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If \mathrm{f(x)=\left\{\begin{array}{cc}a x^2+b, & x \leq 0 \\ 0, & x>0\end{array}\right.}  is differentiable at \mathrm{x=0}, then
 

Option: 1

\mathrm{a=1, b=1}
 


Option: 2

\mathrm{a=1, b=2}
 


Option: 3

\mathrm{a=2, b=0}
 


Option: 4

\mathrm{a=2, b=1}


Answers (1)

best_answer

\text{For continuity at }\mathrm{x=0, b=0}

\text{For differentiability }\mathrm{2 a x=0 \: at\: x=0}

\mathrm{So\: \mathrm{a} \: \text{can be any real number}}

Hence option 3 is correct.

Posted by

vinayak

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