Maths 3 Marks

On differentiating both sides w.r.t. x, we get,
f'(x) = $ 3 x ^ { 2 } - \frac { 3 } { x ^ { 4 } }$
On putting f'(x) = 0, we get,
$ 3 x ^ { 2 } - \frac { 3 } { x ^ { 4 } } = 0$$ \Rightarrow \frac { 3 x ^ { 6 } - 3 } { x ^ { 4 } } = 0$
$ \Rightarrow$ 3x⁶ - 3 = 0 $ \Rightarrow$3x⁶ = 3
$ \Rightarrow$ x⁶ = 1 $ \Rightarrow$ x = $ \pm$1
Now, we find intervals in whch f(x) is strictly increasing or strictly decreasing.

So, f(x) is strictly increasing on intervals (-$ \infty$, -1) and (1, $ \infty$) and it is strictly decreasing on (-1, 1)-{0}.
 Also, f(x) is continuous at x = 1 , -1.
 Hence, f(x) is

1. increasig on intervals (-$ \infty$, -1 ] and [1, $ \infty$).
2. decreasing on interval [-1, 1]- {0}.

$=4\left({\frac{8cosx+4-(2+cosx)^2}{(2+cosx)^2}}\right)$$f'(x)=4\left({\frac{cosx(4-cosx)}{(2+cosx)^2}}\right)$

${since\ - 1 \leqslant \cos x \leqslant 1} $
Hence
$\frac{{\cos x\left( {4 - \cos x} \right)}}{{{{\left( {2 + \cos x} \right)}^2}}} > 0\forall x \in \left( {0,\frac{\pi }{2}} \right)$ and $\left( {\frac{{3\pi }}{2},2\pi } \right)$
$\frac{{\cos x\left( {4 - \cos x} \right)}}{{{{\left( {2 + \cos x} \right)}^2}}} < 0\forall \in \left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right)$
So, f(x) is increasing in $\left( {0,\frac{\pi }{2}} \right)$ and $\left( {\frac{{3\pi }}{2},2\pi } \right)$
and if f(x) is decreasing in $\left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right)$

$ = - \frac { 2 } { \sqrt { 2 } } = - \sqrt { 2 }$

Let  $P = x{y^3}$ [To be maximized] …(ii)

Putting from eq. (i), $x = 60 - y$ in eq. (ii),

$P = \left( {60 - y} \right){y^3} = 60{y^3} - {y^4}$

$\Rightarrow \frac{{dP}}{{dy}} = 180{y^2} - 4{y^3} = 4{y^2}\left( {45 - y} \right)$ …(iii)

Now $\frac{{dP}}{{dy}} = 0$

$\Rightarrow 4{y^2}\left( {45 - y} \right) = 0$

$\Rightarrow y = 0,45$

It is clear that $\frac{{dP}}{{dy}}$ changes sign from positive to negative as y increases through 45.

Therefore, P is maximum when $y = 45.$ 

Hence, $x{y^3}$ is maximum when $x = 60 - 45 = 15$ and y = 45.

According to the question, x + y = 24

$\Rightarrow y = 24 - x$ …(i)

And let z is the product of x and y.

$\Rightarrow $z = xy

$\Rightarrow$ z = x(24 - x) [From eq. (i)]

$\Rightarrow $z = 24x - x²

$\Rightarrow \frac{{dz}}{{dx}} = 24 - 2x$ and $\frac{{{d^2}z}}{{d{x^2}}} = - 2$ 

Now to find turning point, $\frac{{dz}}{{dx}} = 0$

$\Rightarrow 24 - 2x = 0 \Rightarrow x = 12$

At $x = 12,\frac{{{d^2}z}}{{d{x^2}}} = - 2$ [Negative]

$\therefore x = 12$ is a point of local maxima and z is maximum at ​​​​​​​x = 12.

$\therefore $  From eq. (i), ​​​​​​​ y = 24 - 12 = 12

Therefore, the two required numbers are 12 and 12.

$ \Rightarrow \frac{{dy}}{{d\theta }} = \frac{{\left( {2 + \cos \theta } \right).4\cos \theta - 4\sin \theta \left( { - \sin \theta } \right)}}{{{{\left( {2 + \cos \theta } \right)}^2}}} - 1$

$= \frac{{8\cos \theta + 4{{\cos }^2}\theta + 4{{\sin }^2}\theta }}{{{{\left( {2 + \cos \theta } \right)}^2}}} - 1$

$ \Rightarrow \frac{{dy}}{{d\theta }} = \frac{{8\cos \theta + 4\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) - {{\left( {2 + \cos \theta } \right)}^2}}}{{{{\left( {2 + \cos \theta } \right)}^2}}}$

$= \frac{{8\cos \theta + 4 - {{\left( {2 + \cos \theta } \right)}^2}}}{{{{\left( {2 + \cos \theta } \right)}^2}}}$

$\Rightarrow \frac{{dy}}{{d\theta }} = \frac{{\left( {8\cos \theta + 4} \right) - \left( {4 + 4\cos \theta + {{\cos }^2}\theta } \right)}}{{{{\left( {2 + \cos \theta } \right)}^2}}}$

$= \frac{{4\cos \theta - {{\cos }^2}\theta }}{{{{\left( {2 + \cos \theta } \right)}^2}}}$

$= \frac{{\cos \theta \left( {4 - \cos \theta } \right)}}{{{{\left( {2 + \cos \theta } \right)}^2}}}$

Since $0 \leqslant \theta \leqslant \frac{\pi }{2}$ and we have $0 \leqslant \cos \theta \leqslant 1$, therefore $4 - \cos \theta > 0$. 

$\therefore \frac{{dy}}{{dx}} \geqslant 0$ for $0 \leqslant \theta \leqslant \frac{\pi }{2}$

Hence, y is an increasing function of $\theta $ in $\left[ {0,\frac{\pi }{2}} \right]$

Therefore, y is strictly increasing in ( 0, 1 ) and (2, $\infty$).
Also, y is a polynomial function, so it continuous at x = 0, 1 and 2.
 Hence, y is increasing in [ 0, 1] $ \cup $ [2, $\infty$).

$ \Rightarrow \frac{{dx}}{{dt}}$ is negative = –5 cm/minute

$ \Rightarrow \frac{{dx}}{{dt}}$ = – 5 cm/minute  ….(i)

Also, Rate of increase of width y of rectangle is 4 cm/minute

$\Rightarrow \frac{{dy}}{{dt}}$ is positive= 4 cm/minute

$\Rightarrow \frac{{dy}}{{dt}} = 4$ cm/minute   ….(ii)

let z denotes the perimeter of rectangle at any time t

$\therefore x = 2x + 2y$

$\Rightarrow \frac{{dz}}{{dt}} = 2\frac{{dx}}{{dt}} + 2\frac{{dy}}{{dt}}$ (from (i) and (ii) )   

$= 2\left( { - 5} \right) + 2\left( 4 \right) = - 2$ which is negative.             

$\therefore$ Perimeter of the rectangle is decreasing at the rate of 2 cm/min.

(b) Let z denotes the area of rectangle at any time t

$\Rightarrow z = xy$

$ \Rightarrow \frac{{dz}}{{dt}} = x\frac{{dy}}{{dt}} + y\frac{{dx}}{{dt}}$

$= 8\left( 4 \right) + 6\left( { - 5} \right) = 2$ which is positive. (from (i) and (ii))

$\therefore$ Area of the rectangle is increasing at the rate of 2 cm²/min