JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(y=\log _8\left(\frac{1-x^2}{1+x^2}\right),-1\)\(-1 < x< 1 \) Then at \( x=\frac{1}{2},\) the value of \(225\left(y^{\prime}-y^{\prime \prime}\right)\) is equal to
- A \(732\)
- B \(746\)
- C \(742\)
- D \(736\)
Answer & Solution
Correct Answer
(D) \(736\)
Step-by-step Solution
Detailed explanation
\( y=\log _0\left(\frac{1-x^2}{1+x^2}\right) \) \( \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4}\) Again, \(\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}\) Again…
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