NCERT Solutions for Exercise 6.5 Class 12 Maths Chapter 6 - Application of Derivatives

Question:1(ii) Find the maximum and minimum values, if any, of the following functions
given by

Answer:

Given function is,
add and subtract 2 in given equation

Now, for every

Hence, minimum value occurs when

Hence, the minimum value of function occurs at and the minimum value isand it is clear that there is no maximum value of

Question:1(iii) Find the maximum and minimum values, if any, of the following functions
given by

$f(x)=-(x-1{)}^2+10$

Answer:

Given function is,
$f(x)=-(x-1{)}^2+10$
$$\begin{gathered} -(x-1{)}^2\le 0 \\ -(x-1{)}^2+10\le 10 \end{gathered}$$ for every $xϵR$
Hence, maximum value occurs when
$$\begin{gathered} (x-1)=0 \\ x=1 \end{gathered}$$
Hence, maximum value of function $f(x)=-(x-1{)}^2+10$ occurs at x = 1
and the maximum value is
$f(1)=-(1-1{)}^2+10=10$

and it is clear that there is no minimum value of $f(x)=9x^2+12x+2$

Question:1(iv) Find the maximum and minimum values, if any, of the following functions
given by
$g(x)=x^3+1$

Answer:

Given function is,
$g(x)=x^3+1$
value of $x^3$ varies from $-\mathrm{\infty }<x^3<\mathrm{\infty }$
Hence, function $g(x)=x^3+1$ neither has a maximum or minimum value

Question:2(i) Find the maximum and minimum values, if any, of the following functions
given by
$f(x)=|x+2|-1$

Answer:

Given function is
$f(x)=|x+2|-1$
$$\begin{gathered} |x+2|\ge 0 \\ |x+2|-1\ge -1 \end{gathered}$$ $xϵR$
Hence, minimum value occurs when |x + 2| = 0
x = -2
Hence, minimum value occurs at x = -2
and minimum value is
$f(-2)=|-2+2|-1=-1$
It is clear that there is no maximum value of the given function $xϵR$

Question:2(ii) Find the maximum and minimum values, if any, of the following functions
given by
$g(x)=-|x+1|+3$

Answer:

Given function is
$g(x)=-|x+1|+3$
$$\begin{gathered} -|x+1|\le 0 \\ -|x+1|+3\le 3 \end{gathered}$$ $xϵR$
Hence, maximum value occurs when -|x + 1| = 0
x = -1
Hence, maximum value occurs at x = -1
and maximum value is
$g(-1)=-|-1+1|+3=3$
It is clear that there is no minimum value of the given function $xϵR$

Question:2(iii) Find the maximum and minimum values, if any, of the following functions
given by
$h(x)=\sin (2x)+5$

Answer:

Given function is
$h(x)=\sin (2x)+5$
We know that value of sin 2x varies from
$-1\le sin2x\le 1$
$$\begin{gathered} -1+5\le sin2x+5\le 1+5 \\ 4\le sin2x+5\le 6 \end{gathered}$$
Hence, the maximum value of our function $h(x)=\sin (2x)+5$ is 6 and the minimum value is 4

Question:2(iv) Find the maximum and minimum values, if any, of the following functions
given by
$f(x)=|\sin 4x+3|$

Answer:

Given function is
$f(x)=|\sin 4x+3|$
We know that value of sin 4x varies from
$-1\le sin4x\le 1$
$$\begin{gathered} -1+3\le sin4x+3\le 1+3 \\ 2\le sin4x+3\le 4 \\ 2\le |sin4x+3|\le 4 \end{gathered}$$
Hence, the maximum value of our function $f(x)=|\sin 4x+3|$ is 4 and the minimum value is 2

Question:2(v) Find the maximum and minimum values, if any, of the following functions
given by
$h(x)=x+1,xϵ(-1,1)$

Answer:

Given function is
$h(x)=x+1$
It is given that the value of $xϵ(-1,1)$
So, we can not comment about either maximum or minimum value
Hence, function $h(x)=x+1$ has neither has a maximum or minimum value

Question:3(i) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:
$f(x)=x^2$

Answer:

Given function is
$$\begin{gathered} f(x)=x^2 \\ {f}^{{}^{\prime }}(x)=2x \\ {f}^{{}^{\prime }}(x)=0\Rightarrow 2x=0\Rightarrow x=0 \end{gathered}$$
So, x = 0 is the only critical point of the given function
$f^{{}^{\prime }}(0)=0$ So we find it through the 2nd derivative test
$$\begin{gathered} f^{{}^{″}}(x)=2 \\ {f}^{{}^{″}}(0)=2 \\ {f}^{{}^{″}}(0)>0 \end{gathered}$$
Hence, by this, we can say that 0 is a point of minima
and the minimum value is
$f(0)=(0{)}^2=0$

Question:3(ii) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:
$g(x)=x^3-3x$

Answer:

Given function is
$$\begin{gathered} g(x)=x^3-3x \\ {g}^{{}^{\prime }}(x)=3x^2-3 \\ {g}^{{}^{\prime }}(x)=0\Rightarrow 3x^2-3=0\Rightarrow x=\pm 1 \end{gathered}$$
Hence, the critical points are 1 and - 1
Now, by second derivative test
$g^{{}^{″}}(x)=6x$
$g^{{}^{″}}(1)=6>0$
Hence, 1 is the point of minima and the minimum value is
$g(1)=(1{)}^3-3(1)=1-3=-2$
$g^{{}^{″}}(-1)=-6<0$
Hence, -1 is the point of maxima and the maximum value is
$g(1)=(-1{)}^3-3(-1)=-1+3=2$

Question:3(iii) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:
$h(x)=\sin x+\cos x,0<x<\frac{\pi }{2}$

Answer:

Given function is
$$\begin{gathered} h(x)=\sin x+\cos x \\ {h}^{{}^{\prime }}(x)=\cos x-\sin x \\ {h}^{{}^{\prime }}(x)=0 \\ \cos x-\sin x=0 \\ \cos x=\sin x \\ x=\frac{\pi }{4}asxϵ(0,\frac{\pi }{2}) \end{gathered}$$
Now, we use the second derivative test
$$\begin{gathered} h^{{}^{″}}(x)=-\sin x-\cos x \\ {h}^{{}^{″}}(\frac{\pi }{4})=-\sin \frac{\pi }{4}-\cos \frac{\pi }{4} \\ {h}^{{}^{″}}(\frac{\pi }{4})=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \\ {h}^{{}^{″}}(\frac{\pi }{4})=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0 \end{gathered}$$
Hence, $\frac{\pi }{4}$ is the point of maxima and the maximum value is $h\left(\frac{\pi }{4}\right)$ which is $\sqrt{2}$

Question:3(iv) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:

$f(x)=sinx-cosx$

Answer:

Given function is
$$\begin{gathered} h(x)=\sin x-\cos x \\ {h}^{{}^{\prime }}(x)=\cos x+\sin x \\ {h}^{{}^{\prime }}(x)=0 \\ \cos x+\sin x=0 \\ \cos x=-\sin x \\ x=\frac{3\pi }{4}asxϵ(0,2\pi ) \end{gathered}$$
Now, we use second derivative test
$$\begin{gathered} h^{{}^{″}}(x)=-\sin x+\cos x \\ {h}^{{}^{″}}(\frac{3\pi }{4})=-\sin \frac{3\pi }{4}+\cos \frac{3\pi }{4} \\ {h}^{{}^{″}}(3\frac{\pi }{4})=-(\frac{1}{\sqrt{2}})-\frac{1}{\sqrt{2}} \\ {h}^{{}^{″}}(\frac{\pi }{4})=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0 \end{gathered}$$
Hence, $\frac{\pi }{4}$ is the point of maxima and maximum value is $h\left(\frac{3\pi }{4}\right)$ which is $\sqrt{2}$

Question:3(v) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:

$f(x)=x^3-6x^2+9x+15$

Answer:

Givrn function is
$$\begin{gathered} f(x)=x^3-6x^2+9x+15 \\ {f}^{{}^{\prime }}(x)=3x^2-12x+9 \\ {f}^{{}^{\prime }}(x)=0 \\ 3x^2-12x+9=0 \\ 3(x^2-4x+3)=0 \\ {x}^2-4x+3=0 \\ {x}^2-x-3x+3=0 \\ x(x-1)-3(x-1)=0 \\ (x-1)(x-3)=0 \\ x=1andx=3 \end{gathered}$$
Hence 1 and 3 are critical points
Now, we use the second derivative test
$$\begin{gathered} f^{{}^{″}}(x)=6x-12 \\ {f}^{{}^{″}}(1)=6-12=-6<0 \end{gathered}$$
Hence, x = 1 is a point of maxima and the maximum value is
$f(1)=(1{)}^3-6(1{)}^2+9(1)+15=1-6+9+15=19$
$$\begin{gathered} f^{{}^{″}}(x)=6x-12 \\ {f}^{{}^{″}}(3)=18-12=6>0 \end{gathered}$$
Hence, x = 1 is a point of minima and the minimum value is
$f(3)=(3{)}^3-6(3{)}^2+9(3)+15=27-54+27+15=15$

Question:3(vi) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:

$g(x)=\frac{x}{2}+\frac{2}{x},x>0$

Answer:

Given function is
$$\begin{gathered} g(x)=\frac{x}{2}+\frac{2}{x} \\ {g}^{{}^{\prime }}(x)=\frac{1}{2}-\frac{2}{x^2} \\ {g}^{{}^{\prime }}(x)=0 \\ \frac{1}{2}-\frac{2}{x^2}=0 \\ {x}^2=4 \\ x=\pm 2 \end{gathered}$$

( but as $x>0$ we only take the positive value of x i.e. x = 2)
Hence, 2 is the only critical point
Now, we use the second derivative test
$$\begin{gathered} g^{{}^{″}}(x)=\frac{4}{x^3} \\ {g}^{{}^{″}}(2)=\frac{4}{2^3}=\frac{4}{8}=\frac{1}{2}>0 \end{gathered}$$
Hence, 2 is the point of minima and the minimum value is
$$\begin{gathered} g(x)=\frac{x}{2}+\frac{2}{x} \\ g(2)=\frac{2}{2}+\frac{2}{2}=1+1=2 \end{gathered}$$

Question:3(vii) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:

$g(x)=\frac{1}{x^2+2}$

Answer:

Gien function is
$$\begin{gathered} g(x)=\frac{1}{x^2+2} \\ {g}^{{}^{\prime }}(x)=\frac{-2x}{(x^2+2{)}^2} \\ {g}^{{}^{\prime }}(x)=0 \\ \frac{-2x}{(x^2+2{)}^2}=0 \\ x=0 \end{gathered}$$
Hence., x = 0 is only critical point
Now, we use the second derivative test
$$\begin{gathered} g^{{}^{″}}(x)=-\frac{-2(x^2+2{)}^2-(-2x)2(x^2+2)(2x)}{((x^2+2{)}^2{)}^2} \\ {g}^{{}^{″}}(0)=\frac{-2\times 4}{(2{)}^4}=\frac{-8}{16}=-\frac{1}{2}<0 \end{gathered}$$
Hence, 0 is the point of local maxima and the maximum value is
$g(0)=\frac{1}{0^2+2}=\frac{1}{2}$

Question:3(viii) Find the local maxima and local minima, if any, of the following functions. Find
also the local maximum and the local minimum values, as the case may be:

$f(x)=x\sqrt{1-x},0<x<1$

Answer:

Given function is
$f(x)=x\sqrt{1-x}$
$f^{{}^{\prime }}(x)=\sqrt{1-x}+\frac{x(-1)}{2\sqrt{1-x}}$
$$\begin{gathered} =\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}\Rightarrow \frac{2-3x}{2\sqrt{1-x}} \\ {f}^{{}^{\prime }}(x)=0 \\ \frac{2-3x}{2\sqrt{1-x}}=0 \\ 3x=2 \\ x=\frac{2}{3} \end{gathered}$$
Hence, $x=\frac{2}{3}$ is the only critical point
Now, we use the second derivative test
$f^{{}^{″}}(x)=\frac{(-1)(2\sqrt{1-x})-(2-x)(2.\frac{-1}{2\sqrt{1-x}}(-1))}{(2\sqrt{1-x}{)}^2}$
$=\frac{-2\sqrt{1-x}-\frac{2}{\sqrt{1-x}}+\frac{x}{\sqrt{1-x}}}{4(1-x)}$
$=\frac{3x}{4(1-x)\sqrt{1-x}}$
$f^{"}(\frac{2}{3})>0$
Hence, it is the point of minima and the minimum value is
$$\begin{gathered} f(x)=x\sqrt{1-x} \\ f(\frac{2}{3})=\frac{2}{3}\sqrt{1-\frac{2}{3}} \\ f(\frac{2}{3})=\frac{2}{3}\sqrt{\frac{1}{3}} \\ f(\frac{2}{3})=\frac{2}{3\sqrt{3}} \\ f(\frac{2}{3})=\frac{2\sqrt{3}}{9} \end{gathered}$$

Question:4(i) Prove that the following functions do not have maxima or minima:
$f(x)=e^x$

Answer:

Given function is
$f(x)=e^x$
$$\begin{gathered} f^{{}^{\prime }}(x)=e^x \\ {f}^{{}^{\prime }}(x)=0 \\ {e}^x=0 \end{gathered}$$
But exponential can never be 0
Hence, the function $f(x)=e^x$ does not have either maxima or minima

Question:4(ii) Prove that the following functions do not have maxima or minima:

$g(x)=\log x$

Answer:

Given function is
$g(x)=\log x$
$$\begin{gathered} g^{{}^{\prime }}(x)=\frac{1}{x} \\ {g}^{{}^{\prime }}(x)=0 \\ \frac{1}{x}=0 \end{gathered}$$
Since log x deifne for positive x i.e. $x>0$
Hence, by this, we can say that $g^{{}^{\prime }}(x)>0$ for any value of x
Therefore, there is no $cϵR$ such that $g^{{}^{\prime }}(c)=0$
Hence, the function $g(x)=\log x$ does not have either maxima or minima

Question:4(iii) Prove that the following functions do not have maxima or minima:

$h(x)=x^3+x^2+x+1$

Answer:

Given function is
$h(x)=x^3+x^2+x+1$
$$\begin{gathered} h^{{}^{\prime }}(x)=3x^2+2x+1 \\ {h}^{{}^{\prime }}(x)=0 \\ 3x^2+2x+1=0 \\ 2x^2+x^2+2x+1=0 \\ 2x^2+(x+1{)}^2=0 \end{gathered}$$
But, it is clear that there is no $cϵR$ such that $f^{{}^{\prime }}(c)=0$
Hence, the function $h(x)=x^3+x^2+x+1$ does not have either maxima or minima

Question:5(i) Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
$f(x)=x^3,xϵ[-2,2]$

Answer:

Given function is
$f(x)=x^3$
$$\begin{gathered} f^{{}^{\prime }}(x)=3x^2 \\ {f}^{{}^{\prime }}(x)=0 \\ 3x^2=0\Rightarrow x=0 \end{gathered}$$
Hence, 0 is the critical point of the function $f(x)=x^3$
Now, we need to see the value of the function $f(x)=x^3$ at x = 0 and as $xϵ[-2,2]$

we also need to check the value at end points of given range i.e. x = 2 and x = -2
$$\begin{gathered} f(0)=(0{)}^3=0 \\ f(2=(2{)}^3=8 \\ f(-2)=(-2{)}^3=-8 \end{gathered}$$
Hence, maximum value of function $f(x)=x^3$ occurs at x = 2 and value is 8
and minimum value of function $f(x)=x^3$ occurs at x = -2 and value is -8

Question:5(ii) Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:

$f(x)=\sin x+\cos x,xϵ[0,\pi ]$

Answer:

Given function is
$f(x)=\sin x+\cos x$
$$\begin{gathered} f^{{}^{\prime }}(x)=\cos x-\sin x \\ {f}^{{}^{\prime }}(x)=0 \\ \cos x-\sin x=0 \\ \cos =\sin x \\ x=\frac{\pi }{4} \end{gathered}$$ as $xϵ[0,\pi ]$
Hence, $x=\frac{\pi }{4}$ is the critical point of the function $f(x)=\sin x+\cos x$
Now, we need to check the value of function $f(x)=\sin x+\cos x$ at $x=\frac{\pi }{4}$ and at the end points of given range i.e. $x=0andx=\pi$
$f(\frac{\pi }{4})=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$
$f(0)=\sin 0+\cos 0=0+1=1$
$f(\pi )=\sin \pi +\cos \pi =0+(-1)=-1$
Hence, the absolute maximum value of function $f(x)=\sin x+\cos x$ occurs at $x=\frac{\pi }{4}$ and value is $\sqrt{2}$
and absolute minimum value of function $f(x)=\sin x+\cos x$ occurs at $x=\pi$ and value is -1

Question:5(iii) Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
$f(x)=4x-\frac{1}{2}{x}^2,xϵ[-2,\frac{9}{2}]$

Answer:

Given function is
$f(x)=4x-\frac{1}{2}{x}^2$
$$\begin{gathered} f^{{}^{\prime }}(x)=4-x \\ {f}^{{}^{\prime }}(x)=0 \\ 4-x=0 \\ x=4 \end{gathered}$$
Hence, x = 4 is the critical point of function $f(x)=4x-\frac{1}{2}{x}^2$
Now, we need to check the value of function $f(x)=4x-\frac{1}{2}{x}^2$ at x = 4 and at the end points of given range i.e. at x = -2 and x = 9/2
$f(4)=4(4)-\frac{1}{2}(4{)}^2$
$=16-\frac{1}{2}.16=16-8=8$
$f(-2)=4(-2)-\frac{1}{2}.(-2{)}^2=-8-2=-10$
$f(\frac{9}{2})=4(\frac{9}{2})-\frac{1}{2}.{\left(\frac{9}{2}\right)}^2=18-\frac{81}{8}=\frac{63}{8}$
Hence, absolute maximum value of function $f(x)=4x-\frac{1}{2}{x}^2$ occures at x = 4 and value is 8
and absolute minimum value of function $f(x)=4x-\frac{1}{2}{x}^2$ occures at x = -2 and value is -10

Question:5(iv) Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

$f(x)=(x-1{)}^2+3,xϵ[-3,1]$

Answer:

Given function is
$f(x)=(x-1{)}^2+3$
$$\begin{gathered} f^{{}^{\prime }}(x)=2(x-1) \\ {f}^{{}^{\prime }}(x)=0 \\ 2(x-1)=0 \\ x=1 \end{gathered}$$
Hence, x = 1 is the critical point of function $f(x)=(x-1{)}^2+3$
Now, we need to check the value of function $f(x)=(x-1{)}^2+3$ at x = 1 and at the end points of given range i.e. at x = -3 and x = 1
$f(1)=(1-1{)}^2+3=0^2+3=3$

$f(-3)=(-3-1{)}^2+3=(-4{)}^2+3=16+3=19$
$f(1)=(1-1{)}^2+3=0^2+3=3$
Hence, absolute maximum value of function $f(x)=(x-1{)}^2+3$ occurs at x = -3 and value is 19
and absolute minimum value of function $f(x)=(x-1{)}^2+3$ occurs at x = 1 and value is 3

Question:6 . Find the maximum profit that a company can make, if the profit function is
given by $p(x)=41-72x-18x^2$

Answer:

Profit of the company is given by the function
$p(x)=41-72x-18x^2$
$$\begin{gathered} p^{{}^{\prime }}(x)=-72-36x \\ {p}^{{}^{\prime }}(x)=0 \\ -72-36x=0 \\ x=-2 \end{gathered}$$
x = -2 is the only critical point of the function $p(x)=41-72x-18x^2$
Now, by second derivative test
$p^{{}^{″}}(x)=-36<0$
At x = -2 $p^{{}^{″}}(x)<0$
Hence, maxima of function $p(x)=41-72x-18x^2$ occurs at x = -2 and maximum value is
$p(-2)=41-72(-2)-18(-2{)}^2=41+144-72=113$
Hence, the maximum profit the company can make is 113 units

Question:7 . Find both the maximum value and the minimum value of
$3x^4-8x^3+12x^2-48x+25$ on the interval [0, 3].

Answer:

Given function is
$f(x)=3x^4-8x^3+12x^2-48x+{25}^{}$
$$\begin{gathered} f^{{}^{\prime }}(x)=12x^3-24x^2+24x-48 \\ {f}^{{}^{\prime }}(x)=0 \\ 12(x^3-2x^2+2x-4)=0 \\ {x}^3-2x^2+2x-4=0 \end{gathered}$$
Now, by hit and trial let first assume x = 2
$$\begin{gathered} (2{)}^3-2(2{)}^2+2(2)-4 \\ 8-8+4-4=0 \end{gathered}$$
Hence, x = 2 is one value
Now,
$\frac{x^3-2x^2+2x-4}{x-2}=\frac{(x^2+2)(x-2)}{(x-2)}=(x^2+2)$
$x^2=-2$ which is not possible
Hence, x = 2 is the only critical value of function $f(x)=3x^4-8x^3+12x^2-48x+{25}^{}$
Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 0 and x = 3
$=3\times 16-8\times 8+12\times 4-96+25=48-64+48-96+25=-39$

$$\begin{gathered} f(3)=3(3{)}^4-8(3{)}^3+12(3{)}^2-48(3)+25 \\ =3\times 81-8\times 27+12\times 9-144+25 \\ =243-216+108-144+25=16 \end{gathered}$$

$f(0)=3(0{)}^4-8(0{)}^3+12(0{)}^2-48(0)+25=25$
Hence, maximum value of function $f(x)=3x^4-8x^3+12x^2-48x+{25}^{}$ occurs at x = 0 and vale is 25
and minimum value of function $f(x)=3x^4-8x^3+12x^2-48x+{25}^{}$ occurs at x = 2 and value is -39

Question:8 . At what points in the interval $[0,2\pi ]$ does the function $\sin 2x$ attain its maximum value?

Answer:

Given function is
$f(x)=\sin 2x$
$$\begin{gathered} f^{{}^{\prime }}(x)=2\cos 2x \\ {f}^{{}^{\prime }}(x)=0 \\ 2\cos 2x=0 \\ asxϵ[0,2\pi ] \\ 0<x<2\pi \\ 0<2x<4\pi \\ \cos 2x=0at2x=\frac{\pi }{2},2x=\frac{3\pi }{2},2x=\frac{5\pi }{2}and2x=\frac{7\pi }{2} \end{gathered}$$
So, values of x are
$x=\frac{\pi }{4},x=\frac{3\pi }{4},x=\frac{5\pi }{4}andx=\frac{7\pi }{4}$ These are the critical points of the function $f(x)=\sin 2x$
Now, we need to find the value of the function $f(x)=\sin 2x$ at $x=\frac{\pi }{4},x=\frac{3\pi }{4},x=\frac{5\pi }{4}andx=\frac{7\pi }{4}$ and at the end points of given range i.e. at x = 0 and $x=\pi$

$$\begin{gathered} f(x)=\sin 2x \\ f(\frac{\pi }{4})=\sin 2\left(\frac{\pi }{4}\right)=\sin \frac{\pi }{2}=1 \end{gathered}$$

$$\begin{gathered} f(x)=\sin 2x \\ f(\frac{3\pi }{4})=\sin 2\left(\frac{3\pi }{4}\right)=\sin \frac{3\pi }{2}=-1 \end{gathered}$$

$$\begin{gathered} f(x)=\sin 2x \\ f(\frac{5\pi }{4})=\sin 2\left(\frac{5\pi }{4}\right)=\sin \frac{5\pi }{2}=1 \end{gathered}$$

$$\begin{gathered} f(x)=\sin 2x \\ f(\frac{7\pi }{4})=\sin 2\left(\frac{7\pi }{4}\right)=\sin \frac{7\pi }{2}=-1 \end{gathered}$$

$$\begin{gathered} f(x)=\sin 2x \\ f(\pi )=\sin 2(\pi )=\sin 2\pi =0 \end{gathered}$$

$$\begin{gathered} f(x)=\sin 2x \\ f(0)=\sin 2(0)=\sin 0=0 \end{gathered}$$

Hence, at $x=\frac{\pi }{4}andx=\frac{5\pi }{4}$ function $f(x)=\sin 2x$ attains its maximum value i.e. in 1 in the given range of $xϵ[0,2\pi ]$

Question:9 What is the maximum value of the function $\sin x+\cos x$ ?

Answer:

Given function is
$f(x)=\sin x+\cos x$
$$\begin{gathered} f^{{}^{\prime }}(x)=\cos x-\sin x \\ {f}^{{}^{\prime }}(x)=0 \\ \cos x-\sin x=0 \\ \cos =\sin x \\ x=2n\pi +\frac{\pi }{4}wherenϵI \end{gathered}$$
Hence, $x=2n\pi +\frac{\pi }{4}$ is the critical point of the function $f(x)=\sin x+\cos x$
Now, we need to check the value of the function $f(x)=\sin x+\cos x$ at $x=2n\pi +\frac{\pi }{4}$
Value is same for all cases so let assume that n = 0
Now
$f(\frac{\pi }{4})=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$

Hence, the maximum value of the function $f(x)=\sin x+\cos x$ is $\sqrt{2}$

Question:10. Find the maximum value of $2x^3-24x+107$ in the interval [1, 3]. Find the
the maximum value of the same function in [–3, –1].

Answer:

Given function is
$f(x)=2x^3-24x+107$
$$\begin{gathered} f^{{}^{\prime }}(x)=6x^2-24 \\ {f}^{{}^{\prime }}(x)=0 \\ 6(x^2-4)=0 \\ {x}^2-4=0 \\ {x}^2=4 \\ x=\pm 2 \end{gathered}$$

we neglect the value x =- 2 because $xϵ[1,3]$
Hence, x = 2 is the only critical value of function $f(x)=2x^3-24x+107$
Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 1 and x = 3
$$\begin{gathered} f(2)=2(2{)}^3-24(2)+107 \\ =2\times 8-48+107 \\ =16-48+107=75 \end{gathered}$$

$$\begin{gathered} f(3)=2(3{)}^3-24(3)+107 \\ =2\times 27-72+107 \\ =54-72+107=89 \end{gathered}$$

$$\begin{gathered} f(1)=2(1{)}^3-24(1)+107 \\ =2\times 1-24+107 \\ =2-24+107=85 \end{gathered}$$
Hence, maximum value of function $f(x)=2x^3-24x+107$ occurs at x = 3 and vale is 89 when $xϵ[1,3]$
Now, when $xϵ[-3,-1]$
we neglect the value x = 2
Hence, x = -2 is the only critical value of function $f(x)=2x^3-24x+107$
Now, we need to check the value at x = -2 and at the end points of given range i.e. x = -1 and x = -3
$$\begin{gathered} f(-1)=2(-1{)}^3-24(-1)+107 \\ =2\times (-1)+24+107 \\ =-2+24+107=129 \end{gathered}$$

$$\begin{gathered} f(-2)=2(-2{)}^3-24(-2)+107 \\ =2\times (-8)+48+107 \\ =-16+48+107=139 \end{gathered}$$

$$\begin{gathered} f(-3)=2(-3{)}^3-24(-3)+107 \\ =2\times (-27)+72+107 \\ =-54+72+107=125 \end{gathered}$$
Hence, the maximum value of function $f(x)=2x^3-24x+107$ occurs at x = -2 and vale is 139 when $xϵ[-3,-1]$