Let $f:[ - 1,3] \to R$ be defined as $f(x) = \left\{ \begin{array}{l} \left| x \right| + [x], - {\rm{1}} \le x < 1\\ x + \left| x \right|,{\rm{ 1}} \le x < 2\\ x + [x],{\rm{2}} \le x < 3, \end{array} \right.$ Where, $[t]$ denotes the greatest integer less than or equal to $t.$ Then, $f$ is discontinuous at

JEE MATHEMATICS MCQ: CONTINUITY

1 Easy

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Analyzing +4 | -1

Let

f:[ - 1,3] \to R

be defined as

f(x) = \left\{ \begin{array}{l} \left| x \right| + [x], - \mathrm{1} \le x < 1\\ x + \left| x \right|,\mathrm{ 1} \le x < 2\\ x + [x],\mathrm{2} \le x < 3, \end{array} \right.

Where,

[t]

denotes the greatest integer less than or equal to

t.

Then,

f

is discontinuous at

View answer and explanation

Correct option: C

Only three points

Option C (Only three points) is correct because the function is discontinuous at x = 0, x = 1, and x = 3 due to jumps in the function's definition.

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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