Let a and b be real constants such that the function f defined by… - Vidyadip

MCQ

Let $a$ and $b$ be real constants such that the function $f$ defined by $f(x)=\left\{\begin{array}{cc}x^2+3 x+a & , x \leq 1 \\ b x+2 & , x>1\end{array}\right.$ be differentiable on $R$. Then, the value of $\int_{-2}^2 f(x) d x$ equals

A
$\frac{15}{6}$
B
$\frac{19}{6}$
C
21
D
17

✓

Answer

$f$ is continuous$
f^{\prime}(x)=2 x+3, k<1$
$\therefore 4+a=b+2b$
$x>1$
$a = b -2$
$f$ is differentiable
$\therefore b=5$
$\therefore a=3$
$\int_{-2}^1\left(x^2+3 x+3\right) dx+\int_1^2(5 x+2) dx$
$=\left[\frac{x^3}{3}+\frac{3 x^2}{2}+3 x\right]_{-2}^1+\left[\frac{5 x^2}{2}+2 x\right]_1^2$
$=\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right)$
$=6+\frac{3}{2}+12-\frac{5}{2}$
$=17$

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