Question
Discuss the continuity of the function $\text{f(x)}=\begin{cases}2\text{x}-1,&\text{if }\text{ x}<2\\\frac{3\text{x}}{2},&\text{if }\text{ x}\geq2\end{cases}$

Answer

When x < 2, we have
f(x) = 2x - 1
We know that a polynomial function is everywhere continuous.
So, f(x) is continuous for each x < 2
When x > 2, we have
$\text{f(x)}=\frac{3\text{x}}{2}$
The functions 3x and 2 are continuous being the polynomial and constant function, respectively.
Thus, the quotient function $\frac{3\text{x}}{2}$ is continuous at each x > 2
Now,
Let us consider the point x = 2
$\text{f(x)}=\begin{cases}2\text{x}-1,&\text{if }\text{ x}<2\\\frac{3\text{x}}{2},&\text{if }\text{ x}\geq2\end{cases}$
We have
$(\text{LHL at x}= 2)=\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2(2-\text{h})-1)=4-1=3$
$(\text{LHL at x}= 2)=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\frac{3(\text{h}+2)}{2}=3$
Also,
$\text{f}(2)=\frac{3(2)}{2}=3$
$\therefore\ \lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\text{f}(2)$
Thus, f(x) is continuous at x = 2
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

Similar questions

If $\text{A}=\begin{bmatrix}0&1&0\\0&0&1\\\text{p}&\text{q}&\text{r}\end{bmatrix},$ and I is the identity matrix of order $3$, show that $A^3 = pI + qA + rA^2$​​​​​​​.
Find the area bounded by the parabola $y^2 = 4x$ and the line $y = 2x - 4:$
By using vertical strips.
The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, -4, 3). Find the equation of the plane.
A small manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry, then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for the production of each unit of A and B, and the number of man-hours the firm has available per week are as follows:
Gadget
Fondry
Machine-shop
A
B
10
6
5
4
Firm's capacity per week
1000
600
The profit on the sale of A is Rs. 30 per unit as compared with Rs. 20 per unit of B. The problem is to determine the weekly production of gadgets A and B, so that the total profit is maximized. Formulate this problem as a LPP.
Solve the following differential equation:
$\cos^2(\text{x}-2\text{y}) = 1-2\frac{\text{dy}}{\text{dx}}$
$\text{If x = }\sqrt{\text{a}^{\sin^{-1}t},}\text{ y}=\sqrt{\text{a}^{\text{cos}^{-1}}},\text{ show that }\frac{\text{dy}}{\text{dy}}=-\frac{\text{y}}{\text{x}}.$
Find the equation of line passing through the point A(0, 6, -9) and B(-3, -6, 3). If D is the foot of perpendicular drawn from a point C(7, 4, -1) on the line AB, then find the coordiantes of the point D and the equation of line CD.
Differentiate the following functions with respect to x:
$\cos^{-1}\Big(\frac{1-\text{x}^{2\text{n}}}{1+\text{x}^{2\text{n}}}\Big), <\text{x}<\infty$
Differentiate $\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$ with respect to $\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ if 0 < x < 1.
Using integration, find the area of the following region:
$\Big\{(\text{x},\text{y}):\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}\leq1\leq\frac{\text{x}}{3}+\frac{\text{y}}{2}\Big\}$