Question

The value of $x$ at which $g(x)$ becomes zero, is

A. $3$
B. $4$
C. $5$
D. $6$

Answer 1

Hint:We need to find the value of $x$ at which $g(x)$ becomes zero. First, we will find \[g(4)=\int\limits_{0}^{4}{f(t)dt}=\int\limits_{0}^{3}{f(t)dt}+\int\limits_{3}^{4}{f(t)dt}\] .Find the equation of the line joining the points by referring the figure and using $y=mx+b$ . Then after substitution, we will get the value of \[g(4)\] . Then find $g(x)$ using $g(x)=\int\limits_{0}^{x}{f(t)dt}$ . From this, we will get an equation and solve that to get the value of $x$ by equating $g(x)=0$ .

Complete step by step answer:
We need to find the value of $x$ at which $g(x)$ becomes zero.

From the figure, $g(x)$ decreases from $x=3$ .
Let us consider $g(4)$ .
It can be written as \[g(4)=\int\limits_{0}^{4}{f(t)dt}=\int\limits_{0}^{3}{f(t)dt}+\int\limits_{3}^{4}{f(t)dt}...(a)\]
\[\int\limits_{0}^{3}{f(t)dt}\] can be written as
\[\int\limits_{0}^{3}{f(t)dt}=\int\limits_{0}^{1}{f(t)}+\int\limits_{1}^{2}{f(t)}+\int\limits_{2}^{3}{f(t)}\]
We know that the equation of a line joining two points $y=mx+b$ ,where $m$ is the slope that is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .
From figure, for the points $(0,1)$ and $(1,1)$ ,
$m=\dfrac{1-1}{1-0}=0$
Now, to get the value of $b$ , we know that $y=mx+b$ .
$\Rightarrow y=0+b=b$
Let us consider the point $(0,1)$ , then
${{b}_{1}}={{y}_{1}}=1$
Consider the point $(1,1)$ , then
${{b}_{2}}={{y}_{2}}=1$
Hence $b=1$ .
Therefore, $y=mx+b=0+1=1$
Hence \[\int\limits_{0}^{1}{f(t)}=\int\limits_{0}^{1}{1}=\left[ t \right]_{0}^{1}=1\]
From figure, for the points $(1,1)$ and $(2,3)$ ,
$m=\dfrac{3-1}{2-1}=2$
Now, to get the value of $b$ , we know that $y=mx+b$ .
That is, $y=2x+b$
Let us consider the point $(1,1)$ , then
${{b}_{1}}={{y}_{1}}-2{{x}_{1}}=1-2=-1$
Consider the point $(2,3)$ , then
${{b}_{2}}={{y}_{2}}-2{{x}_{2}}=3-4=-1$
Hence $b=-1$ .
Therefore, $y=mx+b=2x-1$
\[\int\limits_{1}^{2}{f(t)}=\int\limits_{1}^{2}{(2t-1)dt}=\left[ {{t}^{2}}-t \right]_{1}^{2}=\left[ 4-2-1+1 \right]=2\]
From figure, for the points $(2,3)$ and $(3,0)$ ,
$m=\dfrac{0-3}{3-2}=-3$
Now, to get the value of $b$ , we know that $y=mx+b$ .
That is, $y=-3x+b$
For the point $(2,3)$ , ${{b}_{1}}={{y}_{1}}+3{{x}_{1}}=3+6=9$
For the point $(3,0)$ , ${{b}_{2}}={{y}_{2}}+3{{x}_{2}}=0+9=9$
Hence $b=9$ .
Therefore, $y=mx+b=-3x+9$
\[\int\limits_{2}^{3}{f(t)}=\int\limits_{2}^{3}{(-3t+9)}=\left[ \dfrac{-3}{2}{{t}^{2}}+9t \right]_{2}^{3}=3\left( -\dfrac{9}{2}+27+\dfrac{4}{2}-6 \right)=\dfrac{3}{2}\]
Hence, \[\int\limits_{0}^{3}{f(t)dt}=\int\limits_{0}^{1}{f(t)}+\int\limits_{1}^{2}{f(t)}+\int\limits_{2}^{3}{f(t)}=1+2+\dfrac{3}{2}=\dfrac{9}{2}...(i)\]
From figure, for the points $(3,0)$ and $(4,-3)$ ,
$m=\dfrac{-3-0}{4-3}=-3$
Now, to get the value of $b$ , we know that $y=mx+b$ .
That is, $y=-3x+b$
For the point $(2,3)$ , ${{b}_{1}}={{y}_{1}}+3{{x}_{1}}=3+6=9$
For the point $(3,0)$ , ${{b}_{2}}={{y}_{2}}+3{{x}_{2}}=0+9=9$
Hence $b=9$ .
Therefore, $y=mx+b=-3x+9$
Hence, \[\int\limits_{3}^{4}{f(t)dt}=\int\limits_{3}^{4}{(-3t+9)dt}=3\left[ -\dfrac{{{t}^{2}}}{2}+3t \right]_{3}^{4}=3\left( -8+12+\dfrac{9}{2}-9 \right)=\dfrac{-3}{2}\]
Substituting these in equation $(a)$ , we get
\[g(4)=\int\limits_{0}^{4}{f(t)dt}=\dfrac{9}{2}-\dfrac{3}{2}=3\]
Now, $g(x)=\int\limits_{0}^{x}{f(t)dt}$
That is, $g(x)=\int\limits_{0}^{4}{f(t)dt}+\int\limits_{4}^{x}{f(t)dt}$ , $4\le x\le 6$
$\Rightarrow g(x)=3+\int\limits_{4}^{x}{-3dt}$
Integrating, we get
$g(x)=3-3[t]_{4}^{x}$
$\Rightarrow g(x)=3-3x+12=-3x+15$
We need to find the value at $g(x)=0$ .
Hence, $-3x+15=0$
$\Rightarrow x=5$
Thus the value at $g(x)=0$ is $x=5$ that lies in $[0,6]$.
Hence, the correct option is A.
Note:
 We need to find the equations of the line joining the corresponding points from the figure. Then only we can proceed to further steps. $y=mx+b$ and $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ are the two main equations. We will be finding the value of $x$ for $4\le x\le 6$.