Question

If \[ \frac{d}{dx} \left[ A\log\left(\frac{\sqrt{1-x^3}+B}{\sqrt{1-x^3}+1}\right) \right] = \frac{1}{x\sqrt{1-x^3}}, \] then \(AB=\)

• A. \(\dfrac{1}{3}\)
• B. \(-\dfrac{1}{3}\)
• C. \(-\dfrac{2}{3}\)
• D. \(\dfrac{2}{3}\)

Hint:

For logarithmic differentiation involving square roots, substitute the square-root expression by a single variable to simplify the derivative.

Answer 1

The Correct Option is B

Step 1: Put \(u=\sqrt{1-x^3}\).
Let \[ u=\sqrt{1-x^3} \] Then, \[ u^2=1-x^3 \] Differentiating, \[ 2u\frac{du}{dx}=-3x^2 \] So, \[ \frac{du}{dx}=-\frac{3x^2}{2u} \]

Step 2: Differentiate the logarithmic expression.
Given expression is \[ A\log\left(\frac{u+B}{u+1}\right) \] Using logarithm property: \[ A\left[\log(u+B)-\log(u+1)\right] \] Differentiating, \[ A\frac{du}{dx} \left[ \frac{1}{u+B}-\frac{1}{u+1} \right] \] \[ = A\frac{du}{dx} \left[ \frac{u+1-u-B}{(u+B)(u+1)} \right] \] \[ = A\frac{du}{dx} \left[ \frac{1-B}{(u+B)(u+1)} \right] \] Substituting \[ \frac{du}{dx}=-\frac{3x^2}{2u}, \] we get \[ -\frac{3Ax^2(1-B)}{2u(u+B)(u+1)} \]

Step 3: Compare with the given derivative.
Given derivative is \[ \frac{1}{xu} \] Therefore, \[ -\frac{3Ax^2(1-B)}{2u(u+B)(u+1)} = \frac{1}{xu} \] Cancelling \(u\), \[ -\frac{3Ax^2(1-B)}{2(u+B)(u+1)} = \frac{1}{x} \] Cross multiplying, \[ -\frac{3A(1-B)x^3}{2} = (u+B)(u+1) \] Since \[ u^2=1-x^3, \] we have \[ x^3=1-u^2 \] Thus, \[ -\frac{3A(1-B)}{2}(1-u^2) = (u+B)(u+1) \]

Step 4: Compare coefficients.
Expanding right side: \[ (u+B)(u+1)=u^2+(B+1)u+B \] For the expression to contain no linear term in \(u\), \[ B+1=0 \] So, \[ B=-1 \] Then, \[ (u+B)(u+1)=(u-1)(u+1)=u^2-1 \] \[ u^2-1=-(1-u^2) \] Now left side becomes \[ -\frac{3A(1-(-1))}{2}(1-u^2) \] \[ =-\frac{3A(2)}{2}(1-u^2) \] \[ =-3A(1-u^2) \] Comparing with \[ -(1-u^2), \] we get \[ -3A=-1 \] \[ A=\frac{1}{3} \]

Step 5: Find \(AB\).
\[ AB=\frac{1}{3}\times (-1) \] \[ AB=-\frac{1}{3} \]

Step 6: Final conclusion.
Therefore, \[ \boxed{-\frac{1}{3}} \]