JEE Main 2022 — Mathematics Calculus

For a function to be continuous $f(x)$f(x) to be $x=0$x=0, we know

$$f(0)=\underset{x\rightarrow 0}{\mathrm{lim}}f(x)$$f(0)=x→0lim​f(x)

For $f(x)=\frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+qx}-9}$f(x)=3729+qx​−97p(729+x)​−3​

limit should be $\frac{0}{0}$00​ from

So, $\underset{x\rightarrow 0}{\mathrm{lim}}(\sqrt[7]{p(729+x)}-3)$x→0lim​(7p(729+x)​−3)

$$\sqrt[7]{p.729}-3=0$$7p.729​−3=0

$$\Rightarrow p\cdot 729={3}^{7}\Rightarrow p=3$$⇒p⋅729=37⇒p=3

Now, $f(0)=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\sqrt[7]{3({3}^{6}+x)}-3}{\sqrt[3]{{3}^{6}+qx}-9}$f(0)=x→0lim​336+qx​−973(36+x)​−3​

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{3[{(1+\frac{x}{{3}^{6}})}^{\frac{1}{7}}-1]}{9[{(1+\frac{qx}{{3}^{6}})}^{\frac{1}{3}}-1]}=\frac{3}{9}\times \frac{\frac{1}{7\cdot {3}^{6}}}{\frac{q}{3\cdot {3}^{6}}}$$=x→0lim​9[(1+36qx​)31​−1]3[(1+36x​)71​−1]​=93​×3⋅36q​7⋅361​​

$$\Rightarrow f(0)=\frac{1}{3}\times \frac{3}{7q}=\frac{1}{7q}$$⇒f(0)=31​×7q3​=7q1​

$$\Rightarrow 7qf(0)-1=0$$⇒7qf(0)−1=0

$\Rightarrow 7\cdot {p}^{2}\cdot qf(0)-{p}^{2}=0$⇒7⋅p2⋅qf(0)−p2=0 (multiplying the equation by ${p}^{2}$p2)

$$\Rightarrow 63qf(0)-{p}^{2}=0$$⇒63qf(0)−p2=0