Ncert Solutions Maths class 12th

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

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.

The area A of the isle with radius r is given by with respect to radius r A = πr².

Then, rate of change of area of the circle $\frac{d·A}{dx}=\frac{d\pi r^2}{dr}$

= 2πr.

When r = 6 cm

$\frac{\mathrm{dA}}{dt}=2\pi \times 6=12\pi .$

Q option (B) is correct.

Given, R (x) = 13x² + 26x + 15.

Marginal revenue is the rate of change of total revenue with respect to the number of units sold Marginal revenue (MR) = $\frac{dR(x)}{dx}$

$=\frac{d}{d}\left(13x^2+26x+15\right)$

= 13 * 2x + 26

= 26x + 26

When x = 7,

MR = 26 * 7 + 26 = 182 + 26 = 208.

Hence, the required marginal reverse = ' 208.

Choose the correct answer fo

Given, c (x) = 0.007 x³- 0.003x² + 15x + 400.

Since the marginal cost is the rate of change of total cost wrt the output we have,

Marginal cost, MC, =$\frac{\mathrm{dC}}{dx}(x)$

= 0.007 * 3x²- 0.003 * 2x + 15.

When x = 17,

Then, MC = 0.007 * 2. (17)² - 0.003 2 (17) + 15.

= 6.069 - 0.102 + 15.

= 20.967

Hence, the required marginal c

Let r cm and h cm be the radius and the height of the cone. Then,

h =$\frac{1}{6}$ r. H = 6h

So, volume, V of the cone =$\frac{1}{3}$ πr²h

$=\frac{1}{3}\pi {(6h)}^2\times h.$ $\frac{4d}{dt}.$

= 12 × h³

Rate of change of volume of the cone wrt the height is

$\frac{\mathrm{dV}}{dh}=\frac{d}{dh}\left(12\pi h^3\right)$ = 12 × π × 3 × h².

As the sand is pouring from the pipe at rate of 12${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

we have

$\frac{d}{dt}=12$

$\Rightarrow \frac{dv}{dh}\times \frac{dh}{dt}=12$

$\Rightarrow 36\pi h^2\frac{dh}{dt}=12.$

$\Rightarrow \frac{dh}{dt}=\frac{12}{36\pi h^2}=\frac{1}{3\pi h^2}.$

${\therefore \frac{dh}{dt}|}_{h=4 }=\frac{1}{3\pi \times {(4)}^2}=\frac{1}{48\pi }.$

Hence, the height is increasing at the r

Given, diameter of the spherical balloon = $\frac{3}{2}$ (2x + 1)

So, radius of the spherical r = $\frac{1}{2}\times \frac{3}{2}(2x+1)$

$=\frac{3}{4}(2x+1)$

Then, volume of the spherical V = $\frac{4}{9}\pi r^3$

$=\frac{4}{3}\pi \times [\frac{3}{4}{\left(\frac{3}{(2x+1)}\right]}^3$

$=\frac{9\pi }{16}{(2x+1)}^3.$

Q Rate of change of volume wrt.tox, $\frac{dV}{dx}=\frac{d}{dx}\left[\frac{\pi }{16}{(2x+1)}^3\right]$

$=\frac{9\pi }{16}\times 3\times {(2x+1)}^2·\frac{d}{dx}(2x+1)$

$=\frac{27\pi }{16}{(2x+1)}^2\times 2=\frac{27\pi }{8}{(2x+1)}^2.$

Let x be the radius of the bubble with volume .V. then,

$\frac{dr}{dt}=\frac{1}{2}$ cm/s

andV = $\frac{4}{3}\pi n^3$

Rate of change of volume $\frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}{\pi }^3\right)$ = $\frac{4}{3}\pi \frac{d}{d{t}^4}{r}^3$

$=\frac{4}{3}\pi \times 3r^2\frac{dr}{dt}.$

= 4πr² ×$\frac{1}{2}$

= 2πr².

${\frac{dv}{dt}|}_{r=cm}$ = 2x (1)² 2π. ${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$