JEE Main 2023 — Mathematics Calculus

Given,

$$f(x)=\int \frac{dx}{(3+4{x}^{2})\sqrt{4-3{x}^{2}}}$$f(x)=∫(3+4x2)4−3x2​dx​

Put $x=\frac{1}{t},dx=-\frac{1}{{t}^{2}}dt$x=t1​,dx=−t21​dt

So, $\displaystyle f(x)=\int \frac{-dt}{{t}^{2}(3+\frac{4}{{t}^{2}})\sqrt{4-\frac{3}{{t}^{2}}}}$f(x)=∫t2(3+t24​)4−t23​​−dt​

$$\Rightarrow f(x)=\int \frac{-tdt}{(3{t}^{2}+4)\sqrt{4{t}^{2}-3}}$$⇒f(x)=∫(3t2+4)4t2−3​−tdt​

Now let, $4{t}^{2}-3={\lambda }^{2}\Rightarrow 8tdt=2\lambda d\lambda$4t2−3=λ2⇒8tdt=2λdλ

$$\Rightarrow f(x)=-\int \frac{\lambda d\lambda }{4\cdot (3(\frac{{\lambda }^{2}+3}{4})+4)\cdot \lambda }$$⇒f(x)=−∫4⋅(3(4λ2+3​)+4)⋅λλdλ​

$$\Rightarrow f(x)=-\int \frac{d\lambda }{3{\lambda }^{2}+9+16}$$⇒f(x)=−∫3λ2+9+16dλ​

$$\Rightarrow f(x)=-\int \frac{d\lambda }{3{\lambda }^{2}+25}$$⇒f(x)=−∫3λ2+25dλ​

$$\Rightarrow f(x)=-\frac{1}{3}\int \frac{d\lambda }{{\lambda }^{2}+\frac{25}{3}}$$⇒f(x)=−31​∫λ2+325​dλ​

$$\Rightarrow f(x)=-\frac{1}{3}\times \frac{\sqrt{3}}{5}{\mathrm{tan}}^{-1}(\frac{\sqrt{3}\lambda }{5})+c$$⇒f(x)=−31​×53​​tan−1(53​λ​)+c

$$\Rightarrow f(x)=-\frac{\sqrt{3}}{15}{\mathrm{tan}}^{-1}(\frac{\sqrt{3}\sqrt{(4-3{x}^{2})}}{5x})+c$$⇒f(x)=−153​​tan−1(5x3​(4−3x2)​​)+c

Now using, $f(0)=0\Rightarrow c=+\frac{\sqrt{3}\pi }{30}$f(0)=0⇒c=+303​π​

Hence, $f(1)=\frac{-\sqrt{3}}{15}{\mathrm{tan}}^{-1}(\frac{\sqrt{3}}{5})+\frac{\sqrt{3}}{15}\times \frac{\pi }{2}$f(1)=15−3​​tan−1(53​​)+153​​×2π​

$$\Rightarrow f(1)=\frac{-\sqrt{3}}{15}({\mathrm{tan}}^{-1}(\frac{\sqrt{3}}{5})-\frac{\pi }{2})$$⇒f(1)=15−3​​(tan−1(53​​)−2π​)

$$\Rightarrow f(1)=\frac{\sqrt{3}}{15}(\frac{\pi }{2}-{\mathrm{tan}}^{-1}(\frac{\sqrt{3}}{5}))$$⇒f(1)=153​​(2π​−tan−1(53​​))

$$\Rightarrow f(1)=\frac{\sqrt{3}}{15}{\mathrm{cot}}^{-1}(\frac{\sqrt{3}}{5})$$⇒f(1)=153​​cot−1(53​​)

$$\Rightarrow f(1)=\frac{\sqrt{3}}{15}{\mathrm{tan}}^{-1}(\frac{5}{\sqrt{3}})$$⇒f(1)=153​​tan−1(3​5​)

$$\Rightarrow f(1)=\frac{1}{5\sqrt{3}}{\mathrm{tan}}^{-1}(\frac{5}{\sqrt{3}})$$⇒f(1)=53​1​tan−1(3​5​)

Now comparing with given value of $f(1)$f(1) we get, \alpha =5&\beta =\sqrt{3}

$$\Rightarrow {\alpha }^{2}+{\beta }^{2}=28$$⇒α2+β2=28