Continuity And Differentiability Question 7
Question 7 - 2024 (30 Jan Shift 2)
Let $a$ and $b$ be real constants such that the function $f$ defined by $f(x)=\begin{cases}x^{2}+3 x+a & , x \leq 1 \\ b x+2 & , x>1\end{cases}$ be differentiable on $R$. Then, the value of $\int _{-2}^{2} f(x) d x$ equals
(1) $\frac{15}{6}$
(2) $\frac{19}{6}$
(3) 21
(4) 17
Show Answer
Answer (4)
Solution
$f$ is continuous
$f^{\prime}(x)=2 x+3, k<1$
$\therefore 4+a=b+2$
$a=b-2$
b, $x>1$
$a=b-2$
$f$ is differentiable
$\therefore b=5$
$\therefore \quad a=3$
$\int _{-2}^{1}\left(x^{2}+3 x+3\right) d x+\int _1^{2}(5 x+2) d x$
$=\left[\frac{x^{3}}{3}+\frac{3 x^{2}}{2}+3 x\right] _{-2}^{1}+\left[\frac{5 x^{2}}{2}+2 x\right] _1^{2}$
$=\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right)$
$=6+\frac{3}{2}+12-\frac{5}{2}=17$
