Let $f: R \rightarrow R$ be a function defined as $f(x)=\left\{\begin{array}{cc}\frac{\sin (a+1)x+\sin 2x}{2x}, & \text{if } x 0\end{array}\right.$ If $f$ is continuous at $x=0$, then the value of $a+b$ is equal to :

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Continuity and Differentiability

Let

f : R \to R

be a function defined as

f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 1 ) x + s i n 2 x}{2 x}, b \frac{x + b x ^{3} - x}{b x ^{5/2}} if x < 0, if x = 0, if x > 0

f

is continuous at

x = 0

, then the value of

a + b

is equal to :

Answer: (D)

Continuity at the origin requires both one-sided limits to equal

f (0) = b

. For

x < 0

x \to 0^{-} lim \frac{sin ( a + 1 ) x + sin 2 x}{2 x} = \frac{( a + 1 ) + 2}{2} = \frac{a + 3}{2}

For

x > 0

, taking

x

out of the numerator,

\frac{x + b x ^{3} - x}{b x ^{5/2}} = \frac{x ( 1 + b x ^{2} - 1 )}{b x ^{5/2}} = \frac{1 + b x ^{2} - 1}{b x ^{2}}

and

1 + u - 1 \sim \frac{u}{2}

u \to 0

with

u = b x^{2}

, so this tends to

\frac{1}{2}

whatever the value of

b

. Equating the right-hand limit to

f (0)

gives

b = \frac{1}{2}

, and then

\frac{a + 3}{2} = \frac{1}{2}

gives

a = - 2

. Hence

a + b = - 2 + \frac{1}{2} = - \frac{3}{2}

a + b = - \frac{3}{2}

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option