Let $$A\,sq.$$  units in the area measure when the radius is $$r$$ units. their $$A = \pi {r^2}$$
Differentiate both side w.r.t $$'t'\,\frac{{dA}}{{dt}} = 2\pi r\frac{{dr}}{{dt}}......\left( {\text{i}} \right)$$
We have, $$\frac{{dA}}{{dt}} = 3c\frac{{dr}}{{dt}}$$
From equation $$\left( {\text{i}} \right),$$ we get $$3c.\frac{{dr}}{{dt}} = 2\pi r.\frac{{dr}}{{dt}} \Rightarrow 3c = 2\pi r$$
Now, $$c = \frac{2}{3}\pi \left( 6 \right) = 4\pi {\text{ when }}r = 6$$

$$\eqalign{ & {\text{From the figure,}} \cr & PC = b\,{\text{cosec}}\,\theta {\text{ and }}AP = a\,\sec \,\theta \cr & AC = PC + AP \cr & {\text{or }}AC = b\,{\text{cosec}}\,\theta {\text{ }} + a\,\sec \,\theta ......\left( 1 \right) \cr & \therefore \,\frac{{d\left( {AC} \right)}}{{d\theta }} = - b\,{\text{cosec}}\,\theta \,\cot \,\theta + a\,\sec \,\theta \,\tan \,\theta \cr & {\text{For minimum length, }}\frac{{d\left( {AC} \right)}}{{d\theta }} = 0 \cr & {\text{or }}a\,\sec \,\theta \,\tan \,\theta = b\,{\text{cosec}}\,\theta \,\cot \,\theta \cr & {\text{or }}\tan \,\theta = {\left( {\frac{b}{a}} \right)^{\frac{1}{3}}} \cr & \therefore \,\,\sin \,\theta = \frac{{{{\left( b \right)}^{\frac{1}{3}}}}}{{\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} }}{\text{ and cos}}\,\theta = \frac{{{{\left( a \right)}^{\frac{1}{3}}}}}{{\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} }}.....\left( 2 \right) \cr & {\text{Also}},{\text{ }}\theta \, \in \left( {0,\,\frac{\pi }{2}} \right) \cr & \mathop {\lim }\limits_{\theta \to 0} \left( {a\,\sec \,\theta + b\,{\text{cosec}}\,\theta } \right) \to \infty \cr & {\text{Therefore, }}\theta = {\tan ^{ - 1}}{\left( {\frac{b}{a}} \right)^{\frac{1}{3}}}{\text{ is a point of minima}}{\text{.}} \cr & {\text{For this value of}}\,\theta , \cr & AC = \frac{{b\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} }}{{{b^{\frac{1}{3}}}}} + \frac{{a\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} }}{{{a^{\frac{1}{3}}}}}\,\,\,\,\,\left[ {{\text{Using }}\left( 1 \right){\text{ and }}\left( 2 \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \left( {{b^{\frac{2}{3}}} + {a^{\frac{2}{3}}}} \right) \cr & \,\,\,\,\,\,\,\,\,\, = {\left( {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \right)^{\frac{3}{2}}} \cr} $$
Hence, the minimum length of the hypotenuse is $${\left( {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \right)^{\frac{3}{2}}}.$$

Let $$N$$ be the number of bacteria at time $$t.$$
Let $${N_0}$$ be the initial original number of bacteria.
$$\eqalign{ & {\text{Then, }}\frac{d}{{dt}}N \propto N\frac{d}{{dt}} = kN \cr & \Rightarrow \frac{{dN}}{N} = kdt \cr & \Rightarrow \int {\frac{{dN}}{N} = k\int {dt} } \cr & \Rightarrow \log \,N = kt + c \cr & {\text{At }}t = 0,\,N = {N_0} \Rightarrow \log \,{N_0} = 0 + c \Rightarrow c = \log \,{N_0} \cr & \therefore \,\log \,N = kt + \log \,{N_0} \Rightarrow \log \frac{N}{{{N_0}}} = kt \cr & {\text{When }}t = 5\,hrs,\,\,N = 2{N_0} \cr & \therefore \,\log \left( {\frac{{2\,{N_0}}}{{{N_0}}}} \right) = 5k\, \Rightarrow k = \frac{{\log \,2}}{5} \cr & \therefore \,\log \frac{N}{{{N_0}}} = \frac{{\log \,2}}{5}t \cr & {\text{When }}t = 25\,hrs,\,\,\log \frac{N}{{{N_0}}} = \frac{{\log \,2}}{5} \times 25 \cr & \Rightarrow \log \frac{N}{{{N_0}}} = 5\,\log \,2\,\log \frac{N}{{{N_0}}} = \log \,{2^5} \cr & \Rightarrow \frac{N}{{{N_0}}} = 32 \cr & \therefore \,N = 32\,{N_0} \cr} $$
Therefore, 32 times the original.

$$\eqalign{ & y = a{x^2} - 6x + b{\text{ passes through }}\left( {0,\,2} \right) \cr & {\text{i}}{\text{.e}}{\text{., }}2 = 0 - 0 + b{\text{ or }}b = 2 \cr & {\text{Again, }}\frac{{dy}}{{dx}} = 2ax - 6 \cr & {\text{At }}x = \frac{3}{2},\,\frac{{dy}}{{dx}} = 3a - 6 \cr & {\text{Since tangent is parallel to }}x{\text{ - axis}} \cr & \frac{{dy}}{{dx}} = 0{\text{ or }}3a - 6 = 0{\text{ or }}a = 2 \cr & {\text{Hence, }}a = 2,\,b = 2 \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{{x^2}}}{{{e^x}}}\,; \cr & f'\left( x \right) = \frac{{2x.{e^x} - {e^x}.{x^2}}}{{{{\left( {{e^x}} \right)}^2}}} \cr & f'\left( x \right) = \frac{{2x - {x^2}}}{{{e^x}}} \cr} $$
as $${e^x}$$ is always positive and for monotonically increasing ;
$$\eqalign{ & 2x - {x^2} > 0 \cr & \Rightarrow {x^2} - 2x < 0 \cr & \Rightarrow x\left( {x - 2} \right) < 0 \cr & \Rightarrow x\, \in \,\left( {0,\,2} \right)\, \cr} $$

$$f\left( x \right) = \left\{ {_{2x + 3,\,\,{\text{if}}\,x > - 1}^{k - 2x,\,\,{\text{if}}\,x\, \leqslant - 1}} \right.$$

This is true where $$k = - 1$$

$$\eqalign{ & {\text{Distance}}\,{\text{of}}\,{\text{origin}}\,{\text{from}}\,\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} \cr & = \sqrt {{a^2} + {b^2} - 2ab\cos \left( {t - \frac{{at}}{b}} \right)} ; \cr & \leqslant \sqrt {{a^2} + {b^2} + 2ab} \left[ {{{\left\{ {\cos \left( {t - \frac{{at}}{b}} \right)} \right\}}_{\min }} = - 1} \right] \cr & = a + b \cr & \therefore {\text{Maximum}}\,{\text{distance}}\,{\text{from}}\,{\text{origin}} = a + b \cr} $$

$$\eqalign{ & {\text{We}}\,{\text{have}}\,P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d \cr & \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx + c \cr & {\text{But }}P'\left( 0 \right) = 0 \Rightarrow c = 0 \cr & \therefore \,P\left( x \right) = {x^4} + a{x^3} + b{x^2} + d \cr & {\text{As}}\,{\text{given}}\,{\text{that}}\,P\left( { - 1} \right) < P\left( a \right) \cr & \Rightarrow \,1 - a + b + d\, < \,1 + a + b + d\, \Rightarrow a > 0 \cr & {\text{Now }}P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx = x\left( {4{x^2}{\text{ }} + {\text{ }}3ax + 2b} \right) \cr & {\text{As }}P'\left( x \right) = 0,{\text{ there is only one solution }}x = 0, \cr & {\text{therefore }}4{x^2} + 3ax + 2b = 0{\text{ should not have any real roots i}}{\text{.e }}D < 0 \cr & \Rightarrow \,9{a^2} - 32b < 0 \Rightarrow b > \frac{{9{a^2}}}{{32}} > 0 \cr & {\text{Hence}}\,a,b > 0 \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx > 0\,\forall x > 0 \cr & \therefore \,P\left( x \right)\,{\text{is}}\,{\text{an}}\,{\text{increasing}}\,{\text{function}}\,{\text{on}}\,\left( {0,1} \right) \cr & \therefore \,P\left( 0 \right) < P\left( a \right) \cr & {\text{Similarly we can prove }}P\left( x \right)\,{\text{is decreasing on}}\,\left( { - 1,0} \right) \cr & \therefore \,P\left( { - 1} \right) > P\left( 0 \right) \cr & {\text{So we can conclude that}} \cr & {\text{Max}}\,P\left( x \right) = P\left( 1 \right)\,{\text{and}}\,{\text{Min}}\,P\left( x \right) = P\left( 0 \right) \cr & \Rightarrow P\left( { - 1} \right)\,{\text{is}}\,{\text{not minimum but }}P\left( 1 \right){\text{ is the maximum of }}P. \cr} $$

Let $$f\left( x \right) = \frac{{\sin \,2x}}{{\sin \left( {x + \frac{\pi }{4}} \right)}} = \sqrt 2 \left\{ {\frac{{{{\left( {\sin \,x + \cos \,x} \right)}^2} - 1}}{{\sin \,x + \cos \,x}}} \right\} = \sqrt 2 \left( {\frac{{{y^2} - 1}}{y}} \right),$$              where $$y = \sin \,x + \cos \,x$$
Let $$\phi \left( y \right) = \sqrt 2 \left( {\frac{{{y^2} - 1}}{y}} \right){\text{ and }}g\left( x \right) = \sin \,x + \cos \,x$$
We have, $$g'\left( x \right) = \cos \,x - \sin \,x$$
For max. or min. $$g'\left( x \right) = 0 \Rightarrow \tan \,x = 1 \Rightarrow x = \frac{\pi }{4}$$
For this value of $$x.$$
$$g''\left( x \right) < 0.$$   Thus, $$g\left( x \right)$$  is max. at $$x = \frac{\pi }{4}$$  and hence the domain of $$g\left( x \right)$$  is $$\left[ {1,\,\sqrt 2 } \right]$$  i.e. $$y$$ lies between $$1$$ and $$\sqrt 2 $$
Now, $$\phi '\left( y \right) = \sqrt 2 \left( {1 + \frac{1}{{{y^2}}}} \right) > 0{\text{ for all }}y\, \in \left[ {1,\,\sqrt 2 } \right]$$
That is $$\phi \left( y \right)$$  is increasing for all $$y\, \in \left[ {1,\,\sqrt 2 } \right]$$
Thus it attains the greatest value at $$\sqrt 2 $$ and is equal to $$\sqrt 2 \left( {\frac{{{{\left( {\sqrt 2 } \right)}^2} - 1}}{{\sqrt 2 }}} \right) = 1$$
Hence, greatest value of $$f\left( x \right)$$  on $$\left[ {0,\,\frac{\pi }{2}} \right] = $$   greatest value of $$\phi \left( y \right)\,{\text{on}}\,\left[ {1,\,\sqrt 2 } \right] = 1.$$

In the neighborhood of $$x = - 2,\,y = {x^2} + x.$$
Hence, the point on curve is $$\left( { - 2,\,2} \right).$$
$$\frac{{dy}}{{dx}} = 2x + 1{\text{ or }}{\left. {\frac{{dy}}{{dx}}} \right|_{x = - 2}} = - 3$$
So, the slope of the normal at $$\left( { - 2,\,2} \right)$$  is $$\frac{1}{3}$$
Hence, the equation of the normal is $$\frac{1}{3}\left( {x + 2} \right) = y - 2{\text{ or }}3y = x + 8.$$