Application of Derivatives MCQ Questions & Answers in Calculus | Maths

$$y = \frac{{ax + b}}{{\left( {x - 1} \right)\left( {x - 4} \right)}} = \frac{{ax + b}}{{{x^2} - 5x + 4}}$$        has turning point at $$P\left( {2,\, - 1} \right).$$
Thus, $$P\left( {2,\, - 1} \right)$$   lies on the curve.
Therefore $$2a + b = 2.....\left( 1 \right)$$
Also, $$\frac{{dy}}{{dx}} = 0{\text{ at }}P\left( {2,\, - 1} \right)$$
Now, $$\frac{{dy}}{{dx}} = \frac{{a\left( {{x^2} - 5x + 4} \right) - \left( {2x - 5} \right)\left( {ax + b} \right)}}{{{{\left( {{x^2} - 5x + 4} \right)}^2}}}$$
At $$P\left( {2,\, - 1} \right),\,\frac{{dy}}{{dx}} = \frac{{ - 2a + 2a + b}}{4} = 0$$
$${\text{or }}b = 0\,\,{\text{or }}a = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{From equation }}\left( 1 \right)} \right]$$

Let curve intersect each other at point $$P\left( {{x_1},{y_1}} \right)$$

Since, point of intersection is on both the curves, then
$$y_1^2 = 6{x_1}\,......\left( {\text{i}} \right)\,\,\,\,\,{\text{and}}\,9x_1^2 + by_1^2 = 16\,......\left( {{\text{ii}}} \right)$$
Now, find the slope of tangent to both the curves at the point of intersection $$P\left( {{x_1},{y_1}} \right)$$
For slope of curves:
curve (i):
$${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_1},{y_1}} \right)}} = {m_1} = \frac{3}{{{y_1}}}$$
curve (ii):
$${\text{and}}\,{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_1},{y_1}} \right)}} = {m_2} = - \frac{{9{x_1}}}{{b{y_1}}}$$
Since, both the curves intersect each other at right angle then,
$$\eqalign{ & {m_1}{m_2} = - 1 \Rightarrow \frac{{27{x_1}}}{{by_{_1}^2}} = 1 \Rightarrow b = 27\frac{{{x_1}}}{{y_{_1}^2}} \cr & \therefore \,\,{\text{from}}\,{\text{equation}}\,\left( {\text{i}} \right),\,b = 27 \times \frac{1}{6} = \frac{9}{2} \cr} $$

$$\eqalign{ & {\text{The equation of tangent to the curve}}\,y = {e^x}\,{\text{at}}\,\left( {c,{e^c}} \right)\,{\text{is}} \cr & y - {e^c} = {e^c}\left( {x - c} \right)\,\,......\left( 1 \right) \cr & {\text{and equation of line joining}}\,\left( {c - 1,{e^{c - 1}}} \right)\,{\text{and}}\,\left( {c + 1,{e^{c + 1}}} \right)\,{\text{is}} \cr & y - {e^{c - 1}} = \frac{{{e^{c + 1}} - {e^{c - 1}}}}{{\left( {c + 1} \right) - \left( {c - 1} \right)}}\left[ {x - \left( {c - 1} \right)} \right] \cr & \Rightarrow y - {e^{c - 1}} = \frac{{{e^c}\left( {e - {e^{ - 1}}} \right)}}{2}\left[ {x - c + 1} \right]\,\,......\left( 2 \right) \cr & {\text{Subtracting equation }}\left( {\text{1}} \right){\text{ from }}\left( {\text{2}} \right){\text{, we get}} \cr & {e^c} - {e^{c - 1}} = {e^c}\left( {x - c} \right)\left[ {\frac{{e - {e^{ - 1}} - 2}}{2}} \right] + {e^c}\left( {\frac{{e - {e^{ - 1}}}}{2}} \right) \cr & \Rightarrow x - c = \frac{{\left[ {1 - {e^{ - 1}} - \left( {\frac{{e - {e^{ - 1}}}}{2}} \right)} \right]}}{{\frac{{e - {e^{ - 1}} - 2}}{2}}} = \frac{{2 - e - {e^{ - 1}}}}{{e - {e^{ - 1}} - 2}} \cr & = \frac{{e + {e^{ - 1}} - 2}}{{2 - \left( {e - {e^{ - 1}}} \right)}} = \frac{{\frac{{e + {e^{ - 1}}}}{2} - 1}}{{1 - \frac{{e - {e^{ - 1}}}}{2}}} = \frac{{ + ve}}{{ - ve}} = - ve \cr & \Rightarrow x - c < 0 \Rightarrow x < c \cr & \therefore {\text{The two lines meet on the left of line}}\,x = c \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{2\left( {1 + h} \right) - 2}}{h} = \mathop {\lim }\limits_{h \to 0} 2 = 2 \cr & \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 - h} \right) - f\left( 1 \right)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}\lambda + {{\left( {1 - h} \right)}^2} - 2}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}\lambda - 1 - 2h + {h^2}}}{{ - h}} = 2{\text{ if }}{\sin ^{ - 1}}\lambda = 1,\,{\text{i}}{\text{.e}}{\text{.,}}\,\lambda = \sin \,1 \cr & \therefore \,f\left( x \right)\,{\text{is differentiable at }}x = 1\,\,{\text{if }}\lambda = {\text{sin}}\,{\text{1}}.{\text{ Then}} \cr & f'\left( x \right) = 2x,\,0 < x < 1{\text{ and }}f'\left( x \right) = 2,\,x \geqslant 1 \cr & \therefore f'\left( {1 - \in } \right) = 2\left( {1 - \in } \right) > 0{\text{ and }}f'\left( {1 + \in } \right) = 2 > 0 \cr} $$
So, $$f\left( x \right)$$  does not have a maximum or a minimum at $$x=1.$$

$$\eqalign{ & f'\left( x \right) = 3{x^2} - 12x + 12 = 3{\left( {x - 2} \right)^2} \cr & \therefore \,f'\left( 2 \right) = 0\,; \cr & f'\left( {2 - \in } \right) = 3{ \in ^2} > 0\,; \cr & f'\left( {2 + \in } \right) = 3{ \in ^2} > 0 \cr} $$
Hence, $$f\left( x \right)$$  has neither a maximum nor a minimum at $$x=2.$$

Let 20 be divided in two parts such that first part $$ = x$$
$$\therefore $$  Second part $$ = 20 – x$$
Now, assume that $$P = {x^3}\left( {20 - x} \right) = 20{x^3} - {x^4}$$
Now, $$\frac{{dP}}{{dx}} = 60{x^2} - 4{x^3}\,;\,{\text{and }}\frac{{{d^2}P}}{{d{x^2}}} = 120x - 12{x^2}$$
Put $$\frac{{dP}}{{dx}} = 0$$   for maxima or minima
$$\eqalign{ & \Rightarrow \frac{{dP}}{{dx}} = 0 \cr & \Rightarrow 4{x^2}\left( {15 - x} \right) = 0 \cr & \Rightarrow x = 0,\,x = 15 \cr & \therefore \,{\left( {\frac{{{d^2}P}}{{d{x^2}}}} \right)_{x = 15}} = 120 \times 15 - 12 \times \left( {225} \right) \cr & = 1800 - 2700 \cr & = - 900 < 0 \cr} $$
$$\therefore \,P$$  is a maximum at $$x = 15.$$
$$\therefore $$  First part $$ = 15$$  and second part $$ = 20 – 15 = 5$$
Required product $$ = 15 \times 5 = 75$$

$$\eqalign{ & {\text{Let }}f\left( x \right) = x\,\ln \,x \cr & f'\left( x \right) = \frac{x}{x} + \ln \,x = 1 + \ln \,x \cr & {\text{Put }}f'\left( x \right) = 0 \Rightarrow 1 + \ln \,x \cr & \Rightarrow \ln \,x = - 1 \Rightarrow x = {e^{ - 1}} \cr & {\text{Now, }}f''\left( x \right) = \frac{1}{x} \cr & {\left. {f''\left( x \right)} \right|_{x = {e^{ - 1}}}} = \frac{1}{{{e^{ - 1}}}} = e > 0 \cr} $$
Hence, $$f\left( x \right)$$  attains minimum value at $$x = {e^{ - 1}}.$$

Since, the equation of curves are
$$y = 10 - {x^2}\,........\left( 1 \right)\,\,\,y = 2 + {x^2}\,.......\left( 2 \right)$$
Adding equation (1) and (2), we get
$$2y = 12 \Rightarrow y = 6$$
Then, from equation (1)
$$x = \pm 2$$
Differentiate equation (1) with respect to $$x$$
$$\frac{{dy}}{{dx}} = - 2x \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {2,6} \right)}} = - 4\,{\text{and}}\,{\left( {\frac{{dy}}{{dx}}} \right)_{\left( { - 2,6} \right)}} = 4$$
Differentiate equation (2) with respect to $$x$$
$$\eqalign{ & \frac{{dy}}{{dx}} = 2x \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {2,6} \right)}} = 4{\text{ and }}{\left( {\frac{{dy}}{{dx}}} \right)_{\left( { - 2,6} \right)}} = - 4 \cr & \operatorname{At} \,\left( {2,6} \right)\tan \theta = \left( {\frac{{\left( { - 4} \right) - \left( 4 \right)}}{{1 + \left( { - 4} \right) \times \left( 4 \right)}}} \right) = \frac{{ - 8}}{{15}} \cr & \operatorname{At} \,\left( { - 2,6} \right),\tan \theta = \frac{{\left( 4 \right) - \left( { - 4} \right)}}{{1 + \left( 4 \right)\left( { - 4} \right)}} = \frac{8}{{ - 15}} \cr & \therefore \,\left| {\tan \theta } \right| = \frac{8}{{15}} \cr} $$

The given curves are
$${C_1}:{y^2} = 4x\,......\left( 1 \right)\,{\text{and}}\,{C_2}:{x^2} + {y^2} - 6x + 1 = 0\,......\left( 2 \right)$$
Solving (1) and (2) we get
$$\eqalign{ & {x^2} + 4x - 6x + 1 = 0 \Rightarrow x = 1\,{\text{and}}\, \Rightarrow y = 2\,{\text{or}}\, - 2 \cr & \therefore {\text{Points of intersection of the two curves are}}\,\left( {1,2} \right)\,{\text{and}}\,\left( {1, - 2} \right) \cr & {\text{For}}\,{{\text{C}}_{\text{1}}},\frac{{dy}}{{dx}} = \frac{2}{y} \cr & \therefore {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,2} \right)}} = 1 = {m_1} \cr & {\text{and}}\,{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1, - 2} \right)}} = - 1 = {m_1}' \cr & {\text{For}}\,{{\text{C}}_{\text{2}}},\frac{{dy}}{{dx}} = \frac{{3 - x}}{y}\,\therefore {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1, - 2} \right)}} = 1 = {m_2} \cr & {\text{and}}\,{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1, - 2} \right)}} = - 1 = {m_2}' \cr & \because {m_1} = {m_2}\,{\text{and}}\,{m_1}' = {m_2}' \cr & \therefore {C_1}\,{\text{and}}\,{C_2}\,{\text{touch each other at two points}}{\text{.}} \cr} $$

$$\eqalign{ & y = {a^{1 - n}}{x^n}{\text{ or }}\frac{{dy}}{{dx}} = {a^{1 - n}}n{x^{n - 1}} \cr & {\text{Sub - normal}} = \left| {y\frac{{dy}}{{dx}}} \right| \cr & = \left| {y{a^{1 - n}}n{x^{n - 1}}} \right| \cr & = \left| {{a^{1 - n}}{x^n}{a^{1 - n}}n{x^{n - 1}}} \right| \cr & = \left| {{a^{2 - 2n}}{x^{2n - 1}}} \right| \cr & {\text{which is constant if }}2n - 1 = 0{\text{ or }}n = \frac{1}{2} \cr} $$