Let $f: R \rightarrow R$ be defined as $$ f(x)=\left\{\begin{array}{cc} 2 \sin \left(-\frac{\pi x}{2}\right), & \text { if } x 1 \end{array}\right. $$ If $f(x)$ is continuous on $R$, then $a+b$ equals :

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Continuity and Differentiability

Let

f : R \to R

be defined as

f (x) = ⎩ ⎨ ⎧ 2 sin (- \frac{π x}{2}), a x^{2} + x + b, sin (π x), if x < - 1 if - 1 \leq x \leq 1 if x > 1

f (x)

is continuous on

R

, then

a + b

equals :

Answer: (C)

Continuity at

x = 1

matches the middle branch value

∣ a + 1 + b ∣

with

lim_{x \to 1^{+}} sin (π x) = 0

, so

∣ a + b + 1 ∣ = 0 \Rightarrow a + b = - 1

Continuity at

x = - 1

matches

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

with

∣ a - 1 + b ∣

, and

a + b = - 1

makes

∣ a + b - 1 ∣ = ∣ - 2 ∣ = 2

, so this requirement is met by the same pair.
Hence

a + b = - 1

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option