QJEEM24Ma69eb: JEE Main 2024 Mathematics - Class 12 JEEM24Ma69eb Solution: Let and be real constants such that the function defined by be…

To determine the integral of the piecewise function $f$ over the interval $[-2, 2]$ , we first ensure that $f$ is differentiable on $\mathbb{R}$ , as given in the problem statement. Differentiability implies continuity, so $f$ must also be continuous at $x=1$ .

The condition for continuity at $x=1$ is:

$x^2 + 3x + a = bx + 2$ at $x=1$ .

This simplifies to:

$1 + 3 + a = b(1) + 2$

$\Rightarrow a = b - 2$

The first derivative of $f$ gives us two different expressions depending on the value of $x$ :

For $x \leq 1$ , $f'(x) = 2x + 3$ ; and for x > 1 , $f'(x) = b$ .

For $f$ to be differentiable at $x=1$ , these derivatives must be equal at that point. Setting $f'(1)$ from both expressions equal to each other gives $2(1) + 3 = b$ , thus $b = 5$ . And from $a = b - 2$ , we have $a = 3$ .

Now, we can calculate the integral of $f$ over the specified interval:

$\int\limits_{-2}^1 (x^2 + 3x + 3) \, dx + \int\limits_{1}^2 (5x + 2) \, dx$

$\begin{aligned}& =\left[\frac{x^3}{3}+\frac{3 x^2}{2}+3 x\right]_{-2}^1+\left[\frac{5 x^2}{2}+2 x\right]_1^2 \\\\& =\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right) \\\\& =6+\frac{3}{2}+12-\frac{5}{2}=17\end{aligned}$