Let $a, b \in R, b \neq 0$. Define a function $f(x)= \begin{cases}a \sin \frac{\pi}{2}(x-1), & \text { for } x \leq 0 \\ \frac{\tan 2 x-\sin 2 x}{b x^{3}}, & \text { for } x>0\end{cases}$ If $f$ is continuous at $x=0$, then $10-a b$ is equal to ____.

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Continuity and Differentiability

Let

a, b \in R, b \neq = 0

. Define a function

f (x) = {a sin \frac{π}{2} (x - 1), \frac{t a n 2 x - s i n 2 x}{b x ^{3}}, for x \leq 0 for x > 0

f

is continuous at

x = 0

, then

10 - ab

is equal to ____.

Answer: 14

For continuity at

x = 0

, we need

lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{+}} f (x)

.

Left limit:

lim_{x \to 0^{-}} a sin \frac{π}{2} (x - 1) = a sin (- \frac{π}{2}) = - a

.

Right limit:

lim_{x \to 0^{+}} \frac{t a n 2 x - s i n 2 x}{b x ^{3}}

. Using Taylor series:

tan 2 x \approx 2 x + \frac{8 x ^{3}}{3}

and

sin 2 x \approx 2 x - \frac{4 x ^{3}}{3}

, so numerator

\approx 4 x^{3}

. Thus the limit is

\frac{4}{b}

.

Equating:

- a = \frac{4}{b}

, so

ab = - 4

.

Therefore

10 - ab = 10 - (- 4) = 14

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Integer answer