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

JEE MATHEMATICS MCQ: CONTINUITY

1 Easy

1 of 1

Analyzing +4 | -1

Let

f:R \to R

be defined as

f(x) = \left\{ \begin{array}{l} 2\sin \left( { - \frac{{\pi x}}{2}} \right),\quad if\;x < - 1\\ \left| {a{x^2} + x + b} \right|,\quad if - 1 \le x \le 1\\ \sin \left( {\pi x} \right),\quad \quad \;if\;x > 1 \end{array} \right.

f(x)

is continuous on

R

, then

a + b

equals

View answer and explanation

Correct option: B

-1

Option B (-1) is correct because the continuity of the function at x=1 and x=-1 leads to the equation a+b=-1.

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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