Class 12th

We have, y = $\frac{4sin\theta }{(2+cos\theta )}-\theta$

Differentiating w rt.Ø we get,

$\frac{dy}{d\theta }=\frac{d}{d\theta }\left[\frac{4sin\theta }{2+cos\theta }-\theta \right]$

$=\frac{(2+cor\theta )\frac{d}{d\theta }(4sin\theta )-4sin\theta \frac{d}{d\theta }(2+cor\theta )}{{(2+cos\theta )}^2}-\frac{d\theta }{d\theta }.$

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

$=\frac{8cos\theta +4co{s}^2\theta +4si{n}^2\theta \cdot }{{(2+cos\theta )}^2.}-1$

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

$=\frac{8cor\theta +4-\left[4+co{s}^2\theta +4cos\theta \right]}{{(2+cor\theta )}^2}.$

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

$=\frac{4cos\theta -co{s}^2\theta }{{(2+cos\theta )}^2}=\frac{cos\theta (4-cos\theta )}{{(2+cos\theta )}^2}$

When $\theta \in \left[0,\frac{\pi }{2}\right]$ we know that, $0\le cos\theta \le 1.$

So, $4-cos\theta >0$

And also, (2 + cos$\theta$)²> 0.

$\therefore \frac{dy}{d\theta }\ge 0\forall \theta \in \left[0,\frac{\pi }{2}\right]$

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

We have, y = [x (x- 2)]².

Differentiating the above w rt. x we get,

$\frac{dy}{dx}=\frac{d}{dx}[x]{(x-2)}^2$

$=2[x(x-2)]\frac{d}{dx}[x(x-2)]$

$=2[x(x-2)]\left[x\frac{d}{dx}(x-2)+(x-2)\frac{dx}{dx}\right]$

= 2 [x (x- 2)] (x + x- 2)

= 2x (x - 2) (2x - 2)

$\frac{dy}{dx}$ = 4x (x - 2) (x - 1).

Now, $\frac{dy}{dx}=0$

$\Rightarrow$ 4x (x - 2) (x - 1) = 0.

i e, x = 0, x = 2, x = 1 divides the real line into

four disjoint interval. $(-\infty ,0],$ [0,1],[1,2] and $\left[2,\infty \right].$

when x $\left[-\infty ,0\right].$

$\frac{dy}{dx}=(-ve)(-ve)(-ve)=$ $(-ve)\le 0,0x=0$

∴f (x) is decreasing in $\left[-\infty ,0\right].$

When $x\in [0,1]$

$\frac{dy}{dx}=(+ve)(-ve)(-ve)$ $=(+ve)\ge 0,x=0x=1$

∴f (x) is increasing in [0,1].

When x $\in [1,2]$

$\frac{dy}{dx}=(+ve)(-ve)(+ve)=(-ve)\le 0,$ $\mathrm{0for}x=1+x=2$

∴f (x) is decreasing.

When $x\in \left(2,\infty \right).$

$\frac{dy}{dx}=(+ve)(+ve)(+ve)=(+ve)\ge \mathrm{0,}0,forx=2$

∴f (x) is in creasing

Hence, f (x) is increasing for x $\in [0,1)x\in [2,\infty )$

We have, y = log $(1+x)-\frac{2x}{2+x},x>-1$

Differentiating the above wrt.x.we get,

$\frac{dy}{dx}=\frac{1}{1+x}\frac{d}{dx}(1+x)-$ $\frac{(2+x)\frac{d}{dx}2x-2x\frac{d}{dx}(2+x)}{{(2+x)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{1+x}-\frac{(2+x)\cdot 2-2x}{{(2+x)}^2}.$

$\Rightarrow \frac{dy}{dx}=\frac{1}{1+x}-\frac{4+2x-2x}{{(2+x)}^2}=\frac{1}{1+x}-\frac{4}{{(2+x)}^2}.$

$=\frac{{(2+x)}^2-4(1+x)}{(1+x){(2+x)}^2}$

$=\frac{4+x^2+4x-4-4x}{(1+x){(2+x)}^2}$

$\frac{dy}{dx}=\frac{x^2}{(1+x){(2+x)}^2}.$

The given domain of the given function isx> -1.

$\Rightarrow$ (x + 1) > 0.

Also, $x^2\ge 0$

(2 + x)²> 0.

Hence, $\frac{dy}{dx}=\frac{(+ve)}{(+v)(+ve)}=(+ve)>0.$

∴ y is an increasing function of x throughout its domain.

We have, f (x) = 2x² 3x

So, f (x) = $\frac{d}{dx}\left(2x^2-3x\right)=4x-3.$

Atf (x) = 0.

$\Rightarrow$ 4x - 3 = 0

i e,  $x=\frac{3}{4}$ divides the real line into two

disjoint interval $\left(-\infty ,\frac{3}{4}\right)\left(\frac{3}{4},\infty \right)$

(a) Now,

f (x) = 4x - 3 > 0 $\forall x\in \left(\frac{3}{4},\infty \right)$

So, f (x) is strictly increasing in $\left(\frac{3}{4},\infty \right)$

(b) Now, f (x) = 4x - 3 < 0 $\forall x\left(-\infty ,\frac{3}{4}\right)$

So, f (x) is strictly decreasing in $\left(-\infty ,\frac{3}{4}\right)$

We have, f (x) = sin x.

So, f (x) = cosx.

(a) when, x ∈$\left(0,\frac{\pi }{2}\right)$ i e, x in 1st quadrat.

f (x) = cos.x> 0

f (x) is strictly increasing in $\left(0,\frac{\pi }{2}\right)$ .

(b) when, x ∈$\left(\frac{\pi }{2},\pi \right)$ in IInd quadrat

f (x) = cosx< 0.

∴f (x) is strictly decreasing (π

(c) When, x ∈ (0, π).

f (x) = cosx is increasing in $\left(0,\frac{\pi }{2}\right)$ and decreasing

in $\left(\frac{\pi }{2},\pi \right)$ and f $\left(\frac{\pi }{2}\right)$ = cos $\frac{\pi }{2}=0.$

∴f (x) is neither increasing not decreasing in (0, π).

We have, f (x) = e^2x

So, f (x) = $\frac{d}{dx}{e}^{2x}$ = $e^{2x}\frac{d}{dx}2x$ = 2e^2x> 0 $\forall x\in R.$

∴f (x) is strictly increasing on R.

We have, .

f (x) = 3x + 17.

So, f (x) = 3 > 0 $\forall x\in R$

∴f (x) is strictly increasing on R.

Given, R (x) = 3x² + 36x + 5.

Marginal revenue,  $\frac{d}{dx}R(x)=\frac{d}{dx}\left(3x^2+36x+5\right)$

= 3 * 2x + 36

= 6x + 36

When x = 15.

$\frac{d}{dx}R\; (x)=6*15+36$ = 90 + 36 = 126

Option (D) is correct.