If $f(x) = \left\{ {\begin{array}{*{20}{l}} { - x - \frac{\pi }{2},}&{x \le - \frac{\pi }{2}}\\ { - \cos x,}&{ - \frac{\pi }{2} 1} \end{array}} \right.$ then which one of the following is not correct?

$f(x)$ is continuous at $x=-\frac{\pi}{2}$

$f(x)$ is not differentiable at $x=0$

$f(x)$ is differentiable at $x=1,-\frac{3}{2}$

JEE MATHEMATICS MCQ: CONTINUITY

1 Easy

1 of 1

Analyzing +4 | -1

f(x) = \left\{ {\begin{array}{llllllllllllllllllll} { - x - \frac{\pi }{2},}&{x \le - \frac{\pi }{2}}\\ { - \cos x,}&{ - \frac{\pi }{2} < x \le 0}\\ {x - 1,}&{0 < x \le 1}\\ {\ln x}&{,x > 1} \end{array}} \right.

then which one of the following is not correct?

View answer and explanation

Correct option: D

f(x)

is discontinuous at

x=0

Option D (

f(x)

is discontinuous at

x=0

) is correct because

f(x)

is actually continuous at

x=0

, as the left-hand limit, right-hand limit, and the function value at

x=0

are all equal to -1.

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