The function f(x)=1|x|,|x|≥1ax2+b,|x|<1 iw If f is continuous and differentiable ther

fx=1|x| , x≤−1ax2+b , -1<x<11x , x≥1Rf1(-1)=limh→0f(-1+h)-f(-1)h=limh→0a(-1+h)+b-1h=limh→0(a+b-1)+ah2-2hhLf-1(-1)=limh→0f(-1-h)-f(-1)-h=limh→0-h (1+h)-h=limh→011+h=1 Since f(x) in differentiable of x=-1∴Rf1(-1)=Lf1(-1)⇒limh→0(a+b-1)+ah2-hhh→0=1 ⇒limh→0a+b-1h+a(h-2)=1⇒a+b-1=0 and -2a=1⇒a+b=1 and a=-12⇒a=-12, b=32

The function f(x)=1|x|,|x|≥1ax2+b,|x|<1 iw If f is continuous and differentiable ther

fx=1|x| , x≤−1ax2+b , -1<x<11x , x≥1Rf1(-1)=limh→0f(-1+h)-f(-1)h=limh→0a(-1+h)+b-1h=limh→0(a+b-1)+ah2-2hhLf-1(-1)=limh→0f(-1-h)-f(-1)-h=limh→0-h (1+h)-h=limh→011+h=1 Since f(x) in differentiable of x=-1∴Rf1(-1)=Lf1(-1)⇒limh→0(a+b-1)+ah2-hhh→0=1 ⇒limh→0a+b-1h+a(h-2)=1⇒a+b-1=0 and -2a=1⇒a+b=1 and a=-12⇒a=-12, b=32

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$The function f (x) = \{\begin{array}{l} \frac{1}{| x |} & , | x | \geq 1 \\ a x^{2} + b & , | x | < 1 \end{array} \begin{array}{l} iw If f is continuous and differentiable ther \end{array}$

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

$a = 1, b = 1$

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

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

answer is C.

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Detailed Solution

$f (x) = \{\begin{cases} \frac{1}{| x |}, x \leq - 1 \\ a x^{2} + b, -1 < x < 1 \\ \frac{1}{x}, x \geq 1 \end{cases}$

$\begin{array}{l} R f^{1} (- 1) = \lim_{h \to 0} \frac{f (- 1 + h) - f (- 1)}{h} = \lim_{h \to 0} \frac{a (- 1 + h) + b - 1}{h} = \lim_{h \to 0} \frac{(a + b - 1) + a (h^{2} - 2 h)}{h} \\ L f^{- 1} (- 1) = \lim_{h \to 0} \frac{f (- 1 - h) - f (- 1)}{- h} = \lim_{h \to 0} \{\frac{- h}{} (\frac{(1 + h)}{- h}\} = \lim_{h \to 0} \frac{1}{1 + h} = 1 \end{array}$

$\begin{array}{l} Since f (x) in differentiable of x = - 1 \\ \begin{array}{l} ∴ R f^{1} (- 1) = L f^{1} (- 1) \\ \Rightarrow \lim_{h \to 0} \underset{h \to 0}{\frac{(a + b - 1) + a (h^{2} - h)}{h}} = 1 \Rightarrow \lim_{h \to 0} \{\frac{a + b - 1}{h} + a (h - 2)\} = 1 \\ \Rightarrow a + b - 1 = 0 and - 2 a = 1 \\ \Rightarrow a + b = 1 and a = \frac{- 1}{2} \\ \Rightarrow a = \frac{- 1}{2}, b = \frac{3}{2} \end{array} \end{array}$