Question

For \(n\in\mathbb{N}\), if \[ y=ax^{n+1}+bx^{-n}, \] then \[ x^2\frac{d^2y}{dx^2} \] is equal to:

• A. \((n-1)y\)
• B. \(n(n-1)y\)
• C. \(n(n+1)y\)
• D. \((n+1)y\)

Hint:

While differentiating power functions, carefully apply the power rule to both positive and negative exponents. Negative exponents are very common in CUET and JEE level differentiation problems.

Answer 1

The Correct Option is C

Concept:
This problem involves second-order differentiation of algebraic power functions. The standard differentiation rule is: \[ \frac{d}{dx}(x^m)=mx^{m-1} \] We differentiate the given function twice and then simplify the obtained expression.

Step 1: Write the given function clearly.
Given: \[ y=ax^{n+1}+bx^{-n} \] where \(a\) and \(b\) are constants.

Step 2: Find the first derivative.
Differentiating term-by-term: \[ \frac{dy}{dx} = a(n+1)x^n+b(-n)x^{-n-1} \] Therefore: \[ \frac{dy}{dx} = a(n+1)x^n-bnx^{-n-1} \]

Step 3: Find the second derivative.
Differentiating again: \[ \frac{d^2y}{dx^2} = a(n+1)n x^{n-1} - bn(-n-1)x^{-n-2} \] Since: \[ -bn(-n-1)=bn(n+1) \] we get: \[ \frac{d^2y}{dx^2} = n(n+1)ax^{n-1} + n(n+1)bx^{-n-2} \] Taking the common factor \(n(n+1)\): \[ \frac{d^2y}{dx^2} = n(n+1)\left(ax^{n-1}+bx^{-n-2}\right) \]

Step 4: Multiply the expression by \(x^2\).
Multiplying both sides by \(x^2\): \[ x^2\frac{d^2y}{dx^2} = n(n+1)\left(ax^{n+1}+bx^{-n}\right) \] But: \[ ax^{n+1}+bx^{-n}=y \] Therefore: \[ x^2\frac{d^2y}{dx^2}=n(n+1)y \]

Step 5: Match the result with the given options.
The obtained expression is: \[ n(n+1)y \] which corresponds to: \[ \boxed{(3)\ n(n+1)y} \]