JEE Main 2024 — Mathematics Calculus

Given,

$$\underset{x\rightarrow 0}{\mathrm{lim}}\frac{3+\alpha sinx+\beta cosx+{\mathrm{log}}_{e}(1-x)}{3{\mathrm{tan}}^{2}x}=\frac{1}{3}$$x→0lim​3tan2x3+αsinx+βcosx+loge​(1−x)​=31​

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{3+\alpha [x-\frac{{x}^{3}}{3!}+\ldots ]+\beta [1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}\ldots ]+(-x-\frac{{x}^{2}}{2}-\frac{{x}^{3}}{3}\ldots )}{3{\mathrm{tan}}^{2}x}=\frac{1}{3}$$⇒x→0lim​3tan2x3+α[x−3!x3​+…]+β[1−2!x2​+4!x4​…]+(−x−2x2​−3x3​…)​=31​

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{(3+\beta )+(\alpha -1)x+(-\frac{1}{2}-\frac{\beta }{2}){x}^{2}+\ldots .}{3{x}^{2}}\times \frac{{x}^{2}}{{\mathrm{tan}}^{2}x}=\frac{1}{3}$$⇒x→0lim​3x2(3+β)+(α−1)x+(−21​−2β​)x2+….​×tan2xx2​=31​

The coefficient of all the powers of $x$x except ${x}^{2}$x2 is $0$0.

$\Rightarrow \beta +3=0,\alpha -1=0$⇒β+3=0,α−1=0 and $\frac{-\frac{1}{2}-\frac{\beta }{2}}{3}=\frac{1}{3}$3−21​−2β​​=31​

$$\Rightarrow \beta =-3,\alpha =1$$⇒β=−3,α=1

$$\Rightarrow 2\alpha -\beta =2+3=5$$⇒2α−β=2+3=5