Ncert Solutions Maths class 12th

We have, f (x) = x 100 + sin x - 1.

So, f (x) = 100x⁹⁹ + cosx.

(A) When x (0,1). We get.

x>0

$\Rightarrow$ x⁹⁹> 0

$\Rightarrow$ 100 x⁹⁹> 0.

Now, 0 radian = 0 degree

and 1 radian = $\frac{180^{°}\times 1}{\pi }=\frac{180^{°}}{\left(\raisebox{1ex}{$22$}\!\left/ \!\raisebox{-1ex}{$7$}\right.\right)}=57^{°}.$

So, cosx> 0 for x∈ (0,1) radian = (0,57)

∴f (x) > 0 for x∈ (0,1).

(B) When x ∈$\left(\frac{\pi }{2},\pi \right)$ we get,

So, x> 1

$\Rightarrow$ x⁹⁹> 1

$\Rightarrow$ 100x⁹⁹> 100. $\left\{\begin{array}{cc}\hfill ?\left(\frac{\pi }{2},\pi \right)=(1.57,3.14)\hfill & \hfill which is great than 1\hfill \end{array}\right\}$

And cosx is negative betw

(A) We have,

f(x) = cosx

So, f(x) = -sin x

When x $\in \left[0,\frac{\pi }{2}\right)$ we know that sin x> 0.

$\Rightarrow$ -sinx< 0. f(x) < 0 $\forall x\in \left(0,\frac{\pi }{2}\right)$

∴f(x) is strictly decreasing on $\left(0,\frac{\pi }{2}\right)$

(B) We have, f(x) = cos 2x

So, f(x) = -2sin 2x

When $x\in \left(0,\frac{\pi }{2}\right)$ we know that sin x> 0.

i e, 0 $\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
=> 0<2x<π 

So, sin 2x> 0 (sinØ is ( +ve) in 1st and 2ndquadrant).

$\Rightarrow$ -2sin 2x< 0.

$\Rightarrow$ f (x) < 0.

∴f

We have, f (x) = x²-x + 1.

So, f (x) = $\frac{d}{dx}\left(x^2-x+1\right)=2x-1$

Atf (x) = 0.

$\Rightarrow$ 2x - 1 = 0

I e, x = $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ divides the real line into two disjoint interval the interval ( -1, 1) into

Two disjoint interval

$\left(-1,\frac{1}{2}\right)\left(\frac{1}{2},1\right)$

And f (x) is strictly increasing in $\left(\frac{1}{2},1\right)$ and strictly decreasing in $\left(-1,\frac{1}{2}\right)$

Hence, f (x) is with a increasing or decrea

We have, f (x) = log x.

So, f (x) = $\frac{d}{dx}(logx)=\frac{1}{x}.$

i e, f (x) $=\frac{1}{x}$ > 0. When $x\ne 0x\in (0,\infty )$

Hence, the logarithmic fx is strictly increasing on $(0,\infty ).$

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)

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.

(a) f (x) = x² + 2x - 5.

f(x) = 2x + 2 = 2 (x + 1).

At, f(x) = 0

2 (x + 1) = 0

$\Rightarrow$ x = -1.

At, x $(-\infty ,-1),$

f(x) = (- ) ve< 0.

So, f (x)is strictly decreasing or $(-\infty ,-1).$

At x ∈$(-1,\infty )$

f(x) = ( + ve) > 1

f(x) is strictly increasing on $(-1,\infty ).$

(b) f(x) = 10 - 6x- 2x²

So, f(x) = - 6 - 4x = - 2 (3 + 2x).

Atf(x) = 0

$\Rightarrow$ 2 (3 + 2x) = 0.

$\Rightarrow$&nb

We have, f (x) = 2x³- 3x²- 36x + 7.

So, f (x) = $\frac{d}{dx}\left(2x^3-3x^2-36x+7\right)=6x^2-6x-36.$

= 6 (x²-x- 6).

= 6 (x²- 3x + 2x- 6)

= 6 [x (x- 3) + 2 (x- 3)]

= 6 (x- 3) (x + 2).

At, f (x) = 0

$\Rightarrow$ 6 (x- 3) (x + 2) = 0.

So, when x- 3 = 0 or x + 2 = 0.

$\Rightarrow$ x = 3 or x = -2.

Hence we an divide the real line into three disjoint internal

I $(-\infty ,-2)\; II(-2,3)andIII(3,\infty )$

At x ∈$(-\infty ,-2),$

f (x) = ( + v

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 $\left(-\infty ,\frac{3}{4}\right)$