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  • Let <img alt="\mathrm{f(x)=\left\{\begin{array}{ll} 3 x^2-4 \sqrt{x}+1 & \text { for } x<1 \\ a x+b & \text { for } x \geq 1 \end{array}\right. \text {. }}" src="https://entrancecor

Let \mathrm{f(x)=\left\{\begin{array}{ll} 3 x^2-4 \sqrt{x}+1 & \text { for } x<1 \\ a x+b & \text { for } x \geq 1 \end{array}\right. \text {. }}

If \mathrm{f(x)} is the continuous and differentiable for all real values in its domain then

 

Option: 1

\mathrm{a=b=4}


Option: 2

\mathrm{a=b=-4}


Option: 3

a=4\: and\: b=-4


Option: 4

\mathrm{a =-4\: and \: b=4}


Answers (1)

best_answer

\mathrm{f(x)= \begin{cases}3 x^2-4 \sqrt{x}+1 & \text { for } x<1 \\ a x+b & \text { fot } x \geq 1\end{cases}}

For continuity at \mathrm{x=1, f\left(1^{+}\right)=\left(1^{-}\right)}

\mathrm{\begin{aligned} & \Rightarrow \quad a+b=3-4+1 \\ & \Rightarrow \quad a+b=0 \end{aligned}}

For differentiability at \mathrm{x=1}

\mathrm{\begin{aligned} & f^{\prime}\left(1^\times\right)=f^{\prime}\left(1^7\right) \\ \Rightarrow & a=6-2=4 \\ \Rightarrow & b=-4 \end{aligned}}

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