Download App

CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Determine if $\text{f(x)}=\begin{cases}\text{x}^2\sin\frac{1}{\text{x}},&\text{ x}\neq0\\0,&\text{x}=0\end{cases}$ is a continuous function?
✓

Answer

The given function f is $\text{f(x)}=\begin{cases}\text{x}^2\sin\frac{1}{\text{x}},&\text{ x}\neq0\\0,&\text{x}=0\end{cases}$
It is evident that f is defind at all points of the real line.
Let c be a real number.
Case I:
If $\text{c}\neq0,$ then $\text{f(c)}=\text{c}^2\sin\frac{1}{\text{c}}$
$\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=\Big(\lim\limits_{{\text{x}}\rightarrow\text{c}}\sin\frac{1}{\text{x}}\Big)=\text{c}^2\sin\frac{1}{\text{c}}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$
So, f is continuouse at all points $\text{x}\neq0$
Case II:
If c = 0, then f(0) = 0
$\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow0^-}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)$
It is know that $-1\leq\sin\frac{1}{\text{x}}\leq1,\text{x}\neq0$
$\Rightarrow-\text{x}^2\leq\text{x}^2\sin\frac{1}{\text{x}}\leq\text{x}^2$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}(-\text{x}^2)\leq\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)\leq\lim\limits_{{\text{x}}\rightarrow0}\text{x}^2$
$\Rightarrow0\leq\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)\leq0$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=0$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}=0$
Similarly, $\lim\limits_{{\text{x}}\rightarrow0^+}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=0$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}=\text{f}(0)=\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}$
So, f is continuous at x = 0
From the above observations, it can be continuouse that f is continuous at every point of the real line.
Thus, f is a continuous function.

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 "5 Marks Questions" from CONTINUITY AND DIFFERENTIABILITY→CONTINUITY AND DIFFERENTIABILITY — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P(3, 2, 1) from the plane 2x - y + z + 1 = 0. Also, find the image of the point in the plane.
→
using interation, find the area of the region bounded by the triangle ABC vertices A, B, C are (-1, 1), (0, 5) and (3, 2) respectivrly.
→
Show that f(x) = |x - 5| is continuous but not differentiable at x = 5.
→
$\text{If (cos x)}^{\text{y}}=\text{(cos y)}^{\text{x}},\text{find }\frac{\text{dy}}{\text{dx}}.$
→
Two bad eggs are accidently mixed up with ten good ones. Three eggs are drawn at random with replacement from this lot. Compute the mean for the number of bad eggs drawn.
→
Find the equation of the plane through the points $(2, 2, -1)$ and $(3, 4, 2)$ and parallel to the line whose direction ratios are $7, 0, 6.$
→
Form the differential equation of all the circle which pass through the origin and whose centres lies in x-axis.
→
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=2\text{x}+\text{x}^2\tan\text{x},\text{ y}(0)=1$
→
Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P(3, 2, 1) from the plane 2x - y + z + 1 = 0. Also, find the image of the point in the plane.
→
Find the value of $\tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)-\tan^{-1}\Big(\frac{\text{x-y}}{\text{x+y}}\Big)$
→