$$\eqalign{ & {\text{We}}\,{\text{are}}\,{\text{given}}\,f\left( x \right) = {\sin ^4}x + {\cos ^4}x \cr & \Rightarrow f'\left( x \right) = 4{\sin ^3}x\cos x - 4{\cos ^3}x\sin x \cr & = - 4\sin x\cos x\left( {{{\cos }^2}x - {{\sin }^2}x} \right) \cr & = - 2.\sin 2x\cos 2x = - \sin 4x \cr & {\text{Now for }}f\left( x \right){\text{ to be increasing function}} \cr & f'\left( x \right) > 0 \Rightarrow - \sin 4x > 0 \Rightarrow \sin 4x < 0 \cr & \Rightarrow \pi < 4x < 2\pi \Rightarrow \frac{\pi }{4} < x < \frac{\pi }{2} \cr & {\text{Since, If}}\,f\left( x \right)\,{\text{increasing}}\,{\text{on}}\,\left( {\frac{\pi }{4},\frac{\pi }{2}} \right) \cr & \frac{\pi }{4} = \frac{{4\pi }}{8} > \frac{{3\pi }}{8} \cr & {\text{It will be increasing on}}\,\left( {\frac{\pi }{4},\frac{{3\pi }}{8}} \right) \cr} $$

Given that
$$\eqalign{ & \frac{{dv}}{{dt}} = 50\,c{m^3}/\min \Rightarrow \frac{d}{{dt}}\left( {\frac{4}{3}\pi {r^3}} \right) = 50 \cr & \Rightarrow 4\pi {r^2}\frac{{dr}}{{dt}} = 50 \cr & \Rightarrow \frac{{dr}}{{dt}} = \frac{{50}}{{4\pi {{\left( {15} \right)}^2}}} = \frac{1}{{18\pi }}cm/\min \,\left( {{\text{here}}\,r = 10 + 5} \right) \cr} $$

$$\eqalign{ & f'\left( x \right) = 3{x^2} - 3 = 3\left( {{x^2} - 1} \right) \cr & {\text{So, }}f'\left( x \right) \leqslant 0{\text{ if }} - 1 \leqslant x \leqslant 1{\text{ and }}f'\left( x \right) > 0{\text{ if }}1 < x \leqslant 4 \cr} $$
$$\therefore f\left( x \right)$$   is decreasing in $$ - 1 \leqslant x \leqslant 1$$   and increasing in $$1 < x \leqslant 4$$
$$\therefore \min f\left( x \right) = f\left( 1 \right),\,\max f\left( x \right) = $$       the greatest among $$\left\{ {f\left( { - 1} \right),\,f\left( 4 \right)} \right\}$$
$$\eqalign{ & {\text{Now,}}\,f\left( 1 \right) = {1^3} - 3.1 = - 2 \cr & f\left( { - 1} \right) = {\left( { - 1} \right)^3} - 3\left( { - 1} \right) = 2 \cr & f\left( 4 \right) = 64 - 12 = 52 \cr} $$
$$\therefore $$ the range of the function $$\left[ { - 2,\,52} \right]$$

The point $$\left( {a,\,b} \right)$$  lies on both the straight line and the given curve $${\left( {\frac{x}{a}} \right)^n} + {\left( {\frac{y}{b}} \right)^n} = 2$$
Differentiating the equation, we get
$$\eqalign{ & \frac{{dy}}{{dx}} = - \frac{{{x^{n - 1}}}}{{{a^n}}}.\frac{{{b^n}}}{{{y^{n - 1}}}} \cr & \therefore \,{\left( {\frac{{dy}}{{dx}}} \right)_{{\text{at }}\left( {a,\,b} \right)}} = - \frac{b}{a} = {\text{the slope of }}\frac{x}{a} + \frac{y}{b} = 2 \cr} $$
Hence, it touches the curve at $$\left( {a,\,b} \right)$$  whatever may be the value of $$n.$$

$$\eqalign{ & 4x + 2\pi r = 2\quad \Rightarrow 2x + \pi r = 1 \cr & S = {x^2} + \pi {r^2} \cr & S = {\left( {\frac{{1 - \pi r}}{2}} \right)^2} + \pi {r^2} \cr & \frac{{dS}}{{dr}} = 2\left( {\frac{{1 - \pi r}}{2}} \right)\left( {\frac{{ - \pi }}{2}} \right) + 2\pi r \cr & \Rightarrow \frac{{ - \pi }}{2} + \frac{{{\pi ^2}r}}{2} + 2\pi r = 0\quad \Rightarrow r = \frac{1}{{\pi + 4}} \cr & \Rightarrow x = \frac{2}{{\pi + 4}}\quad \Rightarrow x = 2r \cr} $$

Equation of a line passing through $$\left( {{x_1},{y_1}} \right)$$  having slope $$m$$ is given by $$y - {y_1} = m\left( {x - {x_1}} \right)$$
Since the line $$PQ$$  is passing through (1, 2) therefore its equation is $$\left( {y - 2} \right) = m\left( {x - 1} \right)$$
where $$m$$ is the slope of the line $$PQ.$$
Now, point $$P\left( {x,0} \right)$$  will also satisfy the equation of $$PQ.$$
$$\eqalign{ & \therefore y - 2 = m\left( {x - 1} \right) \cr & \Rightarrow 0 - 2 = m\left( {x - 1} \right) \cr & \Rightarrow \, - 2 = m\left( {x - 1} \right)\,\, \Rightarrow x - 1 = \frac{{ - 2}}{m} \cr & \Rightarrow x = \frac{{ - 2}}{m} + 1 \cr & {\text{Also, }}OP = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {0 - 0} \right)}^2}} = x = \frac{{ - 2}}{m} + 1 \cr & {\text{Similarly, point }}Q\left( {0,y} \right){\text{ will satisfy equation of }}PQ \cr & \therefore y - 2 = m\left( {x - 1} \right) \cr & \Rightarrow y - 2 = m\left( { - 1} \right) \Rightarrow y = 2 - m{\text{ and }}OQ = y = 2 - m \cr & {\text{Area of }}\Delta POQ = \frac{1}{2}\left( {OP} \right)\left( {OQ} \right) = \frac{1}{2}\left( {1 - \frac{2}{m}} \right)\left( {2 - m} \right) \cr & \left( {\because {\text{ Area of }}\Delta = \frac{1}{2} \times {\text{base}} \times {\text{height}}} \right) \cr & = \frac{1}{2}\left[ {2 - m - \frac{4}{m} + 2} \right] = \frac{1}{2}\left[ {4 - \left( {m + \frac{4}{m}} \right)} \right] \cr & = 2 - \frac{m}{2} - \frac{2}{m} \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{Area}}\, = f\left( m \right) = 2 - \frac{m}{2} - \frac{2}{m} \cr & {\text{Now,}}\,f'\left( m \right) = \frac{{ - 1}}{2} + \frac{2}{{{m^2}}} \cr & {\text{Put}}\,f'(m) = 0\, \Rightarrow {m^2} = 4 \Rightarrow m = \pm 2 \cr & {\text{Now,}}\,f''\left( m \right) = \frac{{ - 4}}{{{m^3}}} \cr & {\left. {f''\left( m \right)} \right|_{m = 2}} = - \frac{1}{2} < 0 \cr & {\left. {f''\left( m \right)} \right|_{m = - 2}} = \frac{1}{2} > 0 \cr} $$
Area will be least at $$m = - 2$$
Hence, slope of $$PQ$$ is $$ - 2$$

Let $$A$$ be the vertex and $$AO$$  the axis of the cone.
Let $$O'A = h$$   be the depth of water in the cone.
$$\eqalign{ & {\text{In}}\Delta AO'C, \cr & \tan \,{30^ \circ } = \frac{{O'C}}{h}{\text{ or }}O'C = \frac{h}{{\sqrt 3 }} = {\text{radius}} \cr & V = {\text{Volume of water in the cone}} \cr & = \frac{1}{3}\pi {\left( {O'C} \right)^2} \times AO' \cr & = \frac{1}{3}\pi \left( {\frac{{{h^2}}}{3}} \right) \times h \cr & = \frac{\pi }{9}{h^3} \cr & {\text{or }}\frac{{dV}}{{dt}} = \frac{\pi }{3}{h^2}\frac{{dh}}{{dt}}......\left( 1 \right) \cr} $$
But given that depth of water increases at the rate of $$1\,cm/s.$$
So, $$\frac{{dh}}{{dt}} = 1\,cm/s......\left( 2 \right)$$
From $$\left( 1 \right)$$ and $$\left( 2 \right),\,\,\frac{{dV}}{{dt}} = \frac{{\pi {h^2}}}{3}$$
When $$h = 24\,cm,$$   the rate of increase of volume is $$\frac{{dV}}{{dt}} = \frac{{\pi {{\left( {24} \right)}^2}}}{3} = 192\,c{m^3}/s.$$

$$\eqalign{ & f\left( x \right) = 2\,\sin \,x + \sin \,2x \cr & f'\left( x \right) = 2\,\cos \,x + 2\,\cos \,2x = 2\left( {\cos \,x + \cos \,2x} \right) \cr & \therefore \,f'\left( x \right) = 0 \Rightarrow 2\,{\cos ^2}x + \cos \,x - 1 = 0 \cr & \cos \,x = \frac{{ - 1 \pm 3}}{4} = - 1,\,\frac{1}{2} \cr & \therefore \,x = \pi ,\,\frac{\pi }{3} \cr & {\text{Now, }}f\left( 0 \right) = 0,\,f\left( {\frac{{3\pi }}{2}} \right) = - 2 \cr & f\left( \pi \right) = 0,\,f\left( {\frac{\pi }{3}} \right) = 2\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2} = \frac{{3\sqrt 3 }}{2} \cr} $$
$$\therefore $$  Difference between greatest value and least value $$ = \frac{{3\sqrt 3 }}{2} + 2$$

$$\eqalign{ & {\text{Let }}f\left( x \right) = {x^{\frac{1}{3}}}\,\, \Rightarrow f'\left( x \right) = \frac{1}{3}{x^{ - \frac{2}{3}}} \cr & {\text{Now, }}f\left( {x + \Delta x} \right) - f\left( x \right) = f'\left( x \right).\Delta x = \frac{{\Delta x}}{{3\left( {{x^{\frac{2}{3}}}} \right)}} \cr & {\text{We may write, }}0.007 = 0.008 - 0.001, \cr & {\text{taking }}x = 0.008{\text{ and }}dx = - 0.001 \cr & {\text{we have, }}f\left( {0.007} \right) - f\left( {0.008} \right) = - \frac{{0.001}}{{3{{\left( {0.008} \right)}^{\frac{2}{3}}}}} \cr & \Rightarrow f\left( {0.007} \right) - {\left( {0.008} \right)^{\frac{1}{3}}} = - \frac{{0.001}}{{3{{\left( {0.2} \right)}^2}}} \cr & \Rightarrow f\left( {0.007} \right) = 0.2 - \frac{{0.001}}{{3\left( {0.04} \right)}} \cr & \Rightarrow f\left( {0.007} \right) = 0.2 - \frac{1}{{120}} \cr & \Rightarrow f\left( {0.007} \right) = \frac{{23}}{{120}} \cr & {\text{Hence, }}{\left( {0.007} \right)^{\frac{1}{3}}} = \frac{{23}}{{120}} \cr} $$

$$\eqalign{ & {\text{Given that }}xy = {r^2} \cr & \Rightarrow y = \frac{{{r^2}}}{x} \cr & {\text{Let }}S = px + qy = px + \frac{{q{r^2}}}{x} \cr & \Rightarrow \frac{{dS}}{{dx}} = p - \frac{{q{r^2}}}{{{x^2}}} \cr & \frac{{dS}}{{dx}} = 0{\text{ for maximum or minimum}}{\text{.}} \cr & {\text{So, }}0 = p - \frac{{q{r^2}}}{{{x^2}}} \cr & \Rightarrow {x^2} = \frac{{q{r^2}}}{p} \cr & \Rightarrow x = \pm \sqrt {\frac{q}{p}} .r \cr & {\text{Now, }}\frac{{{d^2}S}}{{d{x^2}}} = \frac{{2q{r^2}}}{{{x^3}}} \cr & {\text{At }}x = + \sqrt {\frac{q}{p}} .r\,\,\frac{{{d^2}S}}{{d{x^2}}} > 0 \cr & {\text{Hence, }}S{\text{ is minimum at }}x = \sqrt {\frac{q}{p}} .r \cr & \Rightarrow y = \frac{{{r^2}}}{{\sqrt {\frac{q}{p}} .r}} = \sqrt {\frac{p}{q}} .r \cr & {\text{Minimum value of }}px + qy \cr & = p.\sqrt {\frac{q}{p}} .r + q\sqrt {\frac{p}{q}} .r \cr & = \sqrt {pq} \,r + \sqrt {pq} \,r \cr & = 2r\sqrt {pq} \cr} $$