Question

For \(a \in \mathbb{R}\), consider the real valued function defined on \((-1,1)\): \[ f(x)= \begin{cases} \dfrac{(1+x)^{1/3}-(1+2x)^{1/4}}{x}, & x\neq0,\\ a,& x=0. \end{cases} \] If \(f\) is differentiable at \(x=0\), then value of \(a+f'(0)\) is:

• A. \(\frac{19}{72}\)
• B. \(\frac{7}{72}\)
• C. \(\frac{31}{72}\)
• D. \(-\frac16\)

Hint:

For piecewise functions involving powers, first enforce continuity to determine the unknown constant. Then use Taylor/Binomial expansion to obtain the derivative quickly.

Answer 1

The Correct Option is B

Concept: For differentiability at a point, continuity is necessary. Therefore we first determine the value of \(a\) by evaluating the limit of the given expression as \(x\to0\). After obtaining \(a\), we calculate \(f'(0)\) using series expansion. The binomial expansion formula is: \[ (1+t)^n=1+nt+\frac{n(n-1)}{2!}t^2+\frac{n(n-1)(n-2)}{3!}t^3+\cdots \] This helps us simplify complicated expressions near \(x=0\).

Step 1: Expand \((1+x)^{1/3}\) using binomial theorem.
Using \(n=\frac13\), \[ (1+x)^{1/3} = 1+\frac{x}{3} -\frac{x^2}{9} +\frac{5x^3}{81} +\cdots \]

Step 2: Expand \((1+2x)^{1/4}\).
Using \(n=\frac14\), \[ (1+2x)^{1/4} = 1+\frac{x}{2} -\frac{3x^2}{8} +\frac{7x^3}{16} +\cdots \]

Step 3: Subtract the two expansions.
\[ (1+x)^{1/3}-(1+2x)^{1/4} = -\frac{x}{6} +\frac{19x^2}{72} +\cdots \] Hence \[ f(x) = \frac{-\frac{x}{6}+\frac{19x^2}{72}+\cdots}{x} = -\frac16+\frac{19}{72}x+\cdots \] Taking limit as \(x\to0\), \[ a=\lim_{x\to0}f(x) = -\frac16. \]

Step 4: Find \(f'(0)\).
From \[ f(x) = -\frac16+\frac{19}{72}x+\cdots \] the coefficient of \(x\) directly gives \[ f'(0)=\frac{19}{72}. \]

Step 5: Calculate \(a+f'(0)\).
\[ a+f'(0) = -\frac16+\frac{19}{72} \] \[ = -\frac{12}{72}+\frac{19}{72} = \frac{7}{72}. \] Therefore, \[ \boxed{a+f'(0)=\frac{7}{72}} \] Hence the correct answer is \[ \boxed{(B)} \]