Let a and b be real constants such that the function f defined by f ( x ) = { x 2 + 3 x + a , x ? 1 b x + 2 , x > 1 be differentiable on R . Then, the value of - 2 2 f ( x ) ? d x equals [2024]

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Solution

Q.
Let $a$ and $b$ be real constants such that the function $f$ defined by $f (x)$ $= {\begin{matrix} x^{2} + 3 x + a, & x \leq 1 \\ b x + 2, & x > 1 \end{matrix}$ be differentiable on $R .$ Then, the value of $\int_{- 2}^{2} f (x) d x$ equals [2024]

1 $\frac{19}{6}$

2 $21$

3 $17$

4 $\frac{15}{6}$

Ans.
(3)

We have, $f (x)$ $= {\begin{matrix} x^{2} + 3 x + a, & x \leq 1 \\ b x + 2, & x > 1 \end{matrix}$

Since, $f (x)$ be differentiable on R.

So, $f (x)$ be continuous at $x = 1 .$

$∴ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1)$

$\Rightarrow \lim_{h \to 0} f (1 - h) = \lim_{h \to 0} f (1 + h) = f (1)$

$\Rightarrow a + 4 = b + 2 \Rightarrow b = a + 2$ ...(i)

Also, $L f' (1) = R f' (1)$

$\Rightarrow 5 = b$

$∴ a = 3 and b = 5 (Using (i))$

Now, $\int_{- 2}^{2} f (x) d x = \int_{- 2}^{1} (x^{2} + 3 x + 3) d x + \int_{1}^{2} (5 x + 2) d x$

$= {(\frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 3 x)}_{- 2}^{1} + {(\frac{5 x^{2}}{2} + 2 x)}_{1}^{2}$

$= \frac{1}{3} (1 + 8) + \frac{3}{2} (1 - 4) + 3 (1 + 2) + \frac{5}{2} (4 - 1) + 2 (2 - 1)$

$= 3 - \frac{9}{2} + 9 + \frac{15}{2} + 2 = 17$

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