Question

If $f(x) = ax^2 - b , | x | < 1 = \frac{1}{|x|} , |x | \ge 1$ is differentiable at x = 1, then

• A. $a = \frac{1}{2} , b = - \frac{1}{2}$
• B. $a =- \frac{1}{2} , b = \frac{3}{2}$
• C. $a = b = \frac{1}{2} $
• D. $a = b = - \frac{1}{2} $

Answer 1

Answer: (B)

Since f(x) is diff. at x = 1
$\therefore Lt_{x \to1-} \frac{f\left(x\right) - f\left(1\right)}{x-1} = Lt_{x\to1+} \frac{f\left(x\right) -f\left(1\right)}{x-1}$
$ \therefore Lt_{x\to1-} \frac{ax^{2} -b-1}{x-1} = Lt_{x\to1+} \frac{\frac{1}{\left|x\right|} - 1 }{x-1} $
$\Rightarrow Lt_{x \to1} \frac{a\left(x^{2} -1\right)+a-b-1}{x-1} = Lt_{x\to1} \frac{1-x}{x\left(x-1\right)} ....\left(1\right) $
$\Rightarrow a\left(1+1\right) + Lt_{x\to1} \frac{a-b-1}{x-1} = - 1 $
$\Rightarrow 2a + Lt_{x\to1} \frac{a-b-1}{x-1} =- 1$
$ \therefore a-b -1 = 0 $
$\therefore $ (1) gives $ - Lt_{x\to1}a\left(x+1\right)=-1 $
$\Rightarrow 2a =-1 \therefore a = - \frac{1}{2}$
$ \therefore - \frac{1}{2} - b -1= 0 \therefore b = - 3/2$