If f(x) = begin{cases} frac{1}{|x|}, & |x| ge | vedclass

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(B) Given,$f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geq 1 \ ax^2 + b, & -1 < x < 1 \end{cases}$Expanding the definition,we get:$f(x) = \begin{cases

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$a = \frac{-1}{2}, b = \frac{3}{2}$

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(B) Given,$f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geq 1 \ ax^2 + b, & -1 Expanding the definition,we get:$f(x) = \begin{cases} -\frac{1}{x}, & x \leq -1 \ ax^2 + b, & -1 For $f(x)$ to be differentiable $\forall x \in \mathbb{R}$,it must be continuous and differentiable at $x = 1$.Continuity at $x = 1$: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$.$\implies a(1)^2 + b = \frac{1}{1} \implies a + b = 1$ . . . $(1)$Differentiability at $x = 1$: $f'(1^-) = f'(1^+)$.For $x  1$,$f'(x) = -\frac{1}{x^2}$.$\implies 2a(1) = -\frac{1}{(1)^2} \implies 2a = -1 \implies a = -\frac{1}{2}$.Substituting $a = -\frac{1}{2}$ into equation $(1)$:$-\frac{1}{2} + b = 1 \implies b = 1 + \frac{1}{2} = \frac{3}{2}$.Thus,$a = -\frac{1}{2}$ and $b = \frac{3}{2}$.

If $f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geq 1 \\ ax^2 + b, & -1 < x < 1 \end{cases}$ is differentiable $\forall x \in \mathbb{R}$,then one of the values of $a$ and $b$ is-

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