Let $f:R \to R$ be a function defined as $f(x) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\sin (a + 1)x + \sin 2x}}{{2x}},}&{{\rm{if}}}&{x 0} \end{array}} \right.\,$ If $f$ is continuous at $x = 0$ then the value of $a+b$ is equal to

JEE MATHEMATICS MCQ: CONTINUITY

1 Easy

1 of 1

Analyzing +4 | -1

Let

f:R \to R

be a function defined as

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sin (a + 1)x + \sin 2x}}{{2x}},}&{\mathrm{if}}&{x < 0}\\ {b,}&{\mathrm{if}}&{x < 0}\\ {\frac{{\sqrt {x + b{x^3}} - \sqrt x }}{{b{x^{5/2}}}},}&{\mathrm{if}}&{x > 0} \end{array}} \right.\,

f

is continuous at

x = 0

then the value of

a+b

is equal to

View answer and explanation

Correct option: D

- \frac{3}{2}

The correct answer is D (

- \frac{{3}}{{2}}

) because the continuity condition at x=0 requires the left-hand limit, right-hand limit, and function value to be equal, leading to a+b = -3/2.

Practice All Chapter MCQs →

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{{\left\{ {1 + \left| {\sin x} \right|} \right\}}^{\frac{a}{{\left| {\sin x} \right|}}}}, - \frac{\pi }{6} < x < 0}\\ {b,x = 0}\\ {{e^{\frac{{\tan 2x}}{{\tan 3x}}}},0 < x < \frac{\pi }{6}} \end{array}} \right.

x = 0

f(x)=\left\{\begin{array}{rr}1+\cos ^{-1}\{\cot x\}, & x<\frac{\pi}{2} \\ \pi[x]+1, & x \geq \frac{\pi}{2}\end{array}\right.

\{ \}

f:R \to R

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\cos 3x - \cos x}}{{{x^2}}},\mathrm{for}\,\,x \ne 0}\\ {\lambda ,\mathrm{for}\,\,x = 0} \end{array}} \right.

f

x = 0

\lambda

f

f(x) f(y)=

f(x)+f(y)+f(x y)-2

x, y \in {R}

f(2)=5

\lim \limits_{{x \rightarrow 4}} f(x)

\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}\frac{x^2+3 x-10}{x^2+2 x-15} \text {, when } x \neq-5 \\ a, \quad \text { when } x=-5\end{array}\right.

x=-5

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x\sin \frac{1}{x},x \ne 0}\\ {k,x = 0} \end{array}} \right.

x = 0

k