Class 12th

We have, f (x) = x² e^–x

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

$=x^2{e}^{-x}\frac{d}{dx}(-x)+e^{-x}2\cdot x\cancel{\frac{dx}{dx}}$

= -x² e^-x + e^-x 2x.

= x e^-x ( x + 2).

$=\frac{x(2-x)}{e^{ex}}$

If f (x) = 0.

$\Rightarrow \frac{x(2-x)}{e^x}=0.$

$\Rightarrow$ x = 0, x = 2.

Hence, we get there disjoint interval

$[-\infty ,0),(0,2)(2,\infty )$

When, x $(-\infty ,0),$ we have, f (x) = ( -ve) ( + ve) = ( -ve) < 0.

So, f is strictly decreasing.

When x ∈ (0,2), f (x) = ( + ve) ( + ve) = ( + ve) > 0.

So, f is strictly increasing.

And when x ∈$(2,\infty ),$ f (x) = ( +ve) ( -ve) = ( -ve) < 0.

So, f is strictly decreasing.

Hence, option (D) is correct.

We have, f (x) = x³- 3x² 3x- 100.

So, f (x) = 3x²- 6x + 3 = 3 (x²- 2x + 1) = 3 (x- 1)²

For $x\in R.$

(x- 10)²$\ge 0$ , 0 for x = 1.

$\Rightarrow$ 3 (x- 1)² $\ge 0$

$\Rightarrow$ f (x) $\ge 0$

∴f (x) is increasing on $?$

We have, f (x) = log $\left|cosx\right|.$

f (x) = $\frac{1}{cosx}\frac{d}{dx}x=-\frac{sinx}{cosx}=-tanx$

whenx $\left(0,\frac{\pi }{2}\right),$ we get.

tanx> 0 (Ist quadrant).

$\Rightarrow$ tanx< 0

$\Rightarrow$ f (x) < 0.

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

When x ∈$\left(\frac{3\pi }{2},2\pi \right)$ we get,

tanx|< |0 (ivth quadrant).

$\Rightarrow$ -tanx|>| 0

$\Rightarrow$ f (x) > 0

∴f (x) is increasing on $\left(\frac{3\pi }{2},2\pi \right).$

We have, f (x) = log sin x

So, f (x) = $\frac{1}{sinx}\frac{dsinx}{dx}=\frac{cosx}{sinx}=cotx$

When $x\in \left(0,\frac{\pi }{2}\right)$

f (x) = cot x> 0 (Ist quadrat )

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

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

f (x) = cot x< 0. (IIndquadrant ).

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

We have, f (x) = $\stackrel{?}{x}+\frac{1}{x}$

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

So, for every x∈ I, where I is disjoint from [-1,1]$$

f (x) = $\frac{(+ve)(+ve)}{(+ve)}=(+ve)>0whenx>1.$

and f (x) = $\frac{(-ve)(-ve)}{(+ve)}$ = ( + ve) > 0 when x< -1.

∴f (x) is strictly increasing on I .

We have, f (x) = x² + x + 1

So, f (x) = 2x + a

If f (x) is strictly increasing onx $(1,2),f^{\prime }(x)>0.$

So, 1

$\to 2*1<2*x<2*2$

$\Rightarrow 2<2x<4$

$\Rightarrow 2+a<2x+a<4+a.$

$\Rightarrow 2+a<f^{\prime }(x)<4+a.$

The minimum value of f (x) is 2 + a and that of men value is 4 + a.

∴ 2 +a> 0and 4 + a> 0

$\Rightarrow$ a> -2 and a> -4.

$\Rightarrow$ a> -2.

∴The best value of a is -2.

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^{°}*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 between -1 and 0.

So, f (x) = 100x⁹⁹ + cosx> 100 - 1 = 99 > 0.

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

(c) When x ∈$\left(0,\frac{\pi }{2}\right),$ we get,

$\Rightarrow$ x> 0

$\Rightarrow$ x⁹⁹> 0

$\Rightarrow$ 100x⁹⁹> 0

and cosx> 0. (firstquadrant).

I e, f(x) > 0.

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

Hence, option (D) is correct.

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 decreasing on ( -1,1).

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 $(0,\infty ).$