JEE Main 2023 — Mathematics Calculus

Given,

$${\int }_{0}^{1}\frac{1}{(5+2x-2{x}^{2})(1+{e}^{(2-4x)})}dx=\frac{1}{\alpha }{\mathrm{log}}_{e}(\frac{\alpha +1}{\beta })$$∫01​(5+2x−2x2)(1+e(2−4x))1​dx=α1​loge​(βα+1​)

Now let,

$$I=\frac{1}{2}{\int }_{0}^{1}\frac{1}{(\frac{5}{2}-({x}^{2}-x))(1+{e}^{-4(x-\frac{1}{2})})}dx$$I=21​∫01​(25​−(x2−x))(1+e−4(x−21​))1​dx

$$\Rightarrow I=\frac{1}{2}{\int }_{0}^{1}\frac{1}{((\frac{11}{4})-{(x-\frac{1}{2})}^{2})(1+{e}^{-4(x-\frac{1}{2})})}dx$$⇒I=21​∫01​((411​)−(x−21​)2)(1+e−4(x−21​))1​dx

Now let $x-\frac{1}{2}=t\Rightarrow dx=dt$x−21​=t⇒dx=dt

So, integral becomes,

$$I=\frac{1}{2}{\int }_{\frac{-1}{2}}^{\frac{1}{2}}\frac{1}{{(\frac{\sqrt{11}}{2})}^{2}-{t}^{2})(1+{e}^{-4t})}dt$$I=21​∫2−1​21​​(211​​)2−t2)(1+e−4t)1​dt

$$\Rightarrow I=\frac{1}{2}{\int }_{\frac{-1}{2}}^{0}\frac{1}{{(\frac{\sqrt{11}}{2})}^{2}-{t}^{2})(1+{e}^{-4t})}dt+\underset{{I}_{2}}{\underset{⏟}{\frac{1}{2}{\int }_{0}^{\frac{1}{2}}\frac{1}{{(\frac{\sqrt{11}}{2})}^{2}-{t}^{2})(1+{e}^{-4t})}dt}}$$⇒I=21​∫2−1​0​(211​​)2−t2)(1+e−4t)1​dt+I2​⏟21​∫021​​(211​​)2−t2)(1+e−4t)1​dt​​

Now solving $\displaystyle {I}_{2}=\frac{1}{2}{\int }_{\frac{-1}{2}}^{\frac{1}{2}}\frac{1}{{(\frac{\sqrt{11}}{2})}^{2}-{t}^{2})(1+{e}^{-4t})}dt$I2​=21​∫2−1​21​​(211​​)2−t2)(1+e−4t)1​dt,

Let $t=-z$t=−z, we get $dt=-dz$dt=−dz

So, $\displaystyle {I}_{2}=-\frac{1}{2}{\int }_{\frac{1}{2}}^{0}\frac{dz}{{(\frac{\sqrt{11}}{2})}^{2}-{z}^{2})(1+{e}^{4z})}$I2​=−21​∫21​0​(211​​)2−z2)(1+e4z)dz​

$$\Rightarrow {I}_{2}=\frac{1}{2}{\int }_{0}^{\frac{1}{2}}\frac{dt}{{(\frac{\sqrt{11}}{2})}^{2}-{t}^{2})(1+{e}^{4t})}$$⇒I2​=21​∫021​​(211​​)2−t2)(1+e4t)dt​

Now putting the value of ${I}_{2}\text{in}I$I2​inI we get,

$$I=\frac{1}{2}{\int }_{0}^{\frac{1}{2}}\frac{1}{({(\frac{\sqrt{11}}{2})}^{2}-{t}^{2})(1+{e}^{-4t})}+\frac{1}{({(\frac{\sqrt{11}}{2})}^{2}-{t}^{2})(1+{e}^{4t})}dt$$I=21​∫021​​((211​​)2−t2)(1+e−4t)1​+((211​​)2−t2)(1+e4t)1​dt

$$\Rightarrow I=\frac{1}{2}{\int }_{0}^{\frac{1}{2}}\frac{{e}^{4t}}{({(\frac{\sqrt{11}}{2})}^{2}-{t}^{2})({e}^{4t}+1)}+\frac{1}{({(\frac{\sqrt{11}}{2})}^{2}-{t}^{2})(1+{e}^{4t})}dt$$⇒I=21​∫021​​((211​​)2−t2)(e4t+1)e4t​+((211​​)2−t2)(1+e4t)1​dt

$$\Rightarrow I=\frac{1}{2}{\int }_{0}^{\frac{1}{2}}\frac{1}{{(\frac{\sqrt{11}}{2})}^{2}-{t}^{2}}dt$$⇒I=21​∫021​​(211​​)2−t21​dt

$$\Rightarrow I=\frac{1}{2}\times \frac{1}{2(\frac{\sqrt{11}}{2})}{[\mathrm{ln}|\frac{\frac{\sqrt{11}}{2}+t}{\frac{\sqrt{11}}{2}-t}|]}_{0}^{\frac{1}{2}}$$⇒I=21​×2(211​​)1​[ln∣211​​−t211​​+t​∣]021​​

$$\Rightarrow I=\frac{1}{2\sqrt{11}}\mathrm{ln}(\frac{\sqrt{11}+1}{\sqrt{11}-1})=\frac{1}{2\sqrt{11}}\mathrm{ln}(\frac{(\sqrt{11}+1{)}^{2}}{10})$$⇒I=211​1​ln(11​−111​+1​)=211​1​ln(10(11​+1)2​)

$$\Rightarrow I=\frac{1}{\sqrt{11}}\mathrm{ln}(\frac{\sqrt{11}+1}{\sqrt{10}})$$⇒I=11​1​ln(10​11​+1​)

Now on comparing with $\frac{1}{\alpha }{\mathrm{log}}_{e}(\frac{\alpha +1}{\beta })$α1​loge​(βα+1​) we get,

$$\Rightarrow \alpha =\sqrt{11},\beta =\sqrt{10}\Rightarrow {\alpha }^{4}-{\beta }^{4}=21$$⇒α=11​,β=10​⇒α4−β4=21