Question

The differentiable function y = f(x) has a property that the chord joining any two points $A=\left( {{x}_{1}},f\left( {{x}_{1}} \right) \right)$ and $B=\left( {{x}_{2}},f\left( {{x}_{2}} \right) \right)$ always intersects y axis at $\left( 0,2{{x}_{1}}{{x}_{2}} \right)$ . Given that $f\left( 1 \right)=-1$ then find $\int\limits_{0}^{\dfrac{1}{2}}{f\left( x \right)=}$
$\begin{aligned} & a)\dfrac{1}{6} \\\ & b)\dfrac{1}{8} \\\ & c)\dfrac{1}{12} \\\ & d)\dfrac{1}{24} \\\ \end{aligned}$

Answer 1

Now we are given that chord joining any two points $A=\left( {{x}_{1}},f\left( {{x}_{1}} \right) \right)$ and $B=\left( {{x}_{2}},f\left( {{x}_{2}} \right) \right)$ always intersects y axis at $\left( 0,2{{x}_{1}}{{x}_{2}} \right)$ hence we will write equation of chord joining the points $A=\left( {{x}_{1}},f\left( {{x}_{1}} \right) \right)$ and $B=\left( {{x}_{2}},f\left( {{x}_{2}} \right) \right)$ . Now the chord passes through $\left( 0,2{{x}_{1}}{{x}_{2}} \right)$ . Hence we substitute x = 0 and y = $\left( 2{{x}_{1}}{{x}_{2}} \right)$ in the equation similarly we know that f(1) = - 1 . hence if we substitute ${{x}_{1}}=1\Rightarrow f\left( {{x}_{1}} \right)=-1$ hence we will get the function f(x). Now since we have f(x) we can easily find the value of $\int\limits_{0}^{\dfrac{1}{2}}{f\left( x \right)}$

Complete step by step answer:
Now we have that a chord joining any two points $A=\left( {{x}_{1}},f\left( {{x}_{1}} \right) \right)$ and $B=\left( {{x}_{2}},f\left( {{x}_{2}} \right) \right)$ always intersects y axis at $\left( 0,2{{x}_{1}}{{x}_{2}} \right)$.
First let us find the equation of chord.
We know that equation of line passing through point $\left( {{x}_{1}},{{x}_{2}} \right)$ and $\left( {{y}_{1}},{{y}_{2}} \right)$ is given by $\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}$
Hence the equation of chord passing through $A=\left( {{x}_{1}},f\left( {{x}_{1}} \right) \right)$ and $B=\left( {{x}_{2}},f\left( {{x}_{2}} \right) \right)$ is given by
$\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-f\left( {{x}_{1}} \right)}{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}$
Now for all point ${{x}_{1}},{{x}_{2}}$ we have the chord passes through $\left( 0,2{{x}_{1}}{{x}_{2}} \right)$
Now substituting x = 0 and y = $2{{x}_{1}}{{x}_{2}}$ we get.
$\dfrac{-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{2{{x}_{1}}{{x}_{2}}-f\left( {{x}_{1}} \right)}{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}$ For all point ${{x}_{1}},{{x}_{2}}$
Since the equation is true for all values of ${{x}_{1}},{{x}_{2}}$ it is also true for ${{x}_{1}}=1$
Now at ${{x}_{1}}=1$ we have $f\left( {{x}_{1}} \right)=-1$ since at x = 1 then f(x) = -1.
Hence substituting this we get

& \dfrac{-1}{{{x}_{2}}-1}=\dfrac{2{{x}_{2}}-\left( -1 \right)}{f\left( {{x}_{2}} \right)-\left( -1 \right)} \\\ & \Rightarrow \dfrac{-1}{{{x}_{2}}-1}=\dfrac{2{{x}_{2}}+1}{f\left( {{x}_{2}} \right)+1} \\\ \end{aligned}$$ Now cross multiplying the equations we get $$-f\left( {{x}_{2}} \right)-1=\left( 2{{x}_{2}}+1 \right)\left( {{x}_{2}}-1 \right)$$ Now opening the bracket on RHS we get $$\begin{aligned} & -f\left( {{x}_{2}} \right)-1=\left( 2{{x}_{2}}^{2}-2{{x}_{2}}+{{x}_{2}}-1 \right) \\\ & -f\left( {{x}_{2}} \right)=2{{x}_{2}}^{2}-{{x}_{2}} \\\ \end{aligned}$$ Multiplying by – 1 we get $f\left( {{x}_{2}} \right)=-2{{x}_{2}}^{2}+{{x}_{2}}$ Now note that ${{x}_{2}}$ here can be any element since the equation was true for all point ${{x}_{1}},{{x}_{2}}$ Hence we have $f\left( x \right)=-2{{x}^{2}}+x$ Now integrating both side from 0 to $\dfrac{1}{2}$ we get $\int\limits_{0}^{\dfrac{1}{2}}{f\left( x \right)}=-\int\limits_{0}^{\dfrac{1}{2}}{2{{x}^{2}}}+\int\limits_{0}^{\dfrac{1}{2}}{x}$ $$\begin{aligned} & \int\limits_{0}^{\dfrac{1}{2}}{f\left( x \right)}=-2{{\left[ \dfrac{{{x}^{3}}}{3} \right]}^{\dfrac{1}{2}}}_{0}+{{\left[ \dfrac{{{x}^{2}}}{2} \right]}^{\dfrac{1}{2}}}_{0} \\\ & \int\limits_{0}^{\dfrac{1}{2}}{f\left( x \right)}=-2\left[ \dfrac{{{\left( \dfrac{1}{2} \right)}^{3}}}{3}-0 \right]+\dfrac{{{\left( \dfrac{1}{2} \right)}^{2}}}{2}-0 \\\ \end{aligned}$$ Hence we have $$\int\limits_{0}^{\dfrac{1}{2}}{f\left( x \right)}=-2\left( \dfrac{1}{24} \right)+\dfrac{1}{8}$$ Taking LCM we get $$\begin{aligned} & \int\limits_{0}^{\dfrac{1}{2}}{f\left( x \right)}=-\dfrac{2}{24}+\dfrac{3}{24} \\\ & \int\limits_{0}^{\dfrac{1}{2}}{f\left( x \right)}=\dfrac{1}{24} \\\ \end{aligned}$$ **So, the correct answer is “Option D”.** **Note:** Note that we know nothing about the curve but still for any two points $\left( {{x}_{1}},{{x}_{2}} \right)$ and $\left( f\left( {{x}_{1}} \right),f\left( {{x}_{2}} \right) \right)$ we know that the line joining two points is $\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-f\left( {{x}_{1}} \right)}{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}$ . Hence we can still find the equation of chords and use the conditions. Also it is important to note that we had the chord always intersects y axis at $\left( 0,2{{x}_{1}}{{x}_{2}} \right)$ since it always intersect this is true for all point ${{x}_{1}},{{x}_{2}}$ and not just one particular point.