If the derivative of the function f ( x ) = b x 2 + a x + 4 ; x ≥ − 1 a x 2 + b ; x < − 1 ′ , is everywhere continuous, then Options: A . &nbsp a = 2, b = 3 B . &nbsp a = 3, b = 2 C . &nbsp a = - 2, b = - 3 D . &nbsp a = - 3, b = - 2

If the derivative of the function f(x)=bx2+ax+4;x≥−1ax2+b;x<&

Question -

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

Options:

A . a = 2, b = 3

B . a = 3, b = 2

C . a = - 2, b = - 3

D . a = - 3, b = - 2

Answer: Option A
:
A

W e h a v e, f (x) = {\begin{matrix} a x^{2} + b, & x < - 1 b x^{2} + a x + 4, & x \geq - 1 \end{matrix} ∴ f^{^{'}} (x) = {\begin{matrix} 2 a x, & < - 1 2 b x + a, & x \geq - 1 \end{matrix} S i n c e, f (x) i s d i f f e r e n t i a b l e a t x = - 1, t h e r e f o r e i t i s c o n t i n u o u s a t x = - 1 a n d h e n c e, l i m x \to - 1 - f (x) = l i m x \to - 1 + f (x) \Rightarrow a + b = b - a + 4 \Rightarrow a = 2 a n d a l s o, l i m x \to - 1 - f (x) = l i m x \to - 1 + f (x) \Rightarrow - 2 a = - 2 b + a \Rightarrow 3 a = 2 b \Rightarrow b = 3 (∵ a = 2) H e n c e, a = 2, b = 3

Was this answer helpful ?

Next Question