JEE Main 2021 — Mathematics Calculus

We have,

$$\underset{x\rightarrow 0}{\mathrm{lim}}[\frac{\alpha x{e}^{x}-\beta {\mathrm{log}}_{e}(1+x)+\gamma {x}^{2}{e}^{-x}}{x{\mathrm{sin}}^{2}x}]=10$$x→0lim​[xsin2xαxex−βloge​(1+x)+γx2e−x​]=10

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}[\frac{\alpha x(1+x+\frac{{x}^{2}}{2!}+..)-\beta (x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-...)+\gamma {x}^{2}(1-x+\frac{{x}^{2}}{2!}-..)}{{x}^{3}(\frac{{\mathrm{sin}}^{2}x}{{x}^{2}})}]=10$$⇒x→0lim​[x3(x2sin2x​)αx(1+x+2!x2​+..)−β(x−2x2​+3x3​−...)+γx2(1−x+2!x2​−..)​]=10

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}[\frac{\alpha x(1+x+\frac{{x}^{2}}{2})-\beta (x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3})+\gamma {x}^{2}(1-x)}{{x}^{3}}]=10$$⇒x→0lim​[x3αx(1+x+2x2​)−β(x−2x2​+3x3​)+γx2(1−x)​]=10

Neglecting higher powers.

$$\underset{x\rightarrow 0}{\mathrm{lim}}\frac{x(\alpha -\beta )+{x}^{2}(\alpha +\frac{\beta }{2}+\gamma )+{x}^{3}(\frac{\alpha }{2}-\frac{\beta }{3}-\gamma )}{{x}^{3}}=10$$x→0lim​x3x(α−β)+x2(α+2β​+γ)+x3(2α​−3β​−γ)​=10

For limit to exist

$$\alpha -\beta =0$$α−β=0

$$\alpha +\frac{\beta }{2}+\gamma =0$$α+2β​+γ=0

$$\frac{\alpha }{2}-\frac{\beta }{3}-\gamma =10\ldots (i)$$2α​−3β​−γ=10…(i)

$$\beta =\alpha ,\gamma =-\frac{3\alpha }{2}$$β=α,γ=−23α​

Putting in $(i)$(i), we get

$$\frac{\alpha }{2}-\frac{\alpha }{3}+\frac{3\alpha }{2}=10$$2α​−3α​+23α​=10

$$\Rightarrow \frac{\alpha }{6}+\frac{3\alpha }{2}=10$$⇒6α​+23α​=10

$$\Rightarrow \frac{\alpha +9\alpha }{6}=10$$⇒6α+9α​=10

$$\Rightarrow \alpha =6$$⇒α=6

Hence,

$$\alpha =6,\beta =6,\gamma =-9$$α=6,β=6,γ=−9

$$\Rightarrow \alpha +\beta +\gamma =3$$⇒α+β+γ=3