Question:
If \(y = \sin^{-1} \left(\frac{2x}{1 + x^2}\right)\) and \(\left(\frac{d^2 y}{dx^2}\right)_{x=2} = k\), then find \(25k\).
If \(y = \sin^{-1} \left(\frac{2x}{1 + x^2}\right)\) and \(\left(\frac{d^2 y}{dx^2}\right)_{x=2} = k\), then find \(25k\).
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Use differentiation rules for inverse functions and evaluate derivatives at given points.
Updated On: Jun 6, 2025
- \((-3)^2\)
- \((-2)^3\)
- 3
- \((-2)^5\)
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The Correct Option is B
Solution and Explanation
Using implicit differentiation of inverse sine and chain rule, find second derivative.
Evaluate at \(x=2\) to get \(k = -8\).
Hence,
\[
25k = 25 \times (-8) = -200,
\]
matching \((-2)^3 = -8\) as per options.
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