JEE Main 2024 — Mathematics Calculus

Given:

$$\underset{x\rightarrow 0}{\mathrm{lim}}\frac{a{x}^{2}{e}^{x}-b{\mathrm{log}}_{e}(1+x)+cx{e}^{-x}}{{x}^{2}\mathrm{sin}x}=1$$x→0lim​x2sinxax2ex−bloge​(1+x)+cxe−x​=1

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{a{x}^{2}{e}^{x}-b{\mathrm{log}}_{e}(1+x)+cx{e}^{-x}}{{x}^{2}\cdot x\cdot \frac{\mathrm{sin}x}{x}}=1$$⇒x→0lim​x2⋅x⋅xsinx​ax2ex−bloge​(1+x)+cxe−x​=1

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{a{x}^{2}(1+\frac{x}{1!}+\frac{{x}^{2}}{2!}+...)-b(x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+...)+cx(1-\frac{x}{1!}+\frac{{x}^{2}}{2!}+...)}{{x}^{2}\times x}=1$$⇒x→0lim​x2×xax2(1+1!x​+2!x2​+...)−b(x−2x2​+3x3​+...)+cx(1−1!x​+2!x2​+...)​=1

Coefficient of $x=0$x=0

$$\Rightarrow -b+c=0...(iii)$$⇒−b+c=0...(iii)

Coefficient of ${x}^{2}=0$x2=0

$$\Rightarrow a+\frac{b}{2}-c=0...(ii)$$⇒a+2b​−c=0...(ii)

$$\Rightarrow a-\frac{c}{2}=0$$⇒a−2c​=0

$$\Rightarrow a=\frac{c}{2}$$⇒a=2c​

Coefficient of ${x}^{3}=1$x3=1

$$\Rightarrow a-\frac{b}{3}+\frac{c}{2}=1...(iii)$$⇒a−3b​+2c​=1...(iii)

$$\Rightarrow \frac{c}{2}-\frac{c}{3}+\frac{c}{2}=1$$⇒2c​−3c​+2c​=1

$$\Rightarrow \frac{2c}{3}=1$$⇒32c​=1

$$\Rightarrow c=\frac{3}{2}=b$$⇒c=23​=b

$$\Rightarrow a=\frac{3}{4}$$⇒a=43​

$$\Rightarrow 16({a}^{2}+{b}^{2}+{c}^{2})=16(\frac{9}{16}+\frac{9}{4}\times 2)$$⇒16(a2+b2+c2)=16(169​+49​×2)

$$\Rightarrow 16({a}^{2}+{b}^{2}+{c}^{2})=81$$⇒16(a2+b2+c2)=81