JEE MATHEMATICS MCQ: CONTINUITY

1 Hard

1 of 1

Analyzing +4 | -1

Consider the following two statements:
Statement I: The function

f(x)=\frac{x}{1+|x|}

is differentiable for real

x

.
Statement II: The function

g(x)=x^2|x|

is thrice differentiable for all real

x

.
Then

View answer and explanation

Correct option: B

Statement I is true but statement II is false.

Option B (Statement I is true but statement II is false) is correct because f(x) is differentiable for all real x, while g(x) is not thrice differentiable.

More Questions from CONTINUITY

Practice All Chapter MCQs →

If the function

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.

Is continuous at

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.

Where [] denotes greatest integer and

\{ \}

denotes fractional part function, Then value of jump of discontinuity is: If

f:R \to R

is defined by

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.

and if

f

is continuous at

x = 0

, then

\lambda

is equal to Let

f

be a continuous function satisfying

f(x) f(y)=

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

for all

x, y \in {R}

and

f(2)=5

then

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

is If

\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.

is continuous at

x=-5

, then the value of 'a' will be If

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

is continuous at

x = 0

, then the value of

k