JEE Main 2022 — Mathematics Algebra

Given $\alpha ,\beta$α,β are the roots of the equation ${x}^{2}-(5+{3}^{\sqrt{{\mathrm{log}}_{3}5}}-{5}^{\sqrt{{\mathrm{log}}_{5}3}})x+3({3}^{{({\mathrm{log}}_{3}5)}^{\frac{1}{3}}}-{5}^{{({\mathrm{log}}_{5}3)}^{\frac{2}{3}}}-1)=0$x2−(5+3log3​5​−5log5​3​)x+3(3(log3​5)31​−5(log5​3)32​−1)=0

Now ${3}^{\sqrt{{\mathrm{log}}_{3}5}}-{5}^{\sqrt{{\mathrm{log}}_{5}3}}={3}^{\sqrt{{\mathrm{log}}_{3}5}}-{({3}^{{\mathrm{log}}_{3}5})}^{\sqrt{{\mathrm{log}}_{5}3}}$3log3​5​−5log5​3​=3log3​5​−(3log3​5)log5​3​

$$={3}^{\sqrt{{\mathrm{log}}_{3}5}}-{3}^{\sqrt{{\mathrm{log}}_{3}5}}=0$$=3log3​5​−3log3​5​=0

Also ${3}^{{({\mathrm{log}}_{3}5)}^{\frac{1}{3}}}-{5}^{{({\mathrm{log}}_{5}3)}^{\frac{2}{3}}}={5}^{{({\mathrm{log}}_{5}3)}^{\frac{2}{3}}}-{5}^{{({\mathrm{log}}_{5}3)}^{\frac{2}{3}}}$3(log3​5)31​−5(log5​3)32​=5(log5​3)32​−5(log5​3)32​

$$=0$$=0

So, given equation becomes ${x}^{2}-5x-3=0$x2−5x−3=0

i.e. $\alpha +\beta =5;\alpha \beta =-3$α+β=5;αβ=−3

Now if the roots are $\alpha +\frac{1}{\beta }$α+β1​ and $\beta +\frac{1}{\alpha }$β+α1​

i.e., \frac{\alpha \beta +1}{\beta }&\frac{\alpha \beta +1}{\alpha }

i.e., \frac{-2}{\beta }&\frac{-2}{\alpha }

Then let $\frac{-2}{\alpha }=t\Rightarrow \alpha =\frac{-2}{t}$α−2​=t⇒α=t−2​

As ${\alpha }^{2}-5\alpha -3=0$α2−5α−3=0

$$\Rightarrow {(\frac{-2}{t})}^{2}-5(\frac{-2}{t})-3=0$$⇒(t−2​)2−5(t−2​)−3=0

$$\Rightarrow \frac{4}{{t}^{2}}+\frac{10}{t}-3=0$$⇒t24​+t10​−3=0

$$\Rightarrow 3{t}^{2}-10t-4=0$$⇒3t2−10t−4=0

i.e., $3{x}^{2}-10x-4=0$3x2−10x−4=0 is the required equation.