JEE Main 2024 — Mathematics Calculus

Given: f(x)={\begin{matrix}{x}^{2}+3x+a, & x\leq 1 \\ bx+2, & x>1\end{matrix}

Now, for function to be continuous,

$$\Rightarrow {(1)}^{2}+3\times 1+a=b\times 1+2$$⇒(1)2+3×1+a=b×1+2

$$\Rightarrow 4+a=b+2$$⇒4+a=b+2

$$\Rightarrow a-b=-2...(i)$$⇒a−b=−2...(i)

Now, differentiating and putting the value of $x$x,

$$\Rightarrow 2x+3=b$$⇒2x+3=b

$$\Rightarrow 2(1)+3=b$$⇒2(1)+3=b

$$\Rightarrow b=5$$⇒b=5

$$\Rightarrow a=3$$⇒a=3

$$\Rightarrow {\int }_{-2}^{2}f(x)dx={\int }_{-2}^{1}({x}^{2}+3x+a)dx+{\int }_{1}^{2}(bx+2)dx$$⇒∫−22​f(x)dx=∫−21​(x2+3x+a)dx+∫12​(bx+2)dx

$$\Rightarrow {\int }_{-2}^{2}f(x)dx={\int }_{-2}^{1}({x}^{2}+3x+3)dx+{\int }_{1}^{2}(5x+2)dx$$⇒∫−22​f(x)dx=∫−21​(x2+3x+3)dx+∫12​(5x+2)dx

$$\Rightarrow {\int }_{-2}^{2}f(x)dx={[\frac{{x}^{3}}{3}+\frac{3{x}^{2}}{2}+3x]}_{-2}^{1}+{[\frac{5{x}^{2}}{2}+2x]}_{1}^{2}$$⇒∫−22​f(x)dx=[3x3​+23x2​+3x]−21​+[25x2​+2x]12​

$$\Rightarrow {\int }_{-2}^{2}f(x)dx=[(\frac{1}{3}+\frac{3}{2}+3)-(\frac{-8}{3}+6-6)]+[(\frac{5\times 4}{2}+2\times 2)-(\frac{5}{2}+2)]$$⇒∫−22​f(x)dx=[(31​+23​+3)−(3−8​+6−6)]+[(25×4​+2×2)−(25​+2)]

$$\Rightarrow {\int }_{-2}^{2}f(x)dx=[\frac{1}{3}+\frac{3}{2}+3+\frac{8}{3}]+[14-\frac{9}{2}]$$⇒∫−22​f(x)dx=[31​+23​+3+38​]+[14−29​]

$$\Rightarrow {\int }_{-2}^{2}f(x)dx=\frac{15}{2}+14-\frac{9}{2}$$⇒∫−22​f(x)dx=215​+14−29​

$$\Rightarrow {\int }_{-2}^{2}f(x)dx=17$$⇒∫−22​f(x)dx=17