Download App

Differentiability — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiability1 Mark
MCQ
If $\text{f(x)}=\text{x}^2+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)}+....+\frac{\text{x}^2}{(1+\text{x}^2)}+....,$ then at $x = 0, f(x):$
  • A
    Has not limit.
  • ✓
    Is discontinuous.
  • C
    Is continuous but not differentiable.
  • D
    Is differentiable.
✓

Answer

Correct option: B.
Is discontinuous.
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Big(\text{x}^2+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)^2}+....+\frac{\text{x}^2}{(1+\text{x}^2)^\text{n}}+....,\Big)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\text{x}^2\Big(1+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)^2}+....+\frac{\text{x}^2}{(1+\text{x}^2)^\text{n}}+....,\Big)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\text{x}^2\bigg(\frac{1}{1-\frac{1}{1+\text{x}^2}}\bigg)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}(1+\text{x}^2)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}=1$
But, $f(0)=0$
$\text{f}(0)\neq\lim\limits_{\text{x}\rightarrow0}\text{f(x)}$
Function is discontinuous.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from Differentiability→Differentiability — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

The equation $x-y$ $=4$ and $x^2+4 x y+y^2=0$ represent the sides of a/an
→
$\int\sqrt{\frac{\text{x}}{1-\text{x}}}\text{ dx}$ is equal to:
→
If $\text{x}=\text{f}(\text{t})$ and $\text{y}=\text{g}(\text{t}),$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ is equals to:
→
If $f: R \rightarrow R$ is given by $f(x)=x^3+3$, then $f^{-1}(x)$ is equal to:
→
Area bounded by the ellipse $2 x^2+3 y^2-1$ is
→
The value of c in Lagrange's mean value theorem for the function $f(x) = x(x - 2)$ when $\text{x}\in[1,2]$ is:
→
If $\vec{a}=\hat{i}+\hat{j}-2 \hat{k}, \vec{b}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}-\hat{k}$ and $\vec{c}=m \vec{a}+n \vec{b}$, then $m+n=$
→
$\int_0^\pi \frac{x d x}{4 \cos ^2 x+9 \sin ^2 x}=$
→

$
\int \frac{\sin x+\sin ^3 x}{\cos 2 x} d x=A \cos x+B \log f(x)+c \text { (where }
$
$c$ is a constant of integration). Then values of $A, B$ and $f(x)$ are

→
If $\int\frac{\cos8\text{x}+1}{\tan2\text{x}-\cot2\text{x}}\text{ dx}=\text{a}\cos8\text{x}+\text{C},$ then $a =$
→