If…\scriptstyle 3…\scriptstyle 2… is continuous at x = 0, then the ordered pair (p, q) is equal to

Option C. …f(x) = \left\{ {\matrix{ {{{\sin (p + 1)x + \sin x} \over x}} & {x 0} \cr

Limits Continuity and Differentiability

JEE Main 2019 · 10 Apr, Morning shift · Q4138

MathematicsHard179×since 2002Q4138

f(x) = \left\{ {\matrix{ {{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \cr q & {,x = 0} \cr {{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}}}} & {,x > 0} \cr } } \right.

is continuous at x = 0, then the ordered pair (p, q) is equal to

JEE Main 2019 · 10 Apr, Morning shift · Q4138

MathematicsHard179×since 2002Q4138

f(x) = \left\{ {\matrix{ {{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \cr q & {,x = 0} \cr {{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}}}} & {,x > 0} \cr } } \right.

is continuous at x = 0, then the ordered pair (p, q) is equal to