Ncert Solutions Maths class 12th

The given eqⁿ of the curve is $y=x^3-3x^2-9x+7.$

slope of tangent to the given curve, $\frac{dy}{dx}=3x^2-6x-9$

when the tangent is parallel to x-axis $\frac{dy}{dx}=0$

$\Rightarrow 3x^2-6x-9=0$

$\Rightarrow x^2-2x-3=0$

$\Rightarrow x^2+x-3x-3=0$

$\to x(x+1)-3(x+1)=0$

$\to (x+1)(x-3)=0$

$$ x = 3 or x = -1

When x = 3, $y=3^3-3{(3)}^2-9(3)+7=27-27-27+7=-20$

And when x = -1 $y={(-1)}^3-3{(-1)}^2-9(-1)+7=-1-3+9+7=12$

Hence, the required points are $(3,-20)(-1,12)$

The given eqⁿ of the curves are

$x=1-asin\theta y=bco{s}^2\theta$

so,  $\frac{dx}{d\theta }=-acos\theta \frac{dy}{d\theta }=-2bcos\theta sin\theta$

$\therefore \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{-2bcos\theta sin\theta }{acos\theta }=\frac{2b}{a}sin\theta$

Slope of tangent to curve at $\theta =\frac{\pi }{2}$ is ${\frac{dy}{dx}|}_{\theta =\frac{\pi }{2}}$

$=\frac{2b}{a}sin\frac{\pi }{2}$

$=\frac{2b}{a}$

Hence, slope of normal to curve $=\frac{-1}{2b/a}=-\frac{a}{2b}$

The Equation of the given curve are

$x=aco{s}^3\theta y=asi{n}^3\theta$

So,  $\frac{dx}{d\theta }=3aco{s}^2\theta (-sin\theta )=-3aco{s}^2\theta sin\theta .$

and $\frac{dy}{d\theta }=3asi{n}^2\theta cos\theta$

$\therefore \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{3asi{n}^2\theta cos\theta }{-3aco{s}^2\theta sin\theta }=-tan\theta$

So,  ${\frac{dy}{dx}|}_{x=\pi /4}=-tan\frac{\pi }{4}=-1$ which is the slope of the tanget to the curve.

Now, required slope of normal to the curve =$-1$
$\frac{\mathrm{Slopeoftangent}}{}$ 
$\frac{}{}=\frac{-1}{-1}=1$

Slope of tangent to the given curve $y=x^3-3x+2$ is $\frac{dy}{dx}=3x^2-3.$

so,  ${\frac{dy}{dx}|}_{x=3}=3{(3)}^2-3=27-3=24.$

Slope of tangent to the given curve $y=x^3-x+1$ is

$\frac{dy}{dx}=3x^2-1.$

So,  ${\frac{dy}{dx}|}_{x=2}=3{(2)}^2-1=12-1=11.$

The given eqⁿ of the curve is $y=\frac{x-1}{x-2}$

Slope of tangent at x = 10 is given by,

${\frac{dy}{dx}|}_{x=10}={\frac{(x-2)\frac{d}{dx}(x-1)-(x-1)\frac{d}{dx}(x-2)}{{(x-2)}^2}|}_{x=10}$

$={\frac{(x-2)-(x-1)}{{(x-2)}^2}|}_{x=10}={\frac{x-2-x+1}{{(x-2)}^2}|}_{x=10}$

$={\frac{-1}{{(x-2)}^2}|}_{x=10}$

$=\frac{-1}{{(10-2)}^2}=\frac{-1}{8^2}=-\frac{1}{64}$

The given eqⁿ of the curve is $y=3x^4-4x.$

Slope of the tangent at x = 4 is given by

${{\frac{dy}{dx}]}_{x=4}=12x^3-4]}_{x=4}=12{(4)}^3-4=12*64-4$

$=768-4$

= 764

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 $\left(\frac{3\pi }{2},2\pi \right).$