Continuity And Differentiability Question 7
Question 7 - 2024 (30 Jan Shift 2)
Let $a$ and $b$ be real constants such that the function $\mathrm{f}$ defined by $f(x)=\left{\begin{array}{ll}x^{2}+3 x+a & , x \leq 1 \ b x+2 & , x>1\end{array}\right.$ 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
$\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+3, \mathrm{k}<1$
$\therefore 4+a=b+2$
$a=b-2$
b, $x>1$
$\mathrm{a}=\mathrm{b}-2$
$\mathrm{f}$ is differentiable
$\therefore \mathrm{b}=5$
$\therefore \quad \mathrm{a}=3$
$\int_{-2}^{1}\left(x^{2}+3 x+3\right) d x+\int_{1}^{2}(5 x+2) d x$
$=\left[\frac{x^{3}}{3}+\frac{3 x^{2}}{2}+3 x\right]{-2}^{1}+\left[\frac{5 x^{2}}{2}+2 x\right]{1}^{2}$
$=\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right)$
$=6+\frac{3}{2}+12-\frac{5}{2}=17$
