If the derivative of the function f(x)=ax2+b, x<−1bx2+ax+4, x≥−1 is everywhere continuous, then

f(x)=ax2+b, x<−1bx2+ax+4, x≥−1⇒f'x=2ax, x<−12bx+a, x≥−1Since f(x) is differentiable at x=−1,∴fx is continuous at x=−1∴limx→1−f(x)=limx→1+f(x)⇒a+b=b−a+4⇒a=2 ....(1)Also, since f'(x) is continuous, therefore, f'(x) is continuous at x=−1.∴limx→−1−f'(x)=limx→−1+f'(x)⇒−2a=−2b+a ....(2)Using (1) in (2), we get, b=3∴a=2, b=3.

If the derivative of the function f(x)=ax2+b, x<−1bx2+ax+4, x≥−1 is everywhere continuous, then

f(x)=ax2+b, x<−1bx2+ax+4, x≥−1⇒f'x=2ax, x<−12bx+a, x≥−1Since f(x) is differentiable at x=−1,∴fx is continuous at x=−1∴limx→1−f(x)=limx→1+f(x)⇒a+b=b−a+4⇒a=2 ....(1)Also, since f'(x) is continuous, therefore, f'(x) is continuous at x=−1.∴limx→−1−f'(x)=limx→−1+f'(x)⇒−2a=−2b+a ....(2)Using (1) in (2), we get, b=3∴a=2, b=3.

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$If the derivative of the function f (x) = \{\begin{cases} a x^{2} + b, x < - 1 \\ b x^{2} + a x + 4, x \geq - 1 \end{cases} is everywhere continuous, then$

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$a = - 2, b = - 3$

$a = 2, b = 3$

$a = - 3, b = - 2$

$a = 3, b = 2$

answer is A.

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Detailed Solution

$\begin{array}{l} f (x) = \{\begin{cases} a x^{2} + b, x < - 1 \\ b x^{2} + a x + 4, x \geq - 1 \end{cases} \\ \Rightarrow f' (x) = \{\begin{cases} 2 a x, x < - 1 \\ 2 b x + a, x \geq - 1 \end{cases} \\ Since f(x) is differentiable at x = - 1, \\ ∴f (x) is continuous at x = - 1 \\ ∴ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) \Rightarrow a + b = b - a + 4 \Rightarrow a = 2_{_{_{_{.... (1)}}}} \\ Also, since f'(x) is continuous, therefore, f'(x) is continuous at x= - 1. \\ ∴ \lim_{x \to - 1^{-}} f' (x) = \lim_{x \to - 1^{+}} f' (x) \Rightarrow - 2 a = - 2 b + a .... (2) \\ Using (1) in (2), we get, b=3 \\ ∴ a = 2, b = 3. \end{array}$