\frac{2y}{9(1-y^2)^3} \).
2. At \( y=2\sqrt2 \): \( 1-y^2=-7 \), so \( y''=\frac{4\sqrt2}{9\cdot(-343)}=-\frac{4\sqrt2}{7^3\cdot3^2} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Question 10: The general solution of \( \dfrac{d^3y}{dx^3}+y=0 \) is
(a) \( c_1e^{-x}+e^{x/2}\left(a\cos\frac{\sqrt3x}{2}+b\sin\frac{\sqrt3x}{2}\right) \)
(b) \( c_1e^{\sqrt3x/2}+c_2e^{x/2}+c_3e^{-x} \)
(c) \( c_1e^{-x}+e^{-x/2}\left(a\cos\frac{\sqrt3x}{2}+b\sin\frac{\sqrt3x}{2}\right) \)
(d) \( c_1e^{-x}+c_2e^{-\omega x}+c_3e^{-\omega^2x} \)
Show Answer & Explanation

Answer: (a) \( c_1e^{-x}+e^{x/2}\left(a\cos\frac{\sqrt3x}{2}+b\sin\frac{\sqrt3x}{2}\right) \)

Explanation:
1. Auxiliary equation \( D^3+1=0 \) factorises as \( (D+1)(D^2-D+1)=0 \).
2. Roots: \( -1 \) and \( \frac12\pm\frac{\sqrt3}{2}i \).
3. Hence \( y=c_1e^{-x}+e^{x/2}\left(a\cos\frac{\sqrt3x}2+b\sin\frac{\sqrt3x}2\right) \).

Teacher's Note:
1. Solve the cubic auxiliary equation using the sum of cubes identity.
2. The real part of the complex roots (1/2) goes into the exponential.

Question 11: The differential equation \( \dfrac{dy}{dx}=\dfrac{\sqrt{1-y^2}}{y} \) determines a family of circles with
(a) fixed radius 1 and variable centres along the x-axis
(b) variable radii and a fixed centre at (0, -1)
(c) variable radii and a fixed centre at (0, 1)
(d) fixed radius 1 and variable centres along the y-axis
Show Answer & Explanation

Answer: (a) fixed radius 1 and variable centres along the x-axis

Explanation:
1. Separate: \( \frac{y\,dy}{\sqrt{1-y^2}}=dx \), so \( -\sqrt{1-y^2}=x+c \).
2. Squaring: \( (x+c)^2+y^2=1 \): radius 1, centre \( (-c,0) \) on the x-axis.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Question 12: If \( y'=y+1 \) and \( y(0)=1 \), the value of \( y(\log_e2) \) is
(a) 5
(b) 3
(c) 2
(d) 4
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. \( \frac{dy}{y+1}=dx \) gives \( y+1=Ce^x \); \( y(0)=1 \) gives \( C=2 \).
2. At \( x=\ln2 \): \( y+1=4 \), so \( y=3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Question 13: A solution \( y=y(x) \) of \( x\sqrt{x^2-1}\,dy-y\sqrt{y^2-1}\,dx=0 \) satisfies \( y(2)=\frac2{\sqrt3} \). Statement 1: \( y(x)=\sec\left(\sec^{-1}x-\frac\pi6\right) \). Statement 2: \( \dfrac1y=\dfrac{2\sqrt3}{x}-\sqrt{1-\dfrac1{x^2}} \). Which is correct?
(a) Statement 1 is false and Statement 2 is true
(b) Both are true and Statement 2 explains Statement 1
(c) Both are true but Statement 2 does not explain Statement 1
(d) Statement 1 is true and Statement 2 is false
Show Answer & Explanation

Answer: (d) Statement 1 is true and Statement 2 is false

Explanation:
1. Separating: \( \frac{dy}{y\sqrt{y^2-1}}=\frac{dx}{x\sqrt{x^2-1}} \) gives \( \sec^{-1}y=\sec^{-1}x+C \). With \( y(2)=\frac2{\sqrt3} \): \( \frac\pi6=\frac\pi3+C \), so \( C=-\frac\pi6 \): Statement 1 true.
2. Then \( \frac1y=\cos\left(\sec^{-1}x-\frac\pi6\right)=\frac{\sqrt3}{2x}+\frac12\sqrt{1-\frac1{x^2}} \), which differs from Statement 2: false.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Question 14: Let \( f''(x)=-f(x) \) and \( g(x)=f'(x) \). If \( F(x)=\left\{f\left(\frac x2\right)\right\}^2+\left\{g\left(\frac x2\right)\right\}^2 \) and \( F(5)=5 \), then \( F(10) \) is equal to
(a) 0
(b) 5
(c) 15
(d) 10
Show Answer & Explanation

Answer: (b) 5

Explanation:
1. Let \( h(u)=f(u)^2+g(u)^2 \). Then \( h'(u)=2ff'+2gg'=2fg+2g(-f)=0 \), so \( h \) is constant.
2. Hence \( F(10)=h(5)=F(5)=5 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Question 15: The differential equation for the family of curves \( x^2+y^2-2ay=0 \), where \( a \) is an arbitrary constant, is
(a) \( (x^2+y^2)y'=2xy \)
(b) \( 2(x^2-y^2)y'=xy \)
(c) \( 2(x^2+y^2)y'=xy \)
(d) \( (x^2-y^2)y'=2xy \)
Show Answer & Explanation

Answer: (d) \( (x^2-y^2)y'=2xy \)

Explanation:
1. Differentiate: \( 2x+2yy'-2ay'=0 \), so \( a=\frac{x+yy'}{y'} \).
2. Substitute in the curve: \( x^2+y^2-\frac{2y(x+yy')}{y'}=0 \).
3. This gives \( (x^2-y^2)y'=2xy \).

Teacher's Note:
1. Eliminate the constant a using the derivative.
2. Multiply through by y'.

Question 16: Let \( f \) be differentiable on \( (0,\infty) \) with \( f(1)=1 \) and \( \displaystyle\lim_{t\to x}\frac{t^2f(x)-x^2f(t)}{t-x}=1 \) for each \( x>0 \). Then \( f(x) \) is
(a) \( \frac1x \)
(b) \( -\frac1{3x}+\frac{4x^2}{3} \)
(c) \( \frac1{3x}+\frac{2x^2}{3} \)
(d) \( -\frac1x+\frac2{x^2} \)
Show Answer & Explanation

Answer: (c) \( \frac1{3x}+\frac{2x^2}{3} \)

Explanation:
1. The limit is the derivative in \( t \) of \( t^2f(x)-x^2f(t) \) at \( t=x \): \( 2xf(x)-x^2f'(x)=1 \).
2. So \( \left(\frac{f}{x^2}\right)'=-\frac1{x^4} \), \( f=\frac1{3x}+Cx^2 \); \( f(1)=1 \) gives \( C=\frac23 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Question 17: Let \( \displaystyle\int_0^{f(x)}t\cos t\,dt=\int_0^{x^2}\sqrt z\,\sin\sqrt z\,dz \) and \( f(0)=0 \). Then \( 4x\sin x+4\cos x-2x^2\cos x-f(x)\sin f(x)-\cos f(x) \) equals
(a) 4
(b) 3
(c) 2
(d) 1
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Left side: \( \int_0^{f}t\cos t\,dt=f\sin f+\cos f-1 \).
2. Right side (put \( z=w^2 \)): \( \int_0^x2w^2\sin w\,dw=-2x^2\cos x+4x\sin x+4\cos x-4 \).
3. Equating: \( f\sin f+\cos f=-2x^2\cos x+4x\sin x+4\cos x-3 \).
4. So the given expression equals 3.

Teacher's Note:
1. Evaluate both integrals by parts and equate.
2. Rearranging the identity gives the constant directly.

Free study material for Differential Equations

Multiple Choice Questions (MCQs) for JEE Mathematics Differential Equations

Chapter MCQs with Answers for JEE Mathematics

Explore reliable practice questions for Differential Equations tailored for JEE Mathematics learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.

Expert Practice Material for JEE Mathematics

Cross-reference your completed choices with comprehensive NCERT solutions for JEE Mathematics to ensure absolute clarity across all sub-topics in this chapter.

Complete Your Chapter Revision

Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in JEE assessments.

FAQs

Where can I access latest JEE Mathematics Differential Equations MCQs Set 18?

You can get most exhaustive JEE Mathematics Differential Equations MCQs Set 18 for free on StudiesToday.com. These MCQs for JEE Mathematics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics JEE material?

Yes, our JEE Mathematics Differential Equations MCQs Set 18 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in JEE exams?

By solving our JEE Mathematics Differential Equations MCQs Set 18, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for JEE Mathematics Differential Equations MCQs Set 18?

Yes, Mathematics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Mathematics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Mathematics Differential Equations MCQs Set 18 on StudiesToday.com as they provide instant answers and score to help you track your progress in Mathematics.