If \(x = \int_{0}^{y}\displaystyle\frac{dt}{\sqrt{1+9t^2}}\) and \(\displaystyle\frac{d^2y}{dx^2} = \lambda y\), then, \(\lambda\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9
By the fundamental theorem of calculus, differentiating the integral with respect to its upper limit y:
\( \displaystyle \frac{dx}{dy} = \frac{1}{\sqrt{1+9y^2}} \)
\( \displaystyle \frac{dy}{dx} = \sqrt{1+9y^2} \)
Differentiating again with respect to x, using the chain rule:
\( \displaystyle \frac{d^2y}{dx^2} = \frac{18y}{2\sqrt{1+9y^2}} \cdot \frac{dy}{dx} \)
\( \displaystyle = \frac{9y}{\sqrt{1+9y^2}} \cdot \sqrt{1+9y^2} = 9y \)
Comparing with \(\frac{d^2y}{dx^2} = \lambda y\) gives \(\lambda = 9\).
NCERT Exemplar:
Integrals
