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  • Let <img alt="f(x)=\left\{\begin{array}{cc}e^x-a x, & x<0 \\ b(1-x)^2, & x \geq 0\end{array}\right." src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x%29%3D%5Cleft%5C%7B%5Cbegin

Let f(x)=\left\{\begin{array}{cc}e^x-a x, & x<0 \\ b(1-x)^2, & x \geq 0\end{array}\right.. If f is a differentiable function, then the ordered pair (a,b) is

Option: 1

(1,3)
 


Option: 2

(3,1)
 


Option: 3

(-1,-3)
 


Option: 4

(-3,1)


Answers (1)

best_answer

f\left(0^{-}\right)=1, f\left(0^{+}\right)=b. Thus b=1.\\ f^{\prime}(x)=\left\{\begin{array}{cc}e^x-a, & x<0 \\ -2 b(1-x), & x>0\end{array} \quad f^{\prime}\left(0^{-}\right)=1-a, f^{\prime}\left(0^{+}\right)=-2 b\right.\\ \text{Thus}\ 1-a=-2 b \Rightarrow a=1+2 b=3.\\ \text{Hence}\ (a,(B) \equiv(3,1).

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