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  • The coefficients a and b that make the function continuous and differentiable <img alt="\mathrm{ f(x)=\left\{\begin{array}{ccc} \frac{1}{|x|} & ; \text { for }|x| \geq 1 \\ a x

The coefficients a and b that make the function continuous and differentiable
\mathrm{ f(x)=\left\{\begin{array}{ccc} \frac{1}{|x|} & ; \text { for }|x| \geq 1 \\ a x^2+b & ; \text { for }|x|<1 \end{array}\right. }

Option: 1

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


Option: 2

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


Option: 3

\mathrm{a=1, b=-1}


Option: 4

\text { none of these }


Answers (1)

best_answer

\mathrm{f(x)=\left\{\begin{array}{ccc} -\frac{1}{x} & ; & x \leq-1 \\ a x^2+b & ; & -1<x<1 \\ \frac{1}{x} & ; & x \geq 1 \end{array}\right.}                     ...(i)

\mathrm{f^{\prime}(x)=\left\{\begin{array}{ccc} \frac{1}{x^2} & ; & x \leq-1 \\ 2 a x & ; & -1<x<1 \\ -\frac{1}{x^2} & ; & x>1 \end{array}\right.}                             ...(ii)

For the function to be continuous

\mathrm{\mathrm{LHL}=\mathrm{RHL}=} value of function

at \mathrm{x=-1 \Rightarrow 1=a+b=1}                                     ...(iii)

at \mathrm{x=1 \Rightarrow a+b=1=1}                                         ...(iv)

For the function to be differentiable

LHD = RHD

at \mathrm{x=-1 \Rightarrow 1=-2 a}                                                      ...(v)

at \mathrm{x=1 \Rightarrow 2 a=-1}                                                            ...(vi)

From (v) or (vi) \mathrm{\Rightarrow \quad a=-\frac{1}{2}}

Put  \mathrm{{ }^a=-\frac{1}{2}} in (iii) or (iv) \mathrm{\Rightarrow b=\frac{3}{2}}

Posted by

Ritika Harsh

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