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The given eqⁿ of curve is $a{y}^2=x^3$ _____(1)

Differentiating eqⁿ (1) wrt.x we get,

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

$\Rightarrow \frac{dy}{dx}=\frac{3x^2}{2ay}$ , slope of tangent

${\frac{dy}{dx}|}_{(x,y)=\left(a{m}^2,a{m}^3\right)}=\frac{3{\left[a{m}^2\right]}^2}{2a\left[a{m}^3\right]}=\frac{3a^2{m}^4}{2a^2{m}^3}=\frac{3m}{2}$

corresponding slope of normal $=\frac{-1}{(3m/2)}=-\frac{2}{3m}$

Hence, eqⁿ of normal at $\left(a{m}^2,a{m}^3\right)$ is

$y-a{m}^3=-\frac{2}{3m}\left(x-a{m}^2\right).$

$\Rightarrow 3my-3a{m}^4=-2x+2a{m}^2$

$\Rightarrow 2x+3my-3a{m}^4-2a{m}^2=0$

$\Rightarrow 2x+3my-a{m}^2\left(2+3m^2\right)=0$

The given eqⁿof the curve is $x^2+y^2-2x-3=0$ ________ (1)

Differentiating the given curve wrt.x we get,

$2x+2y\frac{dy}{dx}-2=0.$

$\Rightarrow 2y\frac{dy}{dx}=2-2x$

$\to \frac{dy}{dx}=\frac{1-x}{y}$ slope of tangent

Given, tangent is | to x-axis

ie,  $\frac{dy}{dx}=0$

$\Rightarrow \frac{1-x}{y}=0$

$\Rightarrow x=1$

Putting x = 1 in eqⁿ (1) we get,

$1^2+y^2-2\left(1_i\right)-3=0$

$\Rightarrow y^2-4=0$

$\Rightarrow y^2=4$

$\Rightarrow y=\pm 2$

Hence, the required points are (1,2) and (1, 2).

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This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (ii) is the correct answer.

A redox couple is defined as pair of compounds or elements having together the oxidised and reduced forms of it and taking part in an oxidation or reduction half reaction.

The given eqⁿ of the curve is $y=4x^3-2x^5$

Slope of tangent, $\frac{dy}{dx}=12x^2-10x^4.$ ________(1)

Let P(x, y) be the required point at the tangent passing through the origin (0,0)

Then, $\frac{dy}{dx}=\frac{y-0}{x-0}=\frac{y}{x}$ _________(2)

So, from (1) and (2) we get,

$\frac{y}{x}=12x^2-10x^4.$

$\Rightarrow y=12x^3-10x^5.$

Putting this value of y in the eqⁿ of curve we get,

$12x^3-10x^5=4x^3-2x^5.$

$\to 8x^3-8x^5=0$

$\Rightarrow 8x^3\left(1-x^2\right)=0$

$\Rightarrow x^3=0$ or $1-x^2=0$

$\Rightarrow x=0$ or $x=\pm 1.$

When, $x=0,y=4{(0)}^3-2{(0)}^5=0$

$x=1,y=4{(1)}^3-2{(1)}^5=4-2=2$

$x=-1,y=4{(-1)}^3-2{(-1)}^5=-4+2=-2$

The required points are (0,0), (1,2) and (1,2)

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

Slope of tangent,  $\frac{dy}{dx}=3x^2$

As, slope of tangent = y – coordinate of the point.

$\frac{dy}{dx}=y$

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

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

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

$\Rightarrow x=0x=3$

When $x=0,y=0$

and when $x=3·,y=3^3=27.$

The required points are $(0,0)and(3,27)$

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (i) is the correct answer.

Here the oxygen of peroxide, which is present in -1 state, is converted to zero oxidation state in O₂ undergoing oxidation and decreases to -2 oxidation state in H₂O undergoing reduction

The given eqⁿ of the curve is $y=7x^3+11$ .

Slope of tangent $\frac{dy}{dx}=21x^2$

${\frac{dy}{dx}|}_{x=2}=21{(2)}^2=21\times 4=84.$

and ${\frac{dy}{dx}|}_{x=-2}=21{(-2)}^2=21\times 4=84$

The tangent to the given curve at x = 2 and x = -2 are parallel.

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This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (iii) is the correct answer

As permanganate ion changes to MnO₂

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The eqⁿof the given curve is $y=x^2-2x+7$

Slope of tangent, $\frac{dy}{dx}=2x-2$

(a) The line $2x-y+9=0\Rightarrow y=2x+9$ compared to $y=mx+c$ gives,

Slope of line = 2.

If the tangent of the curve is parallel to the line

$\frac{dy}{dx}=slope\; of\; line$

$\to 2x-2=2$

$\to x=\frac{4}{2}\Rightarrow x=2$

When $x=2,y={(2)}^2-2(2)+7=7$

Hence, the point of contact of the tangent is (2, 7)

The eqⁿ of tangent is $y-7=2(x-2)$

$\Rightarrow y-7=2x-4$

$\to 2x-y+3=0$

(b) The line $5y-15x=13\to y=\frac{15}{5}x+\frac{13}{5}\Rightarrow y=3x+\frac{13}{5}$

compared to $y=mx+c$ gives

slope of line = 3

As the tangent to the curve is ⊥ to the line.

$\frac{dy}{dx}=\; -1/slope\; opf\; line$

$\Rightarrow 2x-2=-\frac{1}{3}$

$\Rightarrow 6x-6=-1$

$\to 6x=5$

$\to x=\frac{5}{6}$

When $x=\frac{5}{6}$ we get $y={\left(\frac{5}{6}\right)}^2-2\times \frac{5}{6}+7$

$=\frac{25}{36}-\frac{5}{3}+7$

$=\frac{25-60+252}{36}=\frac{217}{36}$

Hence, the point of contact of the tangent is $\left(\frac{5}{6},\frac{217}{36}\right)$

And eqⁿ of the tangent is

$y-\frac{217}{36}=\frac{-1}{3}\left(x-\frac{5}{6}\right)$

$\Rightarrow 3y-\frac{217}{12}=-x+\frac{5}{6}$

$\Rightarrow x+3y-\frac{217}{12}-\frac{5}{6}=0$

$\Rightarrow x+3y-\frac{217-10}{12}=0$

$\Rightarrow x+3y-\frac{227}{12}=0$

$\Rightarrow 12x+36y-227=0.$

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (i) is the correct answer.

Fluorine is most electronegative element that is why it is best oxidant among halogens