Application of Derivatives

$f(x) = {3^{{{({x^2} - 2)}^3} + 4}},\,x \in R$

$f(x) = {81.3^{{{({x^2} - 2)}^3}}}$

$f'(x) = {81.3^{{{({x^2} - 2)}^3}}}\ln 2.3({x^2} - 2)2x$

$ = (486\ln 2)\left( {{3^{{{({x^2} - 2)}^3}}}({x^2} - 2)x} \right)$

$ \Rightarrow x = 0$ is the local minima.

$f''(x) = (486\ln 2)\left( {{3^{{{({x^2} - 2)}^3}}}\,.\,({x^2} - 2)(5{x^2} - 2 + 6{x^2}\ln 3({x^2} - 2))} \right)$

$f''(x) = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \sqrt 2 $

$f''\left( {{{\sqrt 2 }^ + }} \right) > 0$

$f''\left( {{{\sqrt 2 }^ - }} \right) < 0$

$ \Rightarrow x = \sqrt 2 $ is point of inflection

$f''(x) > 0\,\forall x > \sqrt 2 $

$ \Rightarrow f'(x)$ is increasing for $x > \sqrt 2 $

$f(x) = x{e^{x(1 - x)}},\,x \in R$

$f'(x) = x{e^{x(1 - x)}}\,.\,(1 - 2x) + {e^{x(1 - x)}}$

$ = {e^{x(1 - x)}}[x - 2{x^2} + 1]$

$ = - {e^{x(1 - x)}}[2{x^2} - x - 1]$

$ = - {e^{x(1 - x)}}(2x + 1)(x - 1)$

$\therefore$ $f(x)$ is increasing in $\left( { - {1 \over 2},1} \right)$ and decreasing in $\left( { - \infty ,\, - {1 \over 2}} \right) \cup \left( {1,\infty } \right)$

$f(x) = {{5{x^2}} \over 2} + {\alpha \over {{x^5}}}\,\,\{ x > 0\} $

$f'(x) = 5x - {{5\alpha } \over {{x^6}}} = 0$

$ \Rightarrow x = {(\alpha )^{{1 \over 7}}}$

$f{(x)_{\min }} = {{5{{(\alpha )}^{{2 \over 7}}}} \over 2} + {\alpha \over {{\alpha ^{{5 \over 7}}}}} = 14$

${5 \over 2}{\alpha ^{{2 \over 7}}} + {\alpha ^{{2 \over 7}}} = 14$

${7 \over 2}{\alpha ^{{2 \over 7}}} = 14$

$\alpha = 128$

Equation of line passing through the point (50 + $\alpha$, 0) and (0, 50 + $\alpha$) is

$y - 0 = {{50 + \alpha - 0} \over {0 - (50 + \alpha )}}\left( {x - (50 + \alpha )} \right)$

$ \Rightarrow y = - 1\left( {x - (50 + \alpha )} \right)$

$ \Rightarrow y = (50 + \alpha ) - x$

$ \Rightarrow x + y = 50 + \alpha $

Let $p = x{y^4}$

$ = (50 + \alpha - y){y^4}$

$ = (50 + \alpha ){y^4} - {y^5}$

For maximum or minimum value of p,

${{dp} \over {dy}} = 0$

$ \Rightarrow 4{y^3}(50 + \alpha ) - 5{y^4} = 0$

$ \Rightarrow {y^3}[200 + 4\alpha - 5y] = 0$

$\therefore$ $y = 0$

or

$200 + 4\alpha - 5y = 0$

$ \Rightarrow y = {4 \over 5}(50 + \alpha )$

$\therefore$ $x = 50 + \alpha - y$

$ = 50 + \alpha - {4 \over 5}(50 + \alpha )$

$ = 50 + \alpha \left( {1 - {4 \over 5}} \right)$

$ = {1 \over 5}(50 + \alpha )$

$ \Rightarrow 4x = {4 \over 5}(50 + \alpha ) = y$

$ \Rightarrow y = 4x$

$\therefore$ (x, y) lies on the line y = 4x

Given,

$f(x) = 4{x^3} - 11{x^2} + 8x - 5$

$\therefore$ $f'(x) = 12{x^2} - 22x + 8$

$ = 2(6{x^2} - 11x + 4)$

$ = 2(6{x^2} - 8x - 3x + 4)$

$ = 2\left[ {2x(3x - 4) - 1(3x - 4)} \right]$

$ = 2\left[ {(3x - 4)(2x - 1)} \right]$

$f'(x)$ is positive before ${1 \over 2}$ means slope of f(x) is positive before ${1 \over 2}$ and f'(x) is negative after ${1 \over 2}$ means slope of f(x) is negative after ${1 \over 2}$.

$\therefore$ At ${1 \over 2}$, f(x) is local maximum

Also, $f'(x)$ is negative before ${4 \over 3}$ means slope of f(x) is negative before ${4 \over 3}$ and f'(x) is positive after ${4 \over 3}$ means slope of f(x) is positive after ${4 \over 3}$.

$\therefore$ At ${4 \over 3},\,f(x)$ is local minimum

From wavy curve method,

$f'(x) > 0,\,\forall x \in \left( { - \alpha ,\,{1 \over 2}} \right) \cup \left( {{4 \over 3},\,\alpha } \right)$

$\therefore$ f(x) increasing in $x \in \left( { - \alpha ,\,{1 \over 2}} \right) \cup \left( {{4 \over 3},\,\alpha } \right)$

$f'(x) < 0\,\forall x \in \left( {{1 \over 2},{4 \over 3}} \right)$

$\therefore$ f(x) decreasing in $x \in \left( {{1 \over 2},{4 \over 3}} \right)$

Given,

$f(x) = {( - 3)^{{n_1}}}{(x - 5)^{{n_2}}}$

Differentiating both side with respect to x, we get

$f'(x) = {(x - 5)^{{n_2}}}\left( {{n_1}{{(x - 3)}^{{n_1} - 1}}} \right) + {(x - 3)^{{n_1}}}\left( {{n_2}{{(x - 5)}^{{n_2} - 1}}} \right)$

$ = {(x - 3)^{{n_1}}}{(x - 5)^{{n_2}}}\left[ {{{{n_1}} \over {x - 3}} + {{{n_2}} \over {x - 5}}} \right]$

$ = {(x - 3)^{{n_1}}}{(x - 5)^{{n_2}}}\left[ {{{{n_1}(x - 5) + {n_2}(x - 3)} \over {(x - 3)(x - 5)}}} \right]$

$ = {(x - 3)^{{n_1} - 1}}{(x - 5)^{{n_2} - 1}}[{n_1}(x - 5) + {n_2}(x - 3)]$

Option A :

When n₁ = 3 and n₂ = 4 then

$f'(x) = {(x - 3)^{3 - 1}}{(x - 5)^{4 - 1}}\left[ {3(x - 5) + 4(x - 3)} \right]$

$ = {(x - 3)^2}{(x - 5)^3}(7x - 27)$

Critical point is at x = 3, 5 and ${{27} \over 7}$

When value of f'(x) is less than ${{27} \over 5}$, f'(x) is positive means slope of f(x) graph is positive.

And when value of f'(x) is greater than ${{27} \over 7}$ but less than 5 then f'(x) is negative means slope of f(x) is negative.

So at ${{27} \over 7}$ between 3 and 5, f(x) attain local maxima.

$\therefore$ Option A is correct.

Option B :

When n₁ = 4 and n₂ = 3 then

$f'(x) = {(x - 3)^{4 - 1}}{(x - 5)^{3 - 1}}\left[ {4(x - 5) + 3(x - 3)} \right]$

$ = {(x - 3)^3}{(x - 5)^2}(7x - 29)$

Critical point is at x = 3, 5 and ${{29} \over 7}$

When f'(x) is less than ${{29} \over 7}$ but greater than 3 then f'(x) is negative means slope of graph of f(x) is negative.

And when f'(x) is greater than ${{29} \over 7}$ but less than 5 then f'(x) is positive means slope of graph of f(x) is positive.

So at ${{29} \over 7}$ between 3 and 5, f(x) attain local minima.

$\therefore$ Option (B) is correct.

Option C :

When n₁ = 3 and n₂ = 5 then

$f'(x) = {(x - 3)^{3 - 1}}{(x - 5)^{5 - 1}}\left[ {3(x - 5) + 5(x - 3)} \right]$

$ = {(x - 3)^2}{(x - 5)^4}(8x - 30)$

Critical point is at x = 3, 5 and ${{30} \over 8}$

When f'(x) is less than ${{30} \over 8}$ but greater than 3 then f'(x) is negative means slope of graph of f(x) is negative.

And when f'(x) is greater than ${{30} \over 8}$ but less than 5 then f'(x) is positive means slope of graph of f(x) is positive.

So, at ${{30} \over 8}$ between 3 and 5, f(x) attain local minima.

$\therefore$ Option C is not true.

Similarly you can check option D also.

$4a + 3b = 22$

Total area $ = A = {a^2} + {{\sqrt 3 } \over 4}{b^2}$

$A = {\left( {{{22 - 3b} \over 4}} \right)^2} + {{\sqrt 3 } \over 4}{b^2}$

${{dA} \over {dB}} = 2\left( {{{22 - 3b} \over 4}} \right)\left( {{{ - 3} \over 4}} \right) + {{\sqrt 3 } \over 4}\,.\,2b = 0$

$ \Rightarrow {{\sqrt {3b} } \over 2} = {3 \over 8}(22 - 3b)$

$ \Rightarrow 4\sqrt 3 b = 66 - 9b$

$ \therefore b = {{66} \over {9 + 4\sqrt 3 }}$

$\because$ ${s_1} + {s_2} = k$

$76{x^2} + 3\pi {r^2} = k$

$\therefore$ $152x{{dx} \over {dr}} + 6\pi r = 0$

$\therefore$ ${{dx} \over {dr}} = {{ - 6\pi r} \over {152x}}$

Now $V = 40{x^3} + {2 \over 3}\pi {r^3}$

$\therefore$ ${{dv} \over {dr}} = 120{x^2}\,.\,{{dx} \over {dr}} + 2\pi {r^2} = 0$

$ \Rightarrow 120{x^2}\,.\,\left( {{{ - 6\pi } \over {152}}{r \over x}} \right) + 2\pi {r^2} = 0$

$ \Rightarrow 120\left( {{x \over r}} \right) = 2\pi \left( {{{152} \over {6\pi }}} \right)$

$ \Rightarrow \left( {{x \over r}} \right) = {{152} \over 3}{1 \over {120}} = {{19} \over {45}}$

${\left( {{x \over a}} \right)^n} + {\left( {{y \over b}} \right)^n} = 2$

Differentiating both sides with respect to x, we get

$ \Rightarrow n\,.\,{\left( {{x \over a}} \right)^{n - a}}\,.\,{1 \over a} + n\,.\,{\left( {{y \over b}} \right)^{n - 1}}\,.\,{1 \over b}\,.\,{{dy} \over {dx}} = 0$

$ \Rightarrow {{dy} \over {dx}} = - {{{1 \over a}\,.\,{{\left( {{x \over a}} \right)}^{n - 1}}} \over {{1 \over b}{{\left( {{y \over b}} \right)}^{n - 1}}}}$

$ = - {b \over a}{\left( {{{xb} \over {ya}}} \right)^{n - 1}}$

Now, ${{dy} \over {dx}}$ at (a, b) $ = - {b \over a}{\left[ {{{ab} \over {ba}}} \right]^{n - 1}} = - {b \over a}$

Equation of tangent at (a, b) is,

$(y - b) = - {b \over a}(x - a)$

$ \Rightarrow {{y - b} \over b} = - {{x - a} \over a}$

$ \Rightarrow {y \over b} - 1 = - {x \over a} + 1$

$ \Rightarrow {x \over a} + {y \over b} = 2$

$\therefore$ It is tangent for all value of n.

$f(x) = 2{\cos ^{ - 1}}x + 4{\cot ^{ - 1}}x - 3{x^2} - 2x + 10\,\forall x \in [ - 1,1]$

$ \Rightarrow f'(x) = - {2 \over {\sqrt {1 - {x^2}} }} - {4 \over {1 + {x^2}}} - 6x - 2 < 0\,\forall x \in [ - 1,1]$

So f(x) is decreasing function and range of f(x) is [f(1), f($-$1)], which is [$\pi$ + 5, 5$\pi$ + 9]

Now $4a - b = 4(\pi + 5) - (5\pi + 9)$

$ = 11 - \pi $

$\because$ $V = {1 \over 3}\pi {r^2}h$ and ${r \over h} = {7 \over {35}} = {1 \over 5}$

$ \Rightarrow V = {1 \over {75}}\pi {h^3}$

${{dV} \over {dt}} = {1 \over {25}}\pi {h^2}{{dh} \over {dt}} = 1$

$ \Rightarrow {{dh} \over {dt}} = {{25} \over {\pi {h^2}}}$

Now, $S = \pi rl = \pi \left( {{h \over 5}} \right)\sqrt {{h^2} + {{{h^2}} \over {25}}} = {\pi \over {25}}\sqrt {26} {h^2}$

$ \Rightarrow {{dS} \over {dt}} = {{2\sqrt {26} \pi h} \over {25}}\,.\,{{dh} \over {dt}} = {{2\sqrt {26} } \over h}$

$ \Rightarrow {{dS} \over {d{t_{(h = 10)}}}} = {{\sqrt {26} } \over 5}$

$\because$ ${{dy} \over {dx}} = {{24(1 + \sin t)\cos t} \over {12(1 + \cos 2t)}} = {{1 + \sin t} \over {\cos t}} = \tan \left( {{\pi \over 4} + {t \over 2}} \right)$

$\because$ ${{dy} \over {d{x_{({x_0},{y_0})}}}} = \sqrt 3 = \tan \left( {{\pi \over 4} + {t \over 2}} \right)$

$ \Rightarrow t = {\pi \over 6}$

So, ${y_{0\left( {at\,t = {\pi \over 6}} \right)}} = 12{\left( {1 + \sin {\pi \over 6}} \right)^2} = 27$

$\because$ $ - {{dx} \over {dy}} = {{{x^2}} \over {xy - {x^2}{y^2} - 1}}$

$\therefore$ ${{dy} \over {dx}} = {{{x^2}{y^2} - xy + 1} \over {{x^2}}}$

Let $xy = v \Rightarrow y + x{{dy} \over {dx}} = {{dv} \over {dx}}$

$\therefore$ ${{dv} \over {dx}} - y = {{({v^2} - v + 1)y} \over v}$

$\therefore$ ${{dv} \over {dx}} = {{{v^2} + 1} \over x}$

$\because$ $y(1) = 1 \Rightarrow {\tan ^{ - 1}}(xy) = \ln x + {\tan ^{ - 1}}(1)$

Put $x = e$ and $y = y(e)$ we get

${\tan ^{ - 1}}(e\,.\,y(e)) = 1 + {\tan ^{ - 1}}1$

${\tan ^{ - 1}}(e\,.\,y(e)) - {\tan ^{ - 1}}1 = 1$

$\therefore$ $e(y(e)) = {{1 + \tan (1)} \over {1 - \tan (1)}}$

$\because$ ${f_\lambda }(x) = 4\lambda {x^3} - 36\lambda {x^2} + 36\lambda + 48$

$\therefore$ $f{'_\lambda }(x) = 12(\lambda {x^2} - 6\lambda x + 3)$

For ${f_\lambda }(x)$ increasing : ${(6\lambda )^2} - 12\lambda \le 0$

$\therefore$ $\lambda \in \left[ {0,\,{1 \over 3}} \right]$

$\therefore$ $\lambda^ * = {1 \over 3}$

Now, $f_\lambda ^*(x) = {4 \over 3}{x^3} - 12{x^2} + 36x + 48$

$\therefore$ $f_\lambda ^*(1) + f_\lambda ^*( - 1) = 73{1 \over 2} - 1{1 \over 2} = 72$

We know,

Surface area of balloon (s) = 4$\pi$r²

$\therefore$ ${{ds} \over {dt}} = {d \over {dt}}(4\pi {r^2})$

$ \Rightarrow {{ds} \over {dt}} = 4\pi (2r) \times {{dr} \over {dt}}$

$ \Rightarrow {{ds} \over {dt}} = 8\pi r \times {{dr} \over {dt}}$

Given that, surface area of balloon is increasing in constant rate.

$\therefore$ ${{ds} \over {dt}}$ = constant = k (Assume)

$ \Rightarrow k = 8\pi r\,.\,{{dr} \over {dt}}$

$ \Rightarrow \int {k\,dt = \int {8\pi r\,dr} } $

$ \Rightarrow kt = 8\pi \times {{{r^2}} \over 2} + c$

$ \Rightarrow kt = 4\pi {r^2} + c$ ..... (1)

Given at t = 0, radius r = 3

So, $0 = 4\pi ({3^2}) + c$

$ \Rightarrow c = - 36\pi $

$\therefore$ Equation (1) becomes

$kt = 4\pi {r^2} - 36\pi $

Also given, at t = 5, radius r = 7

$\therefore$ $k(5) = 4\pi {(7)^2} - 36$

$ \Rightarrow k = 32\pi $

$\therefore$ Equation (1) is

$32\pi t = 4\pi {r^2} - 36\pi $

Now at $t = 9$

$ \Rightarrow 32\pi (9) = 4\pi {r^2} - 36\pi $

$ \Rightarrow 8 \times 9 = {r^2} - 9$

$ \Rightarrow {r^2} = 81 \Rightarrow r = 9$

Given curve,

$y = {x^3} + 3{x^2} + 5$

Slope of tangent,

${{dy} \over {dx}} = 3{x^2} + 6x$

Slope of line joined by (x₁, y₁) and (0, 0) is $ = {{{y_0} - 0} \over {{x_1} - 0}}$

$\therefore$ ${{{y_1}} \over {{x_1}}} = 3x_1^2 + 6{x_1}$

$ \Rightarrow {y_1} = 3x_1^3 + 6x_1^2$ ...... (1)

Point (x₁, y₁) is on the line,

$\therefore$ ${y_1} = x_1^3 + 3x_1^2 + 5$ ..... (2)

$\therefore$ $3x_1^3 + 6x_1^2 = x_1^3 + 3x_1^2 + 6$

$ \Rightarrow 2x_1^3 + 3x_1^2 - 5 = 0$

x = 1 satisfy the equation

Now, ${y_1} = x_1^3 + 3x_1^2 + 5$

$ = 1 + 3 + 5$

$ = 9$

$\therefore$ (x₁, y₁) = (1, 9)

Option D does not satisfy point (1, 9).