Download App

Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity5 Marks
Question
Prove that the function $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$ is everywhere continuous.
✓

Answer

When x < 0, we have
$​​\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$
We know that $\sin\text{x}$ as well as the identity function x are everywhere continuous.
So, the quotient function $​​\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$ is continuous at each x < 0
When x > 0, we have f(x) = x + 1, which is a polynomial function.
Therefore, f(x) is continuous at each x > 0
Now, Let us consider the point x = 0
Given, $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(-\text{h})}{-\text{h}}\Big)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(\text{h})}{\text{h}}\Big)=1$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}(\text{h}+1)=1$
Also,
$\text{f}(0)=0+1=1$
$\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\text{f}(0)$
Thus, f(x) is continuous at x = 0
Hence, f(x) is everywhere continuous.

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 "Solve the Following Question.(5 Marks)" from Continuity→Continuity — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

Let $\text{F}(\alpha)=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$ and$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}$
Show that$\big[\text{F}(\alpha)\big]^{-1}=\text{F}(-\alpha)$
→
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\text{x}}+(\sin\text{x})^\text{x}$
→
$\begin{vmatrix}-\text{a}(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^3&2\text{c}^3\\2\text{a}^3&-\text{b}(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^3\\2\text{a}^3&2\text{b}^3&-\text{c}(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{a}^2+\text{b}^2+\text{c}^2)$
→
An amount of Rs. $10,000$ is put into three investments at the rate of $10,12$ and $15 \%$ per annum. The combined income is Rs. $1310$ and the combined income of first and second investment is Rs. $190$ short of the income from the third. Find the investment in each using matrix method.
→
Find the equation of the plane passing through (a, b, c) and parallel to the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2.$
→
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cot^2\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big),$ if 0 < x < 1.
→
$\int\frac{\text{x}-1}{\sqrt{\text{x}+4}}\ \text{dx}$
→
Find the feasible solution of the following inequations graphically.3x + 4y ≥ 12, 4x + 7y ≤ 28, y ≥ 1, x ≥ 0
→
Find the reflection of the point (1, 2, -1) in the plane 3x - 5y + 4z = 5.
→
Differentiate the following functions with respect to x:
$\text{e}^\text{x}\log\sin2\text{x}$
→