Question
Show that $\text{f(x)}=\cos\text{x}^2$ is a continuous function.
✓
Answer
Given, $\text{f(x)}=\cos\big(\text{x}^2\big)$
This function f is defined for every real number and f can be written as the composition of two functions as
f = goh, where $\text{g(x)}=\cos\text{x}$ and h(x) = x2
$\big[\because(\text{goh})(\text{x})=\text{g(h(x))}=\text{g}(\text{x}^2)=\cos(\text{x}^2)=\text{f(x)}\big]$
It has to be first proved that $\text{g(x)}=\cos\text{x}$ and h(x) = x2 are continuous functions.
It is evident that g is defined for every real number.
Let c be a real number.
Then, $\text{g(c)}=\cos\text{c}$
Put x = c + h
If x → c, then h→ 0
$=\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x})=\lim\limits_{{\text{x}}\rightarrow\text{c}}\cos\text{x}$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos(\text{c}+\text{h})$
$=\lim\limits_{{\text{h}}\rightarrow0}\big[\cos\text{c}\cos\text{h}-\sin\text{c}\sin\text{h}\big]$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos\text{c}\cos\text{h}-\lim\limits_{{\text{h}}\rightarrow0}\sin\text{c}\sin\text{h}$
$=\cos\text{c}\cos0-\sin\text{c}\sin0$
$=\cos\text{c}\times1-\sin\text{c}\times0$
$=\cos\text{c}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$
So, $\text{g(x)}=\cos\text{x}$ is a continuous function.
Now,
h(x) = x2
Clearly, h is defined for every real number.
Let k be a real number, then h (k) = k2
$=\lim\limits_{{\text{x}}\rightarrow\text{k}}\text{h(x})=\lim\limits_{{\text{x}}\rightarrow\text{k}}\text{x}^2=\text{k}^2$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{k}}\text{h(x})=\text{h}(\text{k})$
So, h is a continuous function.
It is known that for real valued functions g and h, such that (goh) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then, (fog) is continuous at x = c.
Therefore, $\text{f(x)}=(\text{goh})(\text{x})=\cos(\text{x}^2)$ is a continuous function.
This function f is defined for every real number and f can be written as the composition of two functions as
f = goh, where $\text{g(x)}=\cos\text{x}$ and h(x) = x2
$\big[\because(\text{goh})(\text{x})=\text{g(h(x))}=\text{g}(\text{x}^2)=\cos(\text{x}^2)=\text{f(x)}\big]$
It has to be first proved that $\text{g(x)}=\cos\text{x}$ and h(x) = x2 are continuous functions.
It is evident that g is defined for every real number.
Let c be a real number.
Then, $\text{g(c)}=\cos\text{c}$
Put x = c + h
If x → c, then h→ 0
$=\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x})=\lim\limits_{{\text{x}}\rightarrow\text{c}}\cos\text{x}$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos(\text{c}+\text{h})$
$=\lim\limits_{{\text{h}}\rightarrow0}\big[\cos\text{c}\cos\text{h}-\sin\text{c}\sin\text{h}\big]$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos\text{c}\cos\text{h}-\lim\limits_{{\text{h}}\rightarrow0}\sin\text{c}\sin\text{h}$
$=\cos\text{c}\cos0-\sin\text{c}\sin0$
$=\cos\text{c}\times1-\sin\text{c}\times0$
$=\cos\text{c}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$
So, $\text{g(x)}=\cos\text{x}$ is a continuous function.
Now,
h(x) = x2
Clearly, h is defined for every real number.
Let k be a real number, then h (k) = k2
$=\lim\limits_{{\text{x}}\rightarrow\text{k}}\text{h(x})=\lim\limits_{{\text{x}}\rightarrow\text{k}}\text{x}^2=\text{k}^2$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{k}}\text{h(x})=\text{h}(\text{k})$
So, h is a continuous function.
It is known that for real valued functions g and h, such that (goh) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then, (fog) is continuous at x = c.
Therefore, $\text{f(x)}=(\text{goh})(\text{x})=\cos(\text{x}^2)$ 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.
Explore more
Similar questions
Evaluate the following integrals:
$\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}}{\sqrt[4]{\text{x}+4}+\sqrt[9]{9-\text{x}}}\text{ dx}$
→$\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}}{\sqrt[4]{\text{x}+4}+\sqrt[9]{9-\text{x}}}\text{ dx}$
If a, b, c are the langths of sides, BC, CA and AB of a triangle ABC, prove that $\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}+\overrightarrow{\text{AB}}=\vec{\text{0}}$ and deduce that $\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}.$
→The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.
There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds the she needs atleast 14kg of nitrogen and 14kg of phosphoric acid for her crop. If F1 costs Rs 6/kg and F2 costs Rs 5/kg, determine how much of each type of fertilizer should be used so that the nutrient requirements are met at minimum cost. What is the minimum cost?
For the matrices A and B, verify that (AB)T = BTAT, where $\text{A}=\begin{bmatrix}1&3\\2&4\end{bmatrix},\text{B}=\begin{bmatrix}1&4\\2&5\end{bmatrix}$
→
The relation R = {(1, 1), (2, 2), (3, 3)} on the set {1, 2, 3} is:
- Symmetric only.
- Reflexive only.
- An equivalence relation.
- Transitive only.
Differentiate w.r.t. x the function in Exercise:
$\cot^{-1}\Big[\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}}\Big],\ 0<\text{x}<\frac{\pi}{2}$
→$\cot^{-1}\Big[\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}}\Big],\ 0<\text{x}<\frac{\pi}{2}$
Find the area of region bounded by the triangle whose vertices are (-1, 1), (0, 5) and (3, 2), using integration.
Differentiate the following functions with respect to x:
$(\sin\text{x})^{\cos\text{x}}$
→$(\sin\text{x})^{\cos\text{x}}$