Question

The angle between the tangent lines to the graph of the function $f\left( x \right) = \int\limits_2^x {\left( {2t - 5} \right)} dt$at the point where the graph cuts the x-axis is
(A). $\dfrac{\pi }{6}$
(B). $\dfrac{\pi }{4}$
(C). $\dfrac{\pi }{3}$
(D). $\dfrac{\pi }{2}$

Answer 1

Tangent is a line which touches the curve at only one point.
Solve the given integral to find the function of the curve.
Use differentiation to find the slope of tangent to the curve at the given point.
Try to use the formula of angle between two lines in terms of their slopes to find the required angle in the given question.

Complete step-by-step answer :
Step 1: Solve the definite integral $f\left( x \right)$
Definite integral solving technique:
$\int\limits_a^b {f\left( x \right)} dx = \mathop {\left. {\left[ {F\left( x \right)} \right]} \right|}\nolimits_a^b = F\left( b \right) - F\left( a \right)$
where a and b are limits of integration, a being the lower limit and b the upper limit.

$$f\left( x \right) = \int\limits_2^x {\left( {2t - 5} \right)} dt \\\ \Rightarrow \mathop {\left. {\left( {2\dfrac{{\mathop t\nolimits^2 }}{2} - 5t} \right)} \right|}\nolimits_2^x \\\ \Rightarrow \mathop {\left. {\left( {\mathop t\nolimits^2 - 5t} \right)} \right|}\nolimits_2^x \\\$$

Putting limits into integral

$$\Rightarrow \left( {\mathop x\nolimits^2 - 5x} \right) - \left( {\mathop 2\nolimits^2 - 5 \times 2} \right) \\\ \Rightarrow \left( {\mathop x\nolimits^2 - 5x} \right) - \left( {4 - 10} \right) \\\ \Rightarrow \mathop x\nolimits^2 - 5x + 6 \\\$$

Factorize the expression: $\mathop x\nolimits^2 - 5x + 6$
$f\left( x \right) = \mathop x\nolimits^2 - 3x - 2x + 6 \\\ \Rightarrow x\left( {x - 3} \right) - 2\left( {x - 3} \right) \\\ \Rightarrow \left( {x - 3} \right)\left( {x - 2} \right) \\\$
Step 2: Find the coordinates where the curve cuts the x-axis.
Let y = f\left( x \right)$$$\mathop { = x}\nolimits^2 - 5x + 6$$ \because y = \left( {x - 3} \right)\left( {x - 2} \right)$(from step 1) Given that the curve cuts the x-axis: y coordinate on x-axis is 0$
\Rightarrow y = \left( {x - 3} \right)\left( {x - 2} \right) = 0 \\
\Rightarrow \left( {x - 3} \right) = 0{\text{ and }}\left( {x - 2} \right) = 0 \\
$Hence,$x = 3,2$Thus, the coordinates where the curve cuts the x-axis are$\left( {2,0} \right)$,$\left( {3,0} \right)$. Step 3: Differentiate the function$y = f\left( x \right)$$
\dfrac{{dy}}{{dx}} = f'\left( x \right) = \dfrac{{d\left( {\mathop x\nolimits^2 - 5x + 6} \right)}}{{dx}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = 2x - 5 \\
$Derivative of the function gives the slope of the tangent. Let slope of the tangent to the curve$y = f\left( x \right)$at the point$\left( {2,0} \right)= $$\mathop m\nolimits_1 $$ Thus, the slope of the tangent to the curve y = f\left( x \right)$at the point$\left( {2,0} \right)$is given by:

$${\text{ }}\mathop m\nolimits_1 = \mathop {\dfrac{{dy}}{{dx}}}\nolimits_{\left( {2,0} \right)} = 2x - 5 \\\ \Rightarrow \mathop m\nolimits_1 = 2\left( 2 \right) - 5 \\\ \because {\text{ }}\mathop m\nolimits_1 = - 1 \\\$$

Let slope of the tangent to the curve $y = f\left( x \right)$ at the point $\left( {3,0} \right)$= $\mathop m\nolimits_2$
Thus, the slope of the tangent to the curve $y = f\left( x \right)$ at the point $\left( {3,0} \right)$is given by:

$${\text{ }}\mathop m\nolimits_2 = \mathop {\dfrac{{dy}}{{dx}}}\nolimits_{\left( {3,0} \right)} = 2x - 5 \\\ \Rightarrow \mathop m\nolimits_2 = 2\left( 3 \right) - 5 \\\ \because {\text{ }}\mathop m\nolimits_2 = 1 \\\$$

Step 4: Find the angles between two tangents.
The tan (or tangent) of angle between two lines in terms of their slope is given by:
$\tan \theta = \left| {\dfrac{{\mathop m\nolimits_1 - \mathop m\nolimits_2 }}{{1 + \mathop m\nolimits_1 \mathop m\nolimits_2 }}} \right|$
Where $\theta$is the acute angle between two lines. $\mathop m\nolimits_1$and $\mathop m\nolimits_2$ are the respective slopes of two lines.
Using the above mentioned formula to find angles between two tangents. (we know that tangent is a straight lines)
$\because \tan \theta = \left| {\dfrac{{\left( { - 1} \right) - 1}}{{1 + \left( { - 1} \right)1}}} \right|$ (from step 3: $\mathop m\nolimits_1 = - 1$, $\mathop m\nolimits_2 = 1$)
$\Rightarrow \tan \theta = \infty \\\ \because \theta = \dfrac{\pi }{2} \\\$ ($\because \tan \dfrac{\pi }{2} = \infty$)

Final answer: The angle between tangents to the given curve is $\dfrac{\pi }{2}{\text{ or}}\;{90^ \circ }$. Thus, the correct option is (D).

Note: In similar question equation of the tangent can be asked further. Use the equation of straight line passing through a given point $\left( {\mathop x\nolimits_0 ,\mathop y\nolimits_0 } \right)$ having finite slope $m$:
$y - \mathop y\nolimits_0 = m\left( {x - \mathop x\nolimits_0 } \right)$
Another entity, normal, is associated with the tangent.
The normal line to the curve at a given point, is perpendicular to the tangent at that point.
Thus the slope of the normal line $= - \dfrac{1}{{{\text{slope of tangent}}}}$
The slope of the normal line to the curve $y = f\left( x \right)$ is given by
Slope of normal line $= - \dfrac{1}{{\dfrac{{dy}}{{dx}}}}{\text{ or }} - \dfrac{1}{{f'\left( x \right)}}$