Application of Derivatives

Volume of spherical balloon $ = V = {4 \over 3}\pi {r^3}$

$ \Rightarrow 4500\pi = {{4\pi {r^3}} \over 3}$

( as Given, volume $ = 4500\pi {m^3}$ )

Differentiating both the sides, $w.r.t't'$ we get,

${{dV} \over {dt}} = 4\pi {r^2}\left( {{{dr} \over {dt}}} \right)$

Now, it is given that ${{dV} \over {dt}} = 72\pi $

$\therefore$ After $49$ min, Volume -

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 49 \times 72} \right)\pi $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 3528} \right)\pi $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 972\pi {m^3}$

$ \Rightarrow V = 972\,\,\pi {m^3}$

$\therefore$ $972\pi = {4 \over 3}\pi r{}^3$

$ \Rightarrow {r^3} = 3 \times 243 = 3 \times {3^5} = {3^6} = {\left( {{3^2}} \right)^3} \Rightarrow r = 9$

Also, we have ${{dV} \over {dt}} = 72\pi $

$\therefore$ $72\pi = 4\pi \times 9 \times 9\left( {{{dr} \over {dt}}} \right) \Rightarrow {{dr} \over {dt}} = \left( {{2 \over 9}} \right)$

$f'\left( x \right) = \sqrt x \sin x$

At local maxima or minima, $f'\left( x \right) = 0$

$ \Rightarrow x = 0$ or $sin$ $x=0$

$ \Rightarrow x = 2\pi ,\,\,\pi \in \left( {0,{{5\pi } \over 2}} \right)$

$f''\left( x \right) = \sqrt x \cos \,x + {1 \over {2\sqrt x }}\sin \,x$

$ = {1 \over {2\sqrt x }}\left( {2x\,\cos \,x + \sin \,x} \right)$

At $x = \pi ,$ $f''\left( x \right) < 0$

Hence, local maxima at $x = \pi $

At $x = 2\pi ,\,\,\,f''\left( x \right) > 0$

Hence local minima at $x = 2\pi $

Shortest distance between two curve occurred along -

the common normal

Slope of normal to ${y^2} = x$ at point

$P\left( {{t^2},t} \right)$ is $-2t$ and

slope of line $y - x = 1$ is $1.$

As they are perpendicular to each other

$\therefore$ $\left( { - 2t} \right) = - 1 \Rightarrow t = {1 \over 2}$

$\therefore$ $P\left( {{1 \over 4},{1 \over 2}} \right)$

and shortest distance $ = \left| {{{{1 \over 2} - {1 \over 4} - 1} \over {\sqrt 2 }}} \right|$



So shortest distance between them is ${{3\sqrt 2 } \over 8}$

$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}} = {{{e^x}} \over {{e^{2x}} + 2}}$

$f'\left( x \right) = {{\left( {{e^{2x}} + 2} \right)e{}^x - 2{e^{2x}}.{e^x}} \over {{{\left( {{e^{2x}} + 2} \right)}^2}}}$

$f'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}$

${e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2 $

maximum $f\left( x \right) = {{\sqrt 2 } \over 4} = {1 \over {2\sqrt 2 }}$

$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }}\,\,\forall x \in R$

Since $0 < {1 \over 3} < {1 \over {2\sqrt 2 }} \Rightarrow $ for some $c \in R$

$f\left( c \right) = {1 \over 3}$

Since tangent is parallel to $x$-axis,

$\therefore$ ${{dy} \over {dx}} = 0 \Rightarrow 1 - {8 \over {{x^3}}} = 0 \Rightarrow x = 2 \Rightarrow y = 3$

Equation of tangent is $y - 3 = 0\left( {x - 2} \right) \Rightarrow y = 3$

$f\left( x \right) = \left\{ {\matrix{ {k - 2x,\,\,\,\,if\,\,\,\,x \le - 1} \cr {2x + 3,\,\,\,\,if\,\,\,\,x > - 1} \cr } } \right.$



This is true where $k=-1$

We have $P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$

$ \Rightarrow P'\left( x \right) = 4\,{x^3} + 3a{x^2} + 2bx + c$

But $P'\left( 0 \right) = 0 \Rightarrow c = 0$

$\therefore$ $P\left( x \right) = {x^4} + a{x^3} + b{x^2} + d$

As given that $P\left( { - 1} \right) < P\left( a \right)$

$ \Rightarrow 1 - a + b + d\,\, < \,\,1 + a + b + d \Rightarrow a > 0$

Now $P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx = x\left( {4{x^2} + 3ax + 2b} \right)$

As $P'\left( x \right) = 0,$ there is only one solution $x = 0,$

therefore $4{x^2} + 3ax + 2b = 0$ should not have any real roots i.e. $D < 0$

$ \Rightarrow 9{a^2} - 32b < 0$

$ \Rightarrow b > {{9{a^2}} \over {32}} > 0$

Hence $a,b > 0 \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx > 0$

$\forall x > 0$

$\therefore$ $P(x)$ is an increasing function on $\left( {0,1} \right)$

$\therefore$ $P\left( 0 \right) < P\left( a \right)$

Similarly we can prove $P\left( x \right)$ is decreasing on $\left( { - 1,0} \right)$

$\therefore$ $P\left( { - 1} \right) > P\left( 0 \right)$

So we can conclude that

Max $P\left( x \right) = P\left( 1 \right)$ and Min $P\left( x \right) = P\left( 0 \right)$

$ \Rightarrow P\left( { - 1} \right)$ is not minimum but $P\left( 1 \right)$ is the maximum of $P.$

Let $f\left( x \right) = {x^7} + 14{x^5} + 16{x^3} + 30x - 560$

$ \Rightarrow f'\left( x \right) = 7{x^6} + 70{x^4} + 48{x^2} + 30 > 0,\,\forall x \in R$

$ \Rightarrow f$ is an increasing function on $R$

Also $\mathop {\lim }\limits_{x \to \infty } \,\,f\left( x \right) = \infty $ and $\mathop {\lim }\limits_{x \to - \infty } \,\,f\left( x \right) = - \infty $

$ \Rightarrow $ The curve $y = f\left( x \right)$ crosses $x$-axis only once.

$\therefore$ $f\left( x \right) = 0$ has exactly one real root.

Let $y = {x^3} - px + q$

$ \Rightarrow {{dy} \over {dx}} = 3{x^2} - p$

For ${{dy} \over {dx}} = 0 \Rightarrow 3{x^2} - p = 0$

$ \Rightarrow x = \pm \sqrt {{p \over 3}} $

${{{d^2}y} \over {d{x^2}}} = 6x$

${\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = \sqrt {{p \over 3}} }} = + ve\,\,\,\,$ and

$\,\,\,\,\,\,\,\,\,\,$ ${\left. {\,\,\,{{{d^2}y} \over {d{x^2}}}} \right|_{x = - \sqrt {{p \over 3}} }} = - ve$

$\therefore$ $y$ has ninima at $x = \sqrt {{p \over 3}} $

and maxima at $x = - \sqrt {{p \over 3}} $

Given $f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$

$f'\left( x \right) = {1 \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}.\left( {\cos x - \sin x} \right)$

$ = {{\sqrt 2 .\left( {{1 \over {\sqrt 2 }}\cos x - {1 \over {\sqrt 2 }}\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$

$ = {{\left( {\cos {\pi \over 4}.\cos x - \sin {\pi \over 4}.\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$

$\therefore$ $f'\left( x \right) = {{\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$

if $f'\left( x \right) > O$ then $f\left( x \right)$ is increasing function.

Hence $f(x)$ is increasing, if $ - {\pi \over 2} < x + {\pi \over 4} < {\pi \over 2}$

$ \Rightarrow - {{3\pi } \over 4} < x < {\pi \over 4}$

Hence, $f(x)$ is increasing when $n \in \left( { - {\pi \over 2},{\pi \over 4}} \right)$

Using Lagrange's Mean Value Theorem

Let $f(x)$ be a function defined on $\left[ {a,b} \right]$

then, $f'\left( c \right) = {{f\left( b \right) - f\left( a \right)} \over {b - a}}\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$

$c\,\, \in \left[ {a,b} \right]$

$\therefore$ Given $f\left( x \right) = {\log _e}x$

$\therefore$ $f'\left( x \right) = {1 \over x}$

$\therefore$ equation $(i)$ become ${1 \over c} = {{f\left( 3 \right) - f\left( 1 \right)} \over {3 - 1}}$

$ \Rightarrow {1 \over c} = {{{{\log }_e}3 - {{\log }_e}1} \over 2} = {{{{\log }_e}3} \over 2}$

$ \Rightarrow c = {2 \over {{{\log }_e}3}} \Rightarrow c = 2\,{\log _3}e$

Given that ${p^2} + {q^2} = 1$

$\therefore$ $p = \cos \theta $ and $q = \sin \theta $

Then $p+q$ $ = \cos \theta + \sin \theta $

We know that

$ - \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta \le \sqrt {{a^2} + {b^2}} $

$\therefore$ $ - \sqrt 2 \le \cos \theta + \sin \theta \le \sqrt 2 $

Hence max. value of $p + q$ is $\sqrt 2 $

$f\left( x \right) = {x \over 2} + {2 \over x} \Rightarrow f'\left( x \right) = {1 \over 2} - {2 \over {{x^2}}} = 0$

$ \Rightarrow {x^2} = 4$ or $x=2,-2;$ $\,\,\,\,\,f''\left( x \right) = {4 \over {{x^3}}}$

$f''{\left. {\left( x \right)} \right]_{x = 2}} = + ve \Rightarrow f\left( x \right)$

has local min at $x=2.$

${{dy} \over {dx}} = 2x - 5$

$\therefore$ ${m_1} = {\left( {2x - 5} \right)_{\left( {2,0} \right)}} = - 1,$

${m_2} = {\left( {2x - 5} \right)_{\left( {3,0} \right)}} = 1 \Rightarrow {m_1}{m_2} = - 1$

i.e. the tangents are perpendicular to each other.

Area $ = {1 \over 2}{x^2}\,\sin \,\theta $



Maximum value of $\sin \theta $ is $1$ at $\theta = {\pi \over 2}$

${A_{\max }} = {1 \over 2}{x^2}$

As $\,\,f\left( 1 \right) = - 2\,\,\& \;\,f'\left( x \right) \ge 2\,\forall x \in \left[ {1,6} \right]$

Applying Lagrange's mean value theorem

${{f\left( 6 \right) - f\left( 1 \right)} \over 5} = f'\left( c \right) \ge 2$

$ \Rightarrow f\left( 6 \right) \ge 10 + f\left( 1 \right)$

$ \Rightarrow f\left( 6 \right) \ge 10 - 2$

$ \Rightarrow f\left( 6 \right) \ge 8.$

Clearly function $f\left( x \right) = 3{x^2} - 2x + 1$ is increasing

when $f'\left( x \right) = 6x - 2 \ge 0 \Rightarrow \,\,\,\,\,x \in \left[ {1/3,\left. \infty \right)} \right.$

$\therefore$ $f(x)$ is incorrectly matched with $\left( { - \infty ,{1 \over 3}} \right)$

$x = a\left( {\cos \theta + \theta \sin \theta } \right)$

$ \Rightarrow {{dx} \over {d\theta }} = a\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right)$

$ \Rightarrow {{dx} \over {d\theta }} = a\theta \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$y = a\left( {\sin \theta - \theta \cos \theta } \right)$

${{dy} \over {d\theta }} = a\left[ {\cos \theta - \cos \theta + \theta \sin \theta } \right]$

$ \Rightarrow {{dy} \over {d\theta }} = a\theta \sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

From equations $(1)$ and $(2),$ we get

${{dy} \over {dx}} = \tan \theta \Rightarrow $ Slope of normal $ = - \cot \,\theta $

Equation of normal at

$'\theta '$ is $y - a\left( {\sin \theta - \theta \cos \theta } \right)$

$ = - \cot \theta \left( {x - a\left( {\cos \theta + \theta \sin \theta } \right)} \right.$

$ \Rightarrow y\sin \theta - a{\sin ^2}\theta + a\,\,\theta \cos \theta \sin \theta $

$ = - x\cos \theta + a{\cos ^2}\theta + a\,\theta \sin \theta \cos \theta $

$ \Rightarrow x\cos \theta + y\sin \theta = a$

Clearly this is an equation of straight line -

which is at a constant distance $'a'$ from origin.

Given that

${{dv} \over {dt}} = 50\,c{m^3}/\min $

$ \Rightarrow {d \over {dt}}\left( {{4 \over 3}\pi {r^3}} \right) = 50$

$ \Rightarrow 4\pi {r^2}{{dr} \over {dt}} = 50$

$ \Rightarrow {{dr} \over {dt}} = {{50} \over {4\pi {{\left( {15} \right)}^2}}} = {1 \over {18\pi }}\,\,cm/\min \,\,$

(here $r=10+5)$

Let $f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0$

The other given equation,

$na{}_n{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + .... + {a_1} = 0 = f'\left( x \right)$

Given ${a_1} \ne 0 \Rightarrow f\left( 0 \right) = 0$

Again $f(x)$ has root $\alpha ,\,\,\, \Rightarrow f\left( \alpha \right) = 0$

$\therefore$ $\,\,\,\,\,f\left( 0 \right) = f\left( \alpha \right)$

$\therefore$ By Roll's theorem $f'(x)=0$ has root be-

tween $\left( {0,\alpha } \right)$

Hence $f'(x)$ has a positive root smaller than $\alpha .$

Area of rectangle $ABCD$ $ = 2a\,\cos \,\theta $

$\left( {2b\,\sin \,\theta } \right) = 2ab\,\sin \,2\theta $



$ \Rightarrow $ Area of greatest rectangle is equal to $2ab$

When $\sin \,2\theta = 1.$

The motion of the lizard, which starts from rest and accelerates at a rate of $a = 2 \, \text{cm/s}^2$, can be described by the equation of motion :

$D_l = \frac{1}{2} a t^2$

where $D_l$ is the distance the lizard travels, $a$ is its acceleration, and $t$ is the time.

The insect, moving at a constant speed of $v = 20 \, \text{cm/s}$, has a motion that can be simply described as:

$D_i = v t$

where $D_i$ is the distance the insect travels, $v$ is its velocity, and $t$ is the time.

Since the lizard starts 21 cm behind the insect and needs to catch up, it must travel the distance the insect travels plus an additional 21 cm. Equating these two distances gives us :

$\frac{1}{2} a t^2 = v t + 21$

Substituting the given values into this equation gives :

$t^2 = \frac{(v t + 21) \cdot 2}{a}$

Substituting $a = 2 \, \text{cm/s}^2$ and $v = 20 \, \text{cm/s}$, we simplify to :

$t^2 = 20t + 21$

This simplifies to a quadratic equation :

$t^2 - 20t - 21 = 0$

Solving this quadratic equation for $t$ gives the solutions $t = 21 \, \text{s}$ and $t = -1 \, \text{s}$. Since time cannot be negative, we discard the -1 s solution. Therefore, the lizard will catch the insect after 21 seconds.

So, the correct answer is Option C : 21 s.

${{dx} \over {d\theta }} = - a\sin \theta $ and ${{dy} \over {d\theta }} = a\cos \theta $

$\therefore$ ${{dy} \over {dx}} = - \cot \theta .$

$\therefore$ The slope of the normal at $\theta $ = $ - {1 \over { - \cot \theta }}$$= \tan \theta $

$\therefore$ The equation of the normal at $\theta $ is

$y - a\sin \theta = \tan \theta \left( {x - a - a\cos \theta } \right)$

$ \Rightarrow $ $y - a\sin \theta =$ ${{\sin \theta } \over {\cos \theta }}$ $\left( {x - a - a\cos \theta } \right)$

$ \Rightarrow y\cos \theta - a\sin \theta \cos \theta $

$\,\,\,\,\,\,\,\,\,\,\,$ $ = x\sin \theta - a\sin \theta - a\sin \theta \cos \theta $

$ \Rightarrow x\sin \theta - y\cos \theta = a\sin \theta $

$ \Rightarrow y = \left( {x - a} \right)\tan \theta $

which always passes through $(a, 0)$

$f''\left( x \right) = 6\left( {x - 1} \right).$ Inegrating,

we get $f'\left( x \right) = 3{x^2} - 6x + c$

Slope at $\left( {2,1} \right) = f'\left( 2 \right) = c = 3$

$\left[ {\,\,} \right.$ As slope of tangent at $(2, 1)$ is $3$ $\left. {\,\,} \right]$

$\therefore$ $f'\left( x \right) = 3{x^2} - 6x + 3 = 3{\left( {x - 1} \right)^2}$

Inegrating again, we get

$f\left( x \right) = {\left( {x - 1} \right)^3} + D$

The curve passes through $(2,1)$

$ \Rightarrow 1 = {\left( {2 - 1} \right)^3} + D \Rightarrow D = 0$

$\therefore$ $f\left( x \right) = {\left( {x - 1} \right)^3}$

Let us define a function

$f\left( x \right) = {{ax{}^3} \over 3} + {{b{x^2}} \over 2} + cx$

Being polynomial, it is continuous and differentiable, also,

$f\left( 0 \right) = 0\,$ and $\,\,f\left( 1 \right) = {a \over 3} + {b \over 2} + c$

$ \Rightarrow f\left( 1 \right) = {{2a + 3b + 6c} \over 6} = 0$ (given)

$\therefore$ $f\left( 0 \right) = f\left( 1 \right)$

$\therefore$ $f(x)$ satisfies all conditions of Rolle's theorem

therefore $f'\left( x \right) = 0$ has a root in $\left( {0,1} \right)$

i.e. $a{x^2} + bx + c = 0$ has at lease one root in $(0, 1)$

${y^2} = 18x \Rightarrow 2y{{dy} \over {dx}} = 18 \Rightarrow {{dy} \over {dx}} = {9 \over y}$

Given ${{dy} \over {dx}} = 2 \Rightarrow {9 \over 2} = 2 \Rightarrow y = {9 \over 2}$

Puting in ${y^2} = 18x \Rightarrow x = {9 \over 8}$

$\therefore$ Required point is $\left( {{9 \over 8},{9 \over 2}} \right)$

$f(x) = y = x + {1 \over x}$ or ${{dy} \over {dx}} = 1 - {1 \over {{x^2}}}$

For max. or min, $1 - {1 \over {{x^2}}} = 0 \Rightarrow x = \pm 1$

${{{d^2}y} \over {d{x^2}}} = {2 \over {{x^3}}} $

$ {\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{x = 1}} = 2$ ($+ve$)

$\therefore$ f(x) will be minima at $x=1$.

$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1$

$f'\left( x \right) = 6{x^2} - 18ax + 12{a^2};$

$\,\,\,\,\,\,\,\,\,\,f''\left( x \right) = 12x - 18a$

For max. or min.

$6{x^2} - 18ax + 12{a^2} = 0$

$ \Rightarrow {x^2} - 3ax + 2{a^2} = 0$

$ \Rightarrow x = a$ or $x=2a.$

At $x=a$

$f''(a) = 12a - 18a = -6a < 0$

At $x=a$ maximum As f''($a$) < 0.

At $x=2a$

$f''(a) = 24a - 18a = 6a > 0$

At $x=2a$ minimum As f''(2$a$) > 0.

$\therefore$ $p=a$ and $q=2a$

As per question ${p^2} = q$

$\therefore$ ${a^2} = 2a $

$ \Rightarrow $ $a(a - 2) = 0$

$ \therefore $ $a$ = 0, 2

but $a > 0,$ therefore, $a=2.$

Distance of origin from $\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} $

$ = \sqrt {{a^2} + {b^2} - 2ab\cos \left( {t - {{at} \over b}} \right)} ;$

$ \le \sqrt {{a^2} + {b^2} + 2ab} $ $\left[ {{{\left\{ {\cos \left( {t - {{at} \over b}} \right)} \right\}}_{\min }} = - 1} \right]$

$=a+b$

$\therefore$ Maximum distance from origin $=a+b$

Let $f\left( x \right) = {{a{x^3}} \over 3} + {{b{x^2}} \over 2} + cx \Rightarrow f\left( 0 \right) = 0$ and $f(1)$

$ = {a \over 3} + {b \over 2} + c = {{2a + 3b + 6c} \over 6} = 0$

Also $f(x)$ is continuous and differentiable in $\left[ {0,1} \right]$ and

$\left[ {0,1\left[ {.\,\,} \right.} \right.$ So by Rolle's theorem $f'\left( x \right) = 0.$

i.e $\,\,a{x^2} + bx + c = 0$ has at least one root in $\left[ {0,1} \right].$