If for $\mathrm{p} \neq \mathrm{q} \neq 0$, then function $f(x)=\frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+q x}-9}$ is continuous at $x=0$, then:

(A) $7 \mathrm{pq} \mathrm{f}(0)-1=0$

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Continuity and Differentiability

If for

p \neq = q \neq = 0

, then function

f (x) = \frac{7 p ( 729 + x ) - 3}{3 729 + q x - 9}

is continuous at

x = 0

, then:

Answer: (B)

For continuity at

x = 0

, we need

f (0) = lim_{x \to 0} f (x)

. For the limit to be of

\frac{0}{0}

form:

Numerator at

x = 0

7 729 p - 3 = 0

gives

7 3^{6} p = 3

, so

p = 3

.

Now,

f (0) = lim_{x \to 0} \frac{3 [( 1 + \frac{x}{3 ^{6}} ) ^{1/7} - 1 ]}{9 [( 1 + \frac{q x}{3 ^{6}} ) ^{1/3} - 1 ]}

Using

(1 + u)^{n} \approx 1 + n u

for small

u

f (0) = \frac{3}{9} \times \frac{\frac{1}{7 \cdot 3 ^{6}}}{\frac{q}{3 \cdot 3 ^{6}}} = \frac{1}{3} \times \frac{3}{7 q} = \frac{1}{7 q}

Therefore,

7 q f (0) = 1

, which gives

7 p^{2} q f (0) - p^{2} = 0

, i.e.,

63 q f (0) - p^{2} = 0

(since

p = 3

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option