Question

The angle between the tangent lines to the graph of the function $f\left(x\right)=\underset{2}{\overset{x}{\int }}(2t-5)dt$ at the point where the graph cuts the $x$-axis is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is D $\frac{\pi }{2}$

$f\left(x\right)=\underset{2}{\overset{x}{\int }}(2t-5)dt$

$={[\frac{2t^2}{2}-5t]}_2^x$

$=(x^2-5x)-(4-10)$

$=x^2-5x+6$

$f\left(x\right)=x^2-5x+6$

$y=x^2-5x+6$

Let $x^2-5x+6=0$

$\Rightarrow$ $(x-2)(x-3)=0$

$\Rightarrow$ $x=2$ and $x=3$

Slope of tangent $={\left(\frac{dy}{dx}\right)}_{A(2,0)}=2x-5$

$\Rightarrow$ ${\left(\frac{dy}{dx}\right)}_{A(2,0)}\left(m_1\right)=2\times 2-5=-1$

Slope of tangent $={\left(\frac{dy}{dx}\right)}_{B(3,0)}\left(m_2\right)=2x-5$

$\Rightarrow$ ${\left(\frac{dy}{dx}\right)}_{B(3,0)}=2\times 3-5=1$

$\Rightarrow$ $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\Rightarrow$ $\tan \theta =\left|\frac{-1-1}{1+(-1)\left(1\right)}\right|$

$\Rightarrow$ $\tan \theta =\left|\frac{1+1}{1-1}\right|$

$\Rightarrow$ $\tan \theta =\left|\frac{2}{0}\right|$

$\Rightarrow$ $\tan \theta =\infty$

$\Rightarrow$ $\theta ={\tan }^{-1}\left(\infty \right)$

$\therefore$ $\theta =\frac{\pi }{2}$