Question

Find the values of a and b, if the function f defined by 

\[f\left( x \right) = \begin{cases}x^2 + 3x + a & , & x \leqslant 1 \\ bx + 2 & , & x > 1\end{cases}\] is differentiable at x = 1.

Answer 1

Given that f(x) is differentiable at x = 1. Therefore,  f(x) is continuous at x = 1. 

\[\lim_{x \to 1^-} f\left( x \right) = \lim_{x \to 1^+} f\left( x \right) = f\left( 1 \right)\]
\[ \Rightarrow \lim_{x \to 1} \left( x^2 + 3x + a \right) = \lim_{x \to 1} \left( bx + 2 \right) = 1 + 3 + a\]
\[ \Rightarrow 1 + 3 + a = b + 2\]
\[ \Rightarrow a - b + 2 = 0 . . . . . \left( 1 \right)\]

Again, f(x) is differentiable at x = 1. So,

(LHD at x = 1) = (RHD at x = 1) 

\[\Rightarrow \lim_{x \to 1^-} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1} = \lim_{x \to 1^+} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1}\]

\[\lim_{x \to 1} \frac{\left( x^2 + 3x + a \right) - \left( 4 + a \right)}{x - 1} = \lim_{x \to 1} \frac{\left( bx + 2 \right) - \left( 4 + a \right)}{x - 1}\]
\[ \Rightarrow \lim_{x \to 1} \frac{x^2 + 3x - 4}{x - 1} = \lim_{x \to 1} \frac{\left( bx - 2 - a \right)}{x - 1}\]
\[ \Rightarrow \lim_{x \to 1} \frac{\left( x + 4 \right)\left( x - 1 \right)}{x - 1} = \lim_{x \to 1} \frac{bx - b}{x - 1}\]
\[ \Rightarrow \lim_{x \to 1} \left( x + 4 \right) = \lim_{x \to 1} \frac{b\left( x - 1 \right)}{x - 1}\]
\[ \Rightarrow 5 = b\]

Putting b = 5 in (1), we get
a = 3
Hence, a = 3 and b = 5.