If $f(x) = \,\left\{ {\begin{array}{*{20}{c}} {\frac{{\sin (p + 1)x + \sin x}}{x},}&{x 0} \end{array}} \right.$ is continuous at $x = 0$ then the ordered pair $(p,q)$ is equal to

$\left( { - \frac{1}{2},\frac{3}{2}} \right)$

$\left( {\frac{5}{2},\frac{1}{2}} \right)$

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Analyzing +4 | -1

f(x) = \,\left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sin (p + 1)x + \sin x}}{x},}&{x < 0}\\ {q,}&{x = 0}\\ {\frac{{\sqrt {x + {x^2}} }}{{{x^{3/2}}}},}&{x > 0} \end{array}} \right.

is continuous at

x = 0

then the ordered pair

(p,q)

is equal to

View answer and explanation

Correct option: D

\left( { - \frac{3}{2},\frac{1}{2}} \right)

Option D

\left( { - \frac{3}{2},\frac{1}{2}} \right)

is correct because the function is continuous at x=0 only if the left-hand limit, right-hand limit, and the function's value at x=0 are equal.