NCERT Solutions for Exercise 6.2 Class 12 Maths Chapter 10 - Application of Derivatives

Question 2: Show that the function given by $f(x)=e^{2x}$ is increasing on R.

Answer:

$\text{Let }{x}_1\text{ and }{x}_2\text{ be two numbers in }\text{R}.$
$x_1<x_2\Rightarrow 2x_1<2x_2\Rightarrow e^{2x_1}<e^{2x_2}\Rightarrow f(x_1)<f(x_2)$
Hence, the function $f(x)=e^{2x}$ is strictly increasing in R

Question 3: (a) Show that the function given by f (x) = $\sin x$ is increasing in $(0,\pi /2)$

Answer:

Given f(x) = sinx
$f^{{}^{\prime }}(x)=\cos x$

$\text{Since, }\cos x>0\text{for each }x\in (0,\frac{\pi }{2})$

$f^{{}^{\prime }}(x)>0$
Hence, f(x) = sinx is strictly increasing in $(0,\frac{\pi }{2})$

Question 3: (b) Show that the function given by f (x) = $\sin x$ is

decreasing in $(\frac{\pi }{2},\pi )$

Answer:

f(x) = sin x
$f^{{}^{\prime }}(x)=\cos x$
Since, $\cos x<0$ for each $xϵ(\frac{\pi }{2},\pi )$
So, we have $f^{{}^{\prime }}(x)<0$
Hence, f(x) = sin x is strictly decreasing in $(\frac{\pi }{2},\pi )$

Question 3: (c) Show that the function given by f (x) = $\sin x$ is neither increasing nor decreasing in $(0,\pi )$

Answer:

We know that sin x is strictly increasing in $(0,\frac{\pi }{2})$ and strictly decreasing in $(\frac{\pi }{2},\pi )$
So, by this, we can say that f(x) = sinx is neither increasing or decreasing in range $(0,\pi )$

Question 4: (a). Find the intervals in which the function f given by $f(x)=2x^2-3x$ is increasing

Answer:

$f(x)=2x^2-3x$
$f^{{}^{\prime }}(x)=4x-3$
Now,
$f^{{}^{\prime }}(x)=0$
4x - 3 = 0
$x=\frac{3}{4}$

$\text{So, the range is }(-\mathrm{\infty },\frac{3}{4})\text{and}(\frac{3}{4},\mathrm{\infty })$

So,
$f(x)<0$ when $xϵ(-\mathrm{\infty },\frac{3}{4})$ Hence, f(x) is strictly decreasing in this range
and
$f(x)>0$ when $xϵ(\frac{3}{4},\mathrm{\infty })$ Hence, f(x) is strictly increasing in this range
Hence, $f(x)=2x^2-3x$ is strictly increasing in $xϵ(\frac{3}{4},\mathrm{\infty })$

Question 4: (b) Find the intervals in which the function f given by $f(x)=2x^2-3x$ is
decreasing

Answer:

$f(x)=2x^2-3x$
$f^{{}^{\prime }}(x)=4x-3$
Now,
$f^{{}^{\prime }}(x)=0$
4x - 3 = 0
$x=\frac{3}{4}$

$\text{So, the range is }(-\mathrm{\infty },\frac{3}{4})\text{and}(\frac{3}{4},\mathrm{\infty })$

So,
$f(x)<0$ when $xϵ(-\mathrm{\infty },\frac{3}{4})$ Hence, f(x) is strictly decreasing in this range
and
$f(x)>0$ when $xϵ(\frac{3}{4},\mathrm{\infty })$ Hence, f(x) is strictly increasing in this range
Hence, $f(x)=2x^2-3x$ is strictly decreasing in $xϵ(-\mathrm{\infty },\frac{3}{4})$

Question 5:(a) Find the intervals in which the function f given by $f(x)=2x^3-3x^2-36x+7$ is
increasing

Answer:

It is given that
$f(x)=2x^3-3x^2-36x+7$
So,
$f^{{}^{\prime }}(x)=6x^2-6x-36$
$f^{{}^{\prime }}(x)=0$
$6x^2-6x-36=0\Rightarrow 6(x^2-x-6)$
$x^2-x-6=0$
$x^2-3x+2x-6=0$
$x(x-3)+2(x-3)=0$
$(x+2)(x-3)=0$
x = -2 , x = 3

$\text{So, three ranges are there }(-\mathrm{\infty },-2),(-2,3)\text{and}(3,\mathrm{\infty })$

Function $f^{{}^{\prime }}(x)=6x^2-6x-36$ is positive in interval $(-\mathrm{\infty },-2),(3,\mathrm{\infty })$ and negative in the interval (-2,3)
Hence, $f(x)=2x^3-3x^2-36x+7$ is strictly increasing in $(-\mathrm{\infty },-2)\cup (3,\mathrm{\infty })$
and strictly decreasing in the interval (-2,3)

Question 5:(b) Find the intervals in which the function f given by $f(x)=2x^3-3x^2-36x+7$ is
decreasing

Answer:

We have $f(x)=2x^3-3x^2-36x+7$

Differentiating the function with respect to x, we get :

$f^{\prime }(x)=6x^2-6x-36$

or $=6(x-3)(x+2)$

When $f^{\prime }(x)=0$ , we have :

$0=6(x-3)(x+2)$

or $(x-3)(x+2)=0$

$\text{So, three ranges are there }(-\mathrm{\infty },-2),(-2,3),\text{and }(3,\mathrm{\infty })$
Function $f^{{}^{\prime }}(x)=6x^2-6x-36$ is positive in the interval $(-\mathrm{\infty },-2),(3,\mathrm{\infty })$ and negative in the interval (-2,3)

So, f(x) is decreasing in (-2, 3)

Question 6:(a) Find the intervals in which the following functions are strictly increasing or
decreasing: $x^2+2x-5$

Answer:

f(x) = $x^2+2x-5$
$f^{{}^{\prime }}(x)=2x+2=2(x+1)$
Now,
$$\begin{gathered} f^{{}^{\prime }}(x)=0 \\ 2(x+1)=0 \\ x=-1 \end{gathered}$$

$\text{The range is from }(-\mathrm{\infty },-1)\text{ and }(-1,\mathrm{\infty })$

In interval $(-\mathrm{\infty },-1)$ $f^{{}^{\prime }}(x)=2(x+1)$ is -ve
Hence, function f(x) = $x^2+2x-5$ is strictly decreasing in interval $(-\mathrm{\infty },-1)$
In interval $(-1,\mathrm{\infty })$ $f^{{}^{\prime }}(x)=2(x+1)$ is +ve
Hence, function f(x) = $x^2+2x-5$ is strictly increasing in interval $(-1,\mathrm{\infty })$

Question 6:(b) Find the intervals in which the following functions are strictly increasing or
decreasing

$10-6x-2x^2$

Answer:

Given function is,
$f(x)=10-6x-2x^2$
$f^{{}^{\prime }}(x)=-6-4x$
Now,
$f^{{}^{\prime }}(x)=0$
$6+4x=0$
$x=-\frac{3}{2}$

$\text{So, the range is }(-\mathrm{\infty },-\frac{3}{2})\text{ and }(-\frac{3}{2},\mathrm{\infty })$
In interval $(-\mathrm{\infty },-\frac{3}{2})$ , $f^{{}^{\prime }}(x)=-6-4x$ is +ve
Hence, $f(x)=10-6x-2x^2$ is strictly increasing in the interval $(-\mathrm{\infty },-\frac{3}{2})$
In interval $(-\frac{3}{2},\mathrm{\infty })$ , $f^{{}^{\prime }}(x)=-6-4x$ is -ve
Hence, $f(x)=10-6x-2x^2$ is strictly decreasing in interval $(-\frac{3}{2},\mathrm{\infty })$

Question 6:(c) Find the intervals in which the following functions are strictly increasing or
decreasing:

$-2x^3-9x^2-12x+1$

Answer:

Given function is,
$f(x)=-2x^3-9x^2-12x+1^{}$
$f^{{}^{\prime }}(x)=-6x^2-18x-12$
Now,

$f^{{}^{\prime }}(x)=0$

$-6x^2-18x-12=0$

$-6(x^2+3x+2)=0$

$x^2+3x+2=0$

$x^2+x+2x+2=0$

$x(x+1)+2(x+1)=0$

$(x+2)(x+1)=0$

$x=-2\text{and}x=-1$

So, the range is $(-\mathrm{\infty },-2)$, $(-2,-1)$, and $(-1,\mathrm{\infty })$.
In interval $(-\mathrm{\infty },-2)\cup (-1,\mathrm{\infty })$ , $f^{{}^{\prime }}(x)=-6x^2-18x-12$ is -ve
Hence, $f(x)=-2x^3-9x^2-12x+1^{}$ is strictly decreasing in interval $(-\mathrm{\infty },-2)\cup (-1,\mathrm{\infty })$
In interval (-2,-1) , $f^{{}^{\prime }}(x)=-6x^2-18x-12$ is +ve
Hence, $f(x)=-2x^3-9x^2-12x+1^{}$ is strictly increasing in the interval (-2,-1)

Question 6:(d) Find the intervals in which the following functions are strictly increasing or
decreasing:

$6-9x-x^2$

Answer:

Given function is,
$f(x)=6-9x-x^2$
$f^{{}^{\prime }}(x)=-9-2x$
Now,
$$\begin{gathered} f^{{}^{\prime }}(x)=0 \\ -9-2x=0 \\ 2x=-9 \\ x=-\frac{9}{2} \end{gathered}$$

So, the range is $(-\mathrm{\infty },-\frac{9}{2})$ and $(-\frac{9}{2},\mathrm{\infty })$.
In interval $(-\mathrm{\infty },-\frac{9}{2})$ , $f^{{}^{\prime }}(x)=-9-2x$ is +ve
Hence, $f(x)=6-9x-x^2$ is strictly increasing in interval $(-\mathrm{\infty },-\frac{9}{2})$
In interval $(-\frac{9}{2},\mathrm{\infty })$ , $f^{{}^{\prime }}(x)=-9-2x$ is -ve
Hence, $f(x)=6-9x-x^2$ is strictly decreasing in interval $(-\frac{9}{2},\mathrm{\infty })$

Question 6:(e) Find the intervals in which the following functions are strictly increasing or
decreasing:

$(x+1{)}^3(x-3{)}^3$

Answer:

Given function is,
$f(x)=(x+1{)}^3(x-3{)}^3$
$f^{{}^{\prime }}(x)=3(x+1{)}^2(x-3{)}^3+3(x-3{)}^2(x+1{)}^3$
Now,

$f^{\prime }(x)=0$

$3(x+1{)}^2(x-3{)}^3+3(x-3{)}^2(x+1{)}^3$

$3(x+1{)}^2(x-3{)}^2((x-3)+(x+1))=0$

$(x+1)(x-3)=0\text{or}(2x-2)=0$

$x=-1\text{and}x=3\text{or}x=1$

So, the intervals are $(-\mathrm{\infty },-1)$, $(-1,1)$, $(1,3)$ and $(3,\mathrm{\infty })$.
Our function $f^{{}^{\prime }}(x)=3(x+1{)}^2(x-3{)}^3+3(x-3{)}^2(x+1{)}^3$ is +ve in the interval $(1,3)and(3,\mathrm{\infty })$

Hence, $f(x)=(x+1{)}^3(x-3{)}^3$ is strictly increasing in the interval $(1,3)$ and $(3,\mathrm{\infty })$.

Our function $f^{\prime }(x)=3(x+1{)}^2(x-3{)}^3+3(x-3{)}^2(x+1{)}^3$ is negative in the interval $(-\mathrm{\infty },-1)$ and $(-1,1)$.

Hence, $f(x)=(x+1{)}^3(x-3{)}^3$ is strictly decreasing in the interval $(-\mathrm{\infty },-1)$ and $(-1,1)$.

Question 7: Show that $y=\log (1+x)-\frac{2x}{2+x},x>-1$ is an increasing function of x throughout its domain.

Answer:

Given function is,
$f(x)\Rightarrow y=\log (1+x)-\frac{2x}{2+x}$
$f^{{}^{\prime }}(x)\Rightarrow \frac{dy}{dx}=\frac{1}{1+x}-\frac{2(2+x)-(1)(2x)}{(2+x{)}^2}=\frac{1}{1+x}-\frac{4+2x-2x}{(2+x{)}^2}$
$=\frac{1}{1+x}-\frac{4}{(2+x{)}^2}=\frac{(2+x{)}^2-4(x+1)}{(x+1)(2+x{)}^2}$
$=\frac{4+x^2+4x-4x-4}{(x+1)(2+x{)}^2}=\frac{x^2}{(x+1)(2+x{)}^2}$
$f^{{}^{\prime }}(x)=\frac{x^2}{(x+1)(x+2{)}^2}$
Now, for $x>-1$ , is is clear that $f^{{}^{\prime }}(x)=\frac{x^2}{(x+1)(x+2{)}^2}>0$
Hence, $f(x)\Rightarrow y=\log (1+x)-\frac{2x}{2+x}$ strictly increasing when $x>-1$

Question 8: Find the values of x for which $y=[x(x-2){]}^2$ is an increasing function.

Answer:

Given function is,
$f(x)\Rightarrow y=[x(x-2){]}^2$
$f^{{}^{\prime }}(x)\Rightarrow \frac{dy}{dx}=2[x(x-2)][(x-2)+x]$
$=2(x^2-2x)(2x-2)$
$=4x(x-2)(x-1)$
Now,

$f^{\prime }(x)=0$

$4x(x-2)(x-1)=0$

$x=0,x=2\text{and}x=1$

So, the intervals are $(-\mathrm{\infty },0),(0,1),(1,2)\text{and}(2,\mathrm{\infty })$

In interval $(0,1)\text{and}(2,\mathrm{\infty })$, $f^{\prime }(x)>0$
Hence, $f(x)\Rightarrow y=[x(x-2){]}^2$ is an increasing function in the interval $(0,1)\cup (2,\mathrm{\infty })$

Question 9: Prove that $y=\frac{4\sin \theta }{(2+\cos \theta )}-\theta$ is an increasing function of $\theta in[0,\pi /2]$

Answer:

Given function is,
$f(x)=y=\frac{4\sin \theta }{(2+\cos \theta )}-\theta$

$f^{{}^{\prime }}(x)=\frac{dy}{d\theta }=\frac{4\cos \theta (2+\cos \theta )-(-\sin \theta )4\sin \theta )}{(2+\cos \theta {)}^2}-1$
$=\frac{8\cos \theta +4{\cos }^2\theta +4{\sin }^2\theta -(2+\cos \theta {)}^2}{(2+\cos \theta {)}^2}$
$=\frac{8\cos \theta +4({\cos }^2\theta +{\sin }^2\theta )-4-{\cos }^2\theta -4\cos \theta }{(2+\cos \theta {)}^2}$
$=\frac{8\cos \theta +4-4-{\cos }^2\theta -4\cos \theta }{(2+\cos \theta {)}^2}$
$=\frac{4\cos \theta -{\cos }^2\theta }{(2+\cos \theta {)}^2}$
Now, for $\theta ϵ[0,\frac{\pi }{2}]$

$4\cos \theta \ge {\cos }^2\theta$

$4\cos \theta -{\cos }^2\theta \ge 0$ and $(2+\cos \theta {)}^2>0$
So, $f^{{}^{\prime }}(x)>0for\theta in[0,\frac{\pi }{2}]$
Hence, $f(x)=y=\frac{4\sin \theta }{(2+\cos \theta )}-\theta$ is increasing function in $\theta ϵ[0,\frac{\pi }{2}]$

Question 10: Prove that the logarithmic function is increasing on $(0,\mathrm{\infty })$

Answer:

Let logarithmic function is log x
$f(x)=logx$
$f^{{}^{\prime }}(x)=\frac{1}{x}$
Now, for all values of x in $(0,\mathrm{\infty })$ , $f^{{}^{\prime }}(x)>0$
Hence, the logarithmic function $f(x)=logx$ is increasing in the interval $(0,\mathrm{\infty })$

Question 11: Prove that the function f given by $f(x)=x^2-x+1$ is neither strictly increasing nor decreasing on (– 1, 1).

Answer:

Given function is,
$f(x)=x^2-x+1$
$f^{{}^{\prime }}(x)=2x-1$
Now, for interval $(-1,\frac{1}{2})$ , $f^{{}^{\prime }}(x)<0$ and for interval $(\frac{1}{2},1),f^{{}^{\prime }}(x)>0$
Hence, by this, we can say that $f(x)=x^2-x+1$ is neither strictly increasing nor decreasing in the interval (-1,1)

Question 12: Which of the following functions are decreasing on $0,\pi /2$

$$\begin{gathered} (A)\cos x \\ (B)\cos 2x \\ (C)\cos 3x \\ (D)\tan x \end{gathered}$$

Answer:

(A)
$$\begin{gathered} f(x)=\cos x \\ {f}^{{}^{\prime }}(x)=-\sin x \end{gathered}$$
$f^{{}^{\prime }}(x)<0$ for x in $(0,\frac{\pi }{2})$
Hence, $f(x)=\cos x$ is decreasing function in $(0,\frac{\pi }{2})$

(B)
$$\begin{gathered} f(x)=\cos 2x \\ {f}^{{}^{\prime }}(x)=-2\sin 2x \end{gathered}$$
Now, as
$$\begin{gathered} 0<x<\frac{\pi }{2} \\ 0<2x<\pi \end{gathered}$$
$f^{{}^{\prime }}(x)<0$ for 2x in $(0,\pi )$
Hence, $f(x)=\cos 2x$ is decreasing function in $(0,\frac{\pi }{2})$

(C)
$$\begin{gathered} f(x)=\cos 3x \\ {f}^{{}^{\prime }}(x)=-3\sin 3x \end{gathered}$$
Now, as
$$\begin{gathered} 0<x<\frac{\pi }{2} \\ 0<3x<\frac{3\pi }{2} \end{gathered}$$
$f^{{}^{\prime }}(x)<0$ for $xϵ(0,\frac{\pi }{3})$ and $f^{{}^{\prime }}(x)>0xϵ(\frac{\pi }{3},\frac{\pi }{2})$
Hence, it is clear that $f(x)=\cos 3x$ is neither increasing nor decreasing in $(0,\frac{\pi }{2})$

(D)
$$\begin{gathered} f(x)=\tan x \\ {f}^{{}^{\prime }}(x)={\sec }^{2}x \end{gathered}$$
$f^{{}^{\prime }}(x)>0$ for x in $(0,\frac{\pi }{2})$
Hence, $f(x)=\tan x$ is strictly increasing function in the interval $(0,\frac{\pi }{2})$

So, only (A) and (B) are decreasing functions in $(0,\frac{\pi }{2})$

Question 13: On which of the following intervals is the function f given by $f(x)=x^{100}+\sin x-1$ decreasing ?
(A) (0,1)

(B) $\frac{\pi }{2},\pi$

(C) $0,\frac{\pi }{2}$

(D) None of these

Answer:

(A) Given function is,
$f(x)=x^{100}+\sin x-1$
$f^{{}^{\prime }}(x)=100x^{99}+\cos x$
Now, in interval (0,1)
$f^{{}^{\prime }}(x)>0$
Hence, $f(x)=x^{100}+\sin x-1$ is increasing function in interval (0,1)

(B) Now, in interval $(\frac{\pi }{2},\pi )$

$100x^{99}>0$ but $\cos x<0$
$$\begin{gathered} 100x^{99}>\cos x \\ 100x^{99}-\cos x>0 \end{gathered}$$ , $f^{{}^{\prime }}(x)>0$
Hence, $f(x)=x^{100}+\sin x-1$ is increasing function in interval $(\frac{\pi }{2},\pi )$

(C) Now, in interval $(0,\frac{\pi }{2})$

$100x^{99}>0$ and $\cos x>0$
$$\begin{gathered} 100x^{99}>\cos x \\ 100x^{99}-\cos x>0 \end{gathered}$$ , $f^{{}^{\prime }}(x)>0$
Hence, $f(x)=x^{100}+\sin x-1$ is increasing function in interval $(0,\frac{\pi }{2})$

So, $f(x)=x^{100}+\sin x-1$ is increasing for all cases
Hence, correct answer is None of these

Question 14: For what values of a the function f given by $f(x)=x^2+ax+1$ is increasing on
[1, 2]?

Answer:

Given function is,
$f(x)=x^2+ax+1$
$f^{{}^{\prime }}(x)=2x+a$
Now, we can clearly see that for every value of $a>-2$
$f^{{}^{\prime }}(x)=2x+a$ $>0$
Hence, $f(x)=x^2+ax+1$ is increasing for every value of $a>-2$ in the interval [1,2]

Question 15: Let I be any interval disjoint from [–1, 1]. Prove that the function f given by $f(x)=x+1/x$ is increasing on I.

Answer:

Given function is,
$f(x)=x+1/x$
$f^{{}^{\prime }}(x)=1-\frac{1}{x^2}$
Now,
$$\begin{gathered} f^{{}^{\prime }}(x)=0 \\ 1-\frac{1}{x^2}=0 \\ {x}^2=1 \\ x=\pm 1 \end{gathered}$$

So, intervals are from $(-\mathrm{\infty },-1)$, $(-1,1)$ and $(1,\mathrm{\infty })$
In interval $(-\mathrm{\infty },-1),(1,\mathrm{\infty })$ , $\frac{1}{x^2}<1\Rightarrow 1-\frac{1}{x^2}>0$
$f^{{}^{\prime }}(x)>0$
Hence, $f(x)=x+1/x$ is increasing in interval $(-\mathrm{\infty },-1)\cup (1,\mathrm{\infty })$
In interval (-1,1) , $\frac{1}{x^2}>1\Rightarrow 1-\frac{1}{x^2}<0$
$f^{{}^{\prime }}(x)<0$
Hence, $f(x)=x+1/x$ is decreasing in interval (-1,1)
Hence, the function f given by $f(x)=x+1/x$ is increasing on I disjoint from [–1, 1]

Question 16 : Prove that the function f given by $f(x)=\log \sin x$ is increasing on

Answer:

Given function is,
$f(x)=\log \sin x$
$f^{{}^{\prime }}(x)=\frac{1}{\sin x}\cos x=\cot x$

Now, we know that $\cot x$ is positive in the interval $(0,\frac{\pi }{2})$ and negative in the interval $(\frac{\pi }{2},\pi )$.

$f^{\prime }(x)>0$ in $(0,\frac{\pi }{2})$ and $f^{\prime }(x)<0$ in $(\frac{\pi }{2},\pi )$.
Hence, $f(x)=\log \sin x$ is increasing in the interval $(0,\pi /2)$ and decreasing in interval $(\pi /2,\pi )$

Question 17: Prove that the function f given by f (x) = log |cos x| is decreasing on $(0,\pi /2)$
and increasing on $(3\pi /2,2\pi )$

Answer:

Given function is,
f(x) = log|cos x|
value of cos x is always +ve in both these cases
So, we can write log|cos x| = log(cos x)
Now,
$f^{{}^{\prime }}(x)=\frac{1}{\cos x}(-\sin x)=-\tan x$
We know that in interval $(0,\frac{\pi }{2})$ , $\tan x>0\Rightarrow -\tan x<0$
$f^{{}^{\prime }}(x)<0$
Hence, f(x) = log|cos x| is decreasing in interval $(0,\frac{\pi }{2})$

We know that in interval $(\frac{3\pi }{2},2\pi )$ , $\tan x<0\Rightarrow -\tan x>0$
$f^{{}^{\prime }}(x)>0$
Hence, f(x) = log|cos x| is increasing in interval $(\frac{3\pi }{2},2\pi )$

Question 18: Prove that the function given by $f(x)=x^3-3x^2+3x-100$ is increasing in R.

Answer:

Given function is,
$f(x)=x^3-3x^2+3x-100$
$f^{{}^{\prime }}(x)=3x^2-6x+3$
$=3(x^2-2x+1)=3(x-1{)}^2$
$f^{{}^{\prime }}(x)=3(x-1{)}^2$
We can clearly see that for any value of x in R $f^{{}^{\prime }}(x)>0$
Hence, $f(x)=x^3-3x^2+3x-100$ is an increasing function in R

Question 19: The interval in which $y=x^2{e}^{-x}$ is increasing is

(A) $(-\mathrm{\infty },\mathrm{\infty })$

(B) $(-2,0)$

(C) $(-2,\mathrm{\infty })$

(D) $(0,2)$

Answer:

Given function is,
$f(x)\Rightarrow y=x^2{e}^{-x}$
$f^{{}^{\prime }}(x)\Rightarrow \frac{dy}{dx}=2x{e}^{-x}+-e^{-x}(x^2)$
$x{e}^{-x}(2-x)$
$f^{{}^{\prime }}(x)=x{e}^{-x}(2-x)$
Now, it is clear that $f^{{}^{\prime }}(x)>0$ only in the interval (0,2)
So, $f(x)\Rightarrow y=x^2{e}^{-x}$ is an increasing function for the interval (0,2)
Hence, $(0,2)$ is the answer