WBJEE PYQs for Second Order Derivative with Solutions: Practice WBJEE Previous Year Questions

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Yashodeep Mahurkar

Updated on - Dec 31, 2025

Second Order Derivative is an important topic in the Mathematics section in WBJEE exam. Practising this topic will increase your score overall and make your conceptual grip on WBJEE exam stronger.

This article gives you a full set of WBJEE PYQs for Second Order Derivative with explanations for effective preparation. Practice of WBJEE Mathematics PYQs including Second Order Derivative questions regularly will improve accuracy, speed, and confidence in the WBJEE 2026 exam.

WBJEE PYQs for Second Order Derivative with Solutions

1.
The function \(y=e^{kr}\) satisfies \((\frac{d^2y}{dx^2}+\frac{dy}{dx})(\frac{dy}{dx}-y)=y\frac{dy}{dx}\). It is valid for
- exactly one value of k
- two distinct values of k
- three distinct values of k
- infinitely many values of k
View Solution
2.
If x=sinθ and y=sin kθ, then (1-x²)y₂-xy₁-αy=0, for α=
- k
- -k
- -k²
- k²
View Solution
3.
Given \(\frac{d^2y}{dx^2}=cot\,x\frac{dy}{dx}+4y\,cosec^2x=0\). Changing the independent variable x to z by the substitution z=log tan \(\frac{x}{2}\), the equation is changed to
- \(\frac{d^2y}{dz^2}+\frac{3}{y}=0\)
- \(2\frac{d^2y}{dz^2}+e^y=0\)
- \(\frac{d^2y}{dz^2}-4y=0\)
- \(\frac{d^2y}{dz^2}+4y=0\)
View Solution

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WBJEE PYQs for Second Order Derivative with Solutions

1.The function \(y=e^{kr}\) satisfies \((\frac{d^2y}{dx^2}+\frac{dy}{dx})(\frac{dy}{dx}-y)=y\frac{dy}{dx}\). It is valid for

2.If x=sinθ and y=sin kθ, then (1-x2)y2-xy1-αy=0, for α=

3.Given \(\frac{d^2y}{dx^2}=cot\,x\frac{dy}{dx}+4y\,cosec^2x=0\). Changing the independent variable x to z by the substitution z=log tan \(\frac{x}{2}\), the equation is changed to

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1.
The function \(y=e^{kr}\) satisfies \((\frac{d^2y}{dx^2}+\frac{dy}{dx})(\frac{dy}{dx}-y)=y\frac{dy}{dx}\). It is valid for

2.
If x=sinθ and y=sin kθ, then (1-x²)y₂-xy₁-αy=0, for α=

3.
Given \(\frac{d^2y}{dx^2}=cot\,x\frac{dy}{dx}+4y\,cosec^2x=0\). Changing the independent variable x to z by the substitution z=log tan \(\frac{x}{2}\), the equation is changed to