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JEE MATHEMATICS MCQ: CONTINUITY

1 Easy

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Analyzing +4 | -1

The function

f\left( x \right) = \frac{{1 - \sin x + \cos x}}{{1 + \sin x + \cos x}}

is not defined at

x = \pi

. The value of

f\left( \pi \right)

, so that

f\left( x \right)

is continuous at

x = \pi

, is

View answer and explanation

Correct option: C

- 1

Option C (

-1

) is correct because the limit of the function as

x

approaches

\pi

is

-1

, and for the function to be continuous,

f(\pi)

must equal this limit.

More Questions from CONTINUITY

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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

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

is continuous at

x = 0

\lambda

is equal to Let

f

be a continuous function satisfying

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.

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