JEE MATHEMATICS MCQ: CONTINUITY

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Analyzing +4 | -1

Let

x \neq-1

. For each positive integer

n

, let

\begin{aligned} f_n(x)= & \frac{x}{x+1}+\frac{x^2}{(x+1)\left(x^2+1\right)}+\frac{x^4}{(x+1)\left(x^2+1\right)\left(x^4+1\right)} \\ & +\cdots+\frac{x^{2^{n-1}}}{(x+1)\left(x^2+1\right) \cdots\left(x^{2 x^{2-1}}+1\right)} \end{aligned}

for each positive integer

n

. Suppose

f(x)=\lim \limits_{{n \rightarrow \infty}} f(x)

. Then

View answer and explanation

Correct option: D

f

is continuous for all

x \neq-1

Option D (

f

is continuous for all

x \neq -1

) is correct because by simplifying the expression for

f_n(x)

and taking the limit as n approaches infinity, the resulting function

f(x)

is continuous for all

x \neq -1

More Questions from CONTINUITY

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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