NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 10

NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 10 - Application of Derivatives

Edited By Ramraj Saini | Updated on Dec 03, 2023 08:43 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.3

NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 6 Application of Derivatives are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 6.3 Class 12 Maths chapter 6 is related to the topic normal and tangents. The equations of normals and tangent to a curve and questions related to these are discussed in exercise 6.3 Class 12 Maths. There are twenty-seven questions presented in the Class 12 Maths chapter 6 exercise 6.3. These questions in Class 12th Maths chapter 6 exercise 6.3 are solved by our Mathematics subject matter experts and are reliable and according to the CBSE pattern. The NCERT solutions for Class 12 Maths chapter 6 exercise 6.3 along with all other exercises of the NCERT chapter applications of derivatives gives a good idea of concepts discussed in the NCERT Book.

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12th class Maths exercise 6.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Application of Derivatives Class 12 Chapter 6 Exercise 6.3

Question:1 . Find the slope of the tangent to the curve $y = 3 x ^4 - 4x \: \: at \: \: x \: \: = 4$

Answer:

Given curve is,
$y = 3 x ^4 - 4x$
Now, the slope of the tangent at point x =4 is given by
$\left ( \frac{dy}{dx} \right )_{x=4} = 12x^3 - 4$
$= 12(4)^3-4$
$= 12(64)-4 = 768 - 4 =764$

Question:2 . Find the slope of the tangent to the curve $\frac{x-1}{x-2} , x \neq 2 \: \: at\: \: x = 10$

Answer:

Given curve is,

$y = \frac{x-1}{x-2}$
The slope of the tangent at x = 10 is given by
$\left ( \frac{dy}{dx} \right )_{x=10}= \frac{(1)(x-2)-(1)(x-1)}{(x-2)^2} = \frac{x-2-x+1}{(x-2)^2} = \frac{-1}{(x-2)^2}$
at x = 10
$= \frac{-1}{(10-2)^2} = \frac{-1}{8^2} = \frac{-1}{64}$
hence, slope of tangent at x = 10 is $\frac{-1}{64}$

Question:3 Find the slope of the tangent to curve $y = x ^3 - x +1$ at the point whose x-coordinate is 2.

Answer:

Given curve is,
$y = x ^3 - x +1$
The slope of the tangent at x = 2 is given by
$\left ( \frac{dy}{dx} \right )_{x=2} = 3x^2 - 1 = 3(2)^2 - 1= 3\times 4 - 1 = 12 - 1 = 11$
Hence, the slope of the tangent at point x = 2 is 11

Question:4 Find the slope of the tangent to the curve $y = x ^3 - 3x +2$ at the point whose x-coordinate is 3.

Answer:

Given curve is,
$y = x ^3 - 3x +2$
The slope of the tangent at x = 3 is given by
$\left ( \frac{dy}{dx} \right )_{x=3} = 3x^2 - 3 = 3(3)^2 - 3= 3\times 9 - 3 = 27 - 3 = 24$
Hence, the slope of tangent at point x = 3 is 24

Question:5 Find the slope of the normal to the curve $x = a \cos ^3 \theta , y = a\sin ^3 \theta \: \: at \: \: \theta = \pi /4$

Answer:

The slope of the tangent at a point on a given curve is given by
$\left ( \frac{dy}{dx} \right )$
Now,
$\left ( \frac{dx}{d \theta} \right )_{\theta=\frac{\pi}{4}} = 3a\cos^2 \theta(-\sin \theta) = 3a(\frac{1}{\sqrt2})^2(-\frac{1}{\sqrt2}) = -\frac{3\sqrt2 a}{4}$
Similarly,
$\left ( \frac{dy}{d \theta} \right )_{\theta=\frac{\pi}{4}} = 3a\sin^2 \theta(\cos \theta) = 3a(\frac{1}{\sqrt2})^2(\frac{1}{\sqrt2}) = \frac{3\sqrt2 a}{4}$
$\left ( \frac{dy}{dx} \right ) = \frac{\left ( \frac{dy}{d\theta} \right )}{\left ( \frac{dx}{d\theta} \right )} = \frac{\frac{3\sqrt2 a}{4}}{-\frac{3\sqrt2 a}{4}} = -1$
Hence, the slope of the tangent at $\theta = \frac{\pi}{4}$ is -1
Now,
Slope of normal = $-\frac{1}{slope \ of \ tangent}$ = $-\frac{1}{-1} = 1$
Hence, the slope of normal at $\theta = \frac{\pi}{4}$ is 1

Question:6 Find the slope of the normal to the curve $x = 1- a \sin \theta , y = b \cos ^ 2 \theta \: \: at \: \: \theta = \pi /2$

Answer:

The slope of the tangent at a point on given curves is given by
$\left ( \frac{dy}{dx} \right )$
Now,
$\left ( \frac{dx}{d \theta} \right )_{\theta=\frac{\pi}{2}} = -a(\cos \theta)$
Similarly,
$\left ( \frac{dy}{d \theta} \right )_{\theta=\frac{\pi}{2}} = 2b\cos \theta(-\sin \theta)$
$\left ( \frac{dy}{dx} \right )_{x=\frac{\pi}{2}} = \frac{\left ( \frac{dy}{d\theta} \right )}{\left ( \frac{dx}{d\theta} \right )} = \frac{-2b\cos \theta \sin \theta}{-a\cos \theta} = \frac{2b\sin \theta}{a} = \frac{2b\times1}{a} = \frac{2b}{a}$
Hence, the slope of the tangent at $\theta = \frac{\pi}{2}$ is $\frac{2b}{a}$
Now,
Slope of normal = $-\frac{1}{slope \ of \ tangent}$ = $-\frac{1}{\frac{2b}{a}} = -\frac{a}{2b}$
Hence, the slope of normal at $\theta = \frac{\pi}{2}$ is $-\frac{a}{2b}$

Question:7 Find points at which the tangent to the curve $y = x^3 - 3 x^2 - 9x +7$ is parallel to the x-axis.

Answer:

We are given :

$y = x^3 - 3 x^2 - 9x +7$

Differentiating the equation with respect to x, we get :

$\frac{dy}{dx}\ =\ 3x^2\ -\ 6x\ -\ 9\ +\ 0$

or $=\ 3\left ( x^2\ -\ 2x\ -\ 3 \right )$

or $\frac{dy}{dx}\ =\ 3\left ( x+1 \right )\left ( x-3 \right )$

It is given that tangent is parallel to the x-axis, so the slope of the tangent is equal to 0.

So,

$\frac{dy}{dx}\ =\ 0$

or $0\ =\ 3\left ( x+1 \right )\left ( x-3 \right )$

Thus, Either x = -1 or x = 3

When x = -1 we get y = 12 and if x =3 we get y = -20

So the required points are (-1, 12) and (3, -20).

Question:8 Find a point on the curve $y = ( x-2)^2$ at which the tangent is parallel to the chord joining the points (2, 0) and

(4, 4).

Answer:

Points joining the chord is (2,0) and (4,4)
Now, we know that the slope of the curve with given two points is
$m = \frac{y_2-y_1}{x_2 - x_1} = \frac{4-0}{4-2} = \frac{4}{2} =2$
As it is given that the tangent is parallel to the chord, so their slopes are equal
i.e. slope of the tangent = slope of the chord
Given the equation of the curve is $y = ( x-2)^2$
$\therefore \frac{dy}{dx} = 2(x-2) = 2$
$(x-2) = 1\\ x = 1+2\\ x=3$
Now, when $x=3$ $y=(3- 2)^2 = (1)^2 = 1$
Hence, the coordinates are (3, 1)

Question:9 Find the point on the curve $y = x^3 - 11x + 5$ at which the tangent is $y = x -11$

Answer:

We know that the equation of a line is y = mx + c
Know the given equation of tangent is
y = x - 11
So, by comparing with the standard equation we can say that the slope of the tangent (m) = 1 and value of c if -11
As we know that slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of curve is
$y = x^3 - 11x + 5$
$\frac{dy}{dx} = 3x^2 -11$
$3x^2 -11 = 1\\ 3x^2 = 12 \\ x^2 = 4 \\ x = \pm2$
When x = 2 , $y = 2^3 - 11(2) +5 = 8 - 22+5=-9$
and
When x = -2 , $y = (-2)^3 - 11(22) +5 = -8 + 22+5=19$
Hence, the coordinates are (2,-9) and (-2,19), here (-2,19) does not satisfy the equation y=x-11

Hence, the coordinate is (2,-9) at which the tangent is $y = x -11$

Question:10 Find the equation of all lines having slope –1 that are tangents to the curve $y = \frac{1}{x-1} , x \neq 1$

Answer:

We know that the slope of the tangent of at the point of the given curve is given by $\frac{dy}{dx}$

Given the equation of curve is
$y = \frac{1}{x-1}$
$\frac{dy}{dx} = \frac{-1}{(1-x)^2}$
It is given thta slope is -1
So,
$\frac{-1}{(1-x)^2} = -1 \Rightarrow (1-x)^2 = 1 = 1 - x = \pm 1 \\ \\ x = 0 \ and \ x = 2$
Now, when x = 0 , $y = \frac{1}{x-1} = \frac{1}{0-1} = -1$
and
when x = 2 , $y = \frac{1}{x-1} = \frac{1}{(2-1)} = 1$
Hence, the coordinates are (0,-1) and (2,1)
Equation of line passing through (0,-1) and having slope = -1 is
y = mx + c
-1 = 0 X -1 + c
c = -1
Now equation of line is
y = -x -1
y + x + 1 = 0
Similarly, Equation of line passing through (2,1) and having slope = -1 is
y = mx + c
1 = -1 X 2 + c
c = 3
Now equation of line is
y = -x + 3
y + x - 3 = 0

Question:11 Find the equation of all lines having slope 2 which are tangents to the curve $y = \frac{1}{x-3} , x \neq 3$

Answer:

We know that the slope of the tangent of at the point of the given curve is given by $\frac{dy}{dx}$

Given the equation of curve is
$y = \frac{1}{x-3}$
$\frac{dy}{dx} = \frac{-1}{(x-3)^2}$
It is given that slope is 2
So,
$\frac{-1}{(x-3)^2} = 2 \Rightarrow (x-3)^2 = \frac{-1}{2} = x-3 = \pm \frac{\sqrt-1}{\sqrt2} \\ \\$
So, this is not possible as our coordinates are imaginary numbers
Hence, no tangent is possible with slope 2 to the curve $y = \frac{1}{x-3}$

Question:12 Find the equations of all lines having slope 0 which are tangent to the curve
$y = \frac{1}{x^2 - 2 x +3 }$

Answer:

We know that the slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$

Given the equation of the curve as
$y = \frac{1}{x^2 - 2x + 3}$
$\frac{dy}{dx} = \frac{-(2x-2)}{(x^2-2x+3)^2}$
It is given thta slope is 0
So,
$\frac{-(2x-2)}{(x^2 - 2x +3)^2} = 0 \Rightarrow 2x-2 = 0 = x = 1$
Now, when x = 1 , $y = \frac{1}{x^2-2x+3} = \frac{1}{1^2-2(1)+3} = \frac{1}{1-2+3} =\frac{1}{2}$

Hence, the coordinates are $\left ( 1,\frac{1}{2} \right )$
Equation of line passing through $\left ( 1,\frac{1}{2} \right )$ and having slope = 0 is
y = mx + c
1/2 = 0 X 1 + c
c = 1/2
Now equation of line is
$y = \frac{1}{2}$

Question:13(i) Find points on the curve $\frac{x^2 }{9} + \frac{y^2 }{16} = 1$ at which the tangents are parallel to x-axis

Answer:

Parallel to x-axis means slope of tangent is 0
We know that slope of tangent at a given point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve is
$\frac{x^2 }{9} + \frac{y^2 }{16} = 1 \Rightarrow 9y^2 = 144(1-16x^2)$
$18y\frac{dy}{dx} = -32x$
$\frac{dy}{dx} = \frac{(-32x)}{18y} = 0 \Rightarrow x = 0$
From this, we can say that $x = 0$
Now. when $x = 0$ , $\frac{0^2 }{9} + \frac{y^2 }{16} = 1\Rightarrow \frac{y^2}{16} = 1 \Rightarrow y = \pm 4$
Hence, the coordinates are (0,4) and (0,-4)

Question:13(ii) Find points on the curve $\frac{x^2}{9} + \frac{y^2}{16} = 1$ at which the tangents are parallel to y-axis

Answer:

Parallel to y-axis means the slope of the tangent is $\infty$ , means the slope of normal is 0
We know that slope of the tangent at a given point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve is
$\frac{x^2 }{9} + \frac{y^2 }{16} = 1 \Rightarrow 9y^2 = 144(1-16x^2)$
$18y\frac{dy}{dx} = 144(1-32x)$
$\frac{dy}{dx} = \frac{-32x}{18y} = \infty$
Slope of normal = $-\frac{dx}{dy} = \frac{18y}{32x} = 0$
From this we can say that y = 0
Now. when y = 0, $\frac{x^2 }{9} + \frac{0^2 }{16} \Rightarrow 1 = x = \pm 3$
Hence, the coordinates are (3,0) and (-3,0)

Question:14(i) Find the equations of the tangent and normal to the given curves at the indicated
points:
$y = x^4 - 6x^3 + 13x^2 - 10x + 5 \: \: at\: \: (0, 5)$

Answer:

We know that Slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve
$y = x^4 - 6x^3 + 13x^2 - 10x + 5$
$\frac{dy}{dx}= 4x^3 - 18x^2 + 26x- 10$
at point (0,5)
$\frac{dy}{dx}= 4(0)^3 - 18(0)^2 + 26(0) - 10 = -10$
Hence slope of tangent is -10
Now we know that,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{-10} = \frac{1}{10}$
Now, equation of tangent at point (0,5) with slope = -10 is
$y = mx + c\\ 5 = 0 + c\\ c = 5$
equation of tangent is
$y = -10x + 5\\ y + 10x = 5$
Similarly, the equation of normal at point (0,5) with slope = 1/10 is
$\\y = mx + c \\5 = 0 + c \\c = 5$
equation of normal is
$\\y = \frac{1}{10}x+5 \\ 10y - x = 50$

Question:14(ii) Find the equations of the tangent and normal to the given curves at the indicated
points:
$y = x^4 - 6x^3 + 13x^2 - 10x + 5 \: \: at \: \: (1, 3)$

Answer:

We know that Slope of tangent at a point on given curve is given by $\frac{dy}{dx}$
Given equation of curve
$y = x^4 - 6x^3 + 13x^2 - 10x + 5$
$\frac{dy}{dx}= 4x^3 - 18x^2 + 26x - 10$
at point (1,3)
$\frac{dy}{dx}= 4(1)^3 - 18(1)^2 + 26(1) - 10 = 2$
Hence slope of tangent is 2
Now we know that,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{2}$
Now, equation of tangent at point (1,3) with slope = 2 is
y = 2x + 1
y -2x = 1
Similarly, equation of normal at point (1,3) with slope = -1/2 is
y = mx + c
$3 = \frac{-1}{2}\times 1+ c$
$c = \frac{7}{2}$
equation of normal is
$y = \frac{-1}{2}x+\frac{7}{2} \\ 2y + x = 7$

Question:14(iii) Find the equations of the tangent and normal to the given curves at the indicated
points:

$y = x^3\: \: at \: \: (1, 1)$

Answer:

We know that Slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve
$y = x^3$
$\frac{dy}{dx}= 3x^2$
at point (1,1)
$\frac{dy}{dx}= 3(1)^2 = 3$
Hence slope of tangent is 3
Now we know that,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{3}$
Now, equation of tangent at point (1,1) with slope = 3 is
$y = mx + c\\ 1 = 1 \times 3 + c\\ c = 1 - 3 = -2$
equation of tangent is
$y - 3x + 2 = 0$
Similarly, equation of normal at point (1,1) with slope = -1/3 is
y = mx + c
$1 = \frac{-1}{3}\times 1+ c$
$c = \frac{4}{3}$
equation of normal is
$y = \frac{-1}{3}x+\frac{4}{3} \\ 3y + x = 4$

Question:14(iv) Find the equations of the tangent and normal to the given curves at the indicated points

$y = x^2\: \: at\: \: (0, 0)$

Answer:

We know that Slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve
$y = x^2$
$\frac{dy}{dx}= 2x$
at point (0,0)
$\frac{dy}{dx}= 2(0)^2 = 0$
Hence slope of tangent is 0
Now we know that,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{0} = -\infty$
Now, equation of tangent at point (0,0) with slope = 0 is
y = 0
Similarly, equation of normal at point (0,0) with slope = $-\infty$ is

$\\y = x \times -\infty + 0\\ x = \frac{y}{-\infty}\\ x=0$

Question:14(v) Find the equations of the tangent and normal to the given curves at the indicated points:

$x = \cos t , y = \sin t \: \: at \: \: t = \pi /4$

Answer:

We know that Slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve
$x = \cos t , y = \sin t$
Now,
$\frac{dx}{dt} = -\sin t$ and $\frac{dy}{dt} = \cos t$
Now,
$\left ( \frac{dy}{dx} \right )_{t=\frac{\pi}{4}} = \frac{ \frac{dy}{dt}}{ \frac{dx}{dt}} = \frac{\cos t}{-\sin t} = -\cot t = =- \cot \frac{\pi}{4} = -1$
Hence slope of the tangent is -1
Now we know that,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{-1} = 1$
Now, the equation of the tangent at the point $t = \frac{\pi}{4}$ with slope = -1 is
$x= \cos \frac{\pi}{4} = \frac{1}{\sqrt2}$ and

$y= \sin \frac{\pi}{4} = \frac{1}{\sqrt2}$
equation of the tangent at

$t = \frac{\pi}{4}$ i.e. $\left ( \frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right )$ is

$y- y_1 = m(x-x_1)\\ y-\frac{1}{\sqrt2} = -1(x- \frac{1}{\sqrt2})\\ \sqrt2y + \sqrt2x = 2\\ y + x = \sqrt2$
Similarly, the equation of normal at $t = \frac{\pi}{4}$ with slope = 1 is
$x= \cos \frac{\pi}{4} = \frac{1}{\sqrt2}$ and

$y= \sin \frac{\pi}{4} = \frac{1}{\sqrt2}$
equation of the tangent at

$t = \frac{\pi}{4}$ i.e. $\left ( \frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right )$ is
$\\y- y_1 = m(x-x_1)\\ y-\frac{1}{\sqrt2} = 1(x- \frac{1}{\sqrt2})\\ \sqrt2y - \sqrt2x = 0\\ y - x = 0\\ x=y$

Question:15(a) Find the equation of the tangent line to the curve $y = x^2 - 2x +7$ which is parallel to the line $2x - y + 9 = 0$

Answer:

Parellel to line $2x - y + 9 = 0$ means slope of tangent and slope of line is equal
We know that the equation of line is
y = mx + c
on comparing with the given equation we get slope of line m = 2 and c = 9
Now, we know that the slope of tangent at a given point to given curve is given by $\frac{dy}{dx}$
Given equation of curve is
$y = x^2 - 2x +7$
$\frac{dy}{dx} = 2x - 2 = 2\\ \\ x = 2$
Now, when x = 2 , $y = (2)^2 - 2(2) +7 =4 - 4 + 7 = 7$
Hence, the coordinates are (2,7)
Now, equation of tangent paasing through (2,7) and with slope m = 2 is
y = mx + c
7 = 2 X 2 + c
c = 7 - 4 = 3
So,
y = 2 X x+ 3
y = 2x + 3
So, the equation of tangent is y - 2x = 3

Question:15(b) Find the equation of the tangent line to the curve $y = x^2 -2x +7$ which is perpendicular to the line $5y - 15x = 13.$

Answer:

Perpendicular to line $5y - 15x = 13.\Rightarrow y = 3x + \frac{13}{5}$ means $slope \ of \ tangent = \frac{-1}{slope \ of \ line}$
We know that the equation of the line is
y = mx + c
on comparing with the given equation we get the slope of line m = 3 and c = 13/5
$slope \ of \ tangent = \frac{-1}{slope \ of \ line} = \frac{-1}{3}$
Now, we know that the slope of the tangent at a given point to given curve is given by $\frac{dy}{dx}$
Given the equation of curve is
$y = x^2 - 2x +7$
$\frac{dy}{dx} = 2x - 2 = \frac{-1}{3}\\ \\ x = \frac{5}{6}$
Now, when $x = \frac{5}{6}$ , $y = (\frac{5}{6})^2 - 2(\frac{5}{6}) +7 = \frac{25}{36} - \frac{10}{6} + 7 = \frac{217}{36}$
Hence, the coordinates are $(\frac{5}{6} ,\frac{217}{36})$
Now, the equation of tangent passing through (2,7) and with slope $m = \frac{-1}{3}$ is
$y = mx+ c\\ \frac{217}{36}= \frac{-1}{3}\times \frac{5}{6} + c\\ c = \frac{227}{36}$
So,
$y = \frac{-1}{3}x+\frac{227}{36}\\ 36y + 12x = 227$
Hence, equation of tangent is 36y + 12x = 227

Question:16 Show that the tangents to the curve $y = 7x^3 + 11$ at the points where x = 2 and x = – 2 are parallel .

Answer:

Slope of tangent = $\frac{dy}{dx} = 21x^2$
When x = 2
$\frac{dy}{dx} = 21x^2 = 21(2)^{2} = 21 \times4 = 84$
When x = -2
$\frac{dy}{dx} = 21x^2 = 21(-2)^{2} = 21 \times4 = 84$
Slope is equal when x= 2 and x = - 2
Hence, we can say that both the tangents to curve $y = 7x^3 + 11$ is parallel

Question:17 Find the points on the curve $y = x ^3$ at which the slope of the tangent is equal to the y-coordinate of the point.

Answer:

Given equation of curve is $y = x ^3$
Slope of tangent = $\frac{dy}{dx} = 3x^2$
it is given that the slope of the tangent is equal to the y-coordinate of the point
$3x^2 = y$
We have $y = x ^3$
$3x^2 = x^3\\ 3x^2 - x^3=0\\ x^2(3-x)=0\\ x= 0 \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ x = 3$
So, when x = 0 , y = 0
and when x = 3 , $y = x^3 = 3^3 = 27$

Hence, the coordinates are (3,27) and (0,0)

Question:18 For the curve $y = 4x ^ 3 - 2x ^5$ , find all the points at which the tangent passes
through the origin.

Answer:

Tangent passes through origin so, (x,y) = (0,0)
Given equtaion of curve is $y = 4x ^ 3 - 2x ^5$
Slope of tangent =

$\frac{dy}{dx} = 12x^2 - 10x^4$
Now, equation of tangent is
$Y-y= m(X-x)$
at (0,0) Y = 0 and X = 0
$-y= (12x^3-10x^4)(-x)$
$y= 12x^3-10x^5$
and we have $y = 4x ^ 3 - 2x ^5$
$4x^3-2x^5= 12x^3-10x^5$
$8x^5 - 8x^3=0\\ 8x^3(x^2-1)=0\\ x=0\ \ \ \ \ \ and \ \ \ \ \ \ \ x = \pm1$
Now, when x = 0,

$y = 4(0) ^ 3 - 2(0) ^5 = 0$
when x = 1 ,

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

$y = 4(-1) ^ 3 - 2(-1) ^5 = -4-(-2)=-4+2=-2$
Hence, the coordinates are (0,0) , (1,2) and (-1,-2)

Question:19 Find the points on the curve $x^2 + y^2 - 2x - 3 = 0$ at which the tangents are parallel
to the x-axis.

Answer:

parellel to x-axis means slope is 0
Given equation of curve is
$x^2 + y^2 - 2x - 3 = 0$
Slope of tangent =
$-2y\frac{dy}{dx} = 2x -2\\ \frac{dy}{dx} = \frac{1-x}{y} = 0\\ x= 1$
When x = 1 ,

$-y^2 = x^2 -2x-3= (1)^2-2(1)-3 = 1-5=-4$
$y = \pm 2$
Hence, the coordinates are (1,2) and (1,-2)

Question:20 Find the equation of the normal at the point $( am^2 , am^3 )$ for the curve $ay ^2 = x ^3.$

Answer:

Given equation of curve is
$ay ^2 = x ^3\Rightarrow y^2 = \frac{x^3}{a}$
Slope of tangent

$2y\frac{dy}{dx} = \frac{3x^2 }{a} \Rightarrow \frac{dy}{dx} = \frac{3x^2}{2ya}$
at point $( am^2 , am^3 )$
$\frac{dy}{dx} = \frac{3(am^2)^2}{2(am^3)a} = \frac{3a^2m^4}{2a^2m^3} = \frac{3m}{2}$
Now, we know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-2}{3m}$
equation of normal at point $( am^2 , am^3 )$ and with slope $\frac{-2}{3m}$
$y-y_1=m(x-x_1)\\ y-am^3 = \frac{-2}{3m}(x-am^2)\\ 3ym - 3am^4 = -2(x-am^2)\\ 3ym +2x= 3am^4+2am^2$
Hence, the equation of normal is $3ym +2x= 3am^4+2am^2$

Question:21 Find the equation of the normals to the curve $y = x^3 + 2x + 6$ which are parallel
to the line $x + 14y + 4 = 0.$

Answer:

Equation of given curve is
$y = x^3 + 2x + 6$
Parellel to line $x + 14y + 4 = 0 \Rightarrow y = \frac{-x}{14} -\frac{4}{14}$ means slope of normal and line is equal
We know that, equation of line
y= mx + c
on comparing it with our given equation. we get,
$m = \frac{-1}{14}$
Slope of tangent = $\frac{dy}{dx} = 3x^2+2$
We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-1}{3x^2+2}$
$\frac{-1}{3x^2+2} = \frac{-1}{14}$
$3x^2+2 = 14\\ 3x^2 = 12 \\ x^2 = 4\\ x = \pm 2$
Now, when x = 2, $y = (2)^3 + 2(2) + 6 = 8+4+6 =18$
and
When x = -2 , $y = (-2)^3 + 2(-2) + 6 = -8-4+6 =-6$
Hence, the coordinates are (2,18) and (-2,-6)
Now, the equation of at point (2,18) with slope $\frac{-1}{14}$
$y-y_1=m(x-x_1)\\ y-18=\frac{-1}{14}(x-2)\\ 14y - 252 = -x + 2\\ x+14y = 254$
Similarly, the equation of at point (-2,-6) with slope $\frac{-1}{14}$

$y-y_1=m(x-x_1)\\ y-(-6)=\frac{-1}{14}(x-(-2))\\ 14y + 84 = -x - 2\\ x+14y + 86= 0$
Hence, the equation of the normals to the curve $y = x^3 + 2x + 6$ which are parallel
to the line $x + 14y + 4 = 0.$

are x +14y - 254 = 0 and x + 14y +86 = 0

Question:22 Find the equations of the tangent and normal to the parabola $y ^2 = 4 ax$ at the point $(at ^2, 2at).$

Answer:

Equation of the given curve is
$y ^2 = 4 ax$

Slope of tangent = $2y\frac{dy}{dx} = 4a \Rightarrow \frac{dy}{dx} = \frac{4a}{2y}$
at point $(at ^2, 2at).$
$\frac{dy}{dx}= \frac{4a}{2(2at)} = \frac{4a}{4at} = \frac{1}{t}$
Now, the equation of tangent with point $(at ^2, 2at).$ and slope $\frac{1}{t}$ is
$y-y_1=m(x-x_1)\\ y-2at=\frac{1}{t}(x-at^2)\\ yt - 2at^2 = x - at^2\\ x-yt +at^2 = 0$

We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = -t$
Now, the equation of at point $(at ^2, 2at).$ with slope -t
$y-y_1=m(x-x_1)\\ y-2at=(-t)(x-at^2)\\ y - 2at = -xt + at^3\\ xt+y -2at -at^3 = 0$
Hence, the equations of the tangent and normal to the parabola

$y ^2 = 4 ax$ at the point $(at ^2, 2at).$ are
$x-yt+at^2=0\ \ \ \ and \ \ \ \ xt+y -2at -at^3 = 0 \ \ respectively$

Question:23 Prove that the curves $x = y^2$ and xy = k cut at right angles* $if \: \: 8k ^ 2 = 1.$

Answer:

Let suppose, Curve $x = y^2$ and xy = k cut at the right angle
then the slope of their tangent also cut at the right angle
means,
$\left ( \frac{dy}{dx} \right )_a \times \left ( \frac{dy}{dx} \right )_b = -1$ -(i)
$2y\left ( \frac{dy}{dx} \right )_a = 1 \Rightarrow \left ( \frac{dy}{dx} \right )_a = \frac{1}{2y}$
$\left ( \frac{dy}{dx} \right )_b = \frac{-k}{x^2}$
Now these values in equation (i)
$\frac{1}{2y} \times \frac{-k}{x^2} = -1\\ -k = -2yx^2\\ k =2(xy)(x)\\ k = 2k(k^{\frac{2}{3}}) \ \ \ \ \left ( x = y^2 \Rightarrow y^2y = k \Rightarrow y = k^{\frac{1}{3}} \ and \ x = k^{\frac{2}{3}} \right ) \\ 2(k^{\frac{2}{3}}) = 1\\ \left ( 2(k^{\frac{2}{3}}) \right )^3 = 1^3\\ 8k^2 = 1$
Hence proved

Question:24 Find the equations of the tangent and normal to the hyperbola
$\frac{x^2 }{a^2} - \frac{y^2 }{b^2 }= 1$ at the point $(x_0 , y_0 )$

Answer:

Given equation is
$\frac{x^2 }{a^2} - \frac{y^2 }{b^2 }= 1 \Rightarrow y^2a^2 = x^2b^2 -a^2b^2$
Now ,we know that
slope of tangent = $2ya^2\frac{dy}{dx} = 2xb^2 \Rightarrow \frac{dy}{dx} = \frac{xb^2}{ya^2}$
at point $(x_0 , y_0 )$
$\frac{dy}{dx} = \frac{x_0b^2}{y_0a^2}$
equation of tangent at point $(x_0 , y_0 )$ with slope $\frac{xb^2}{ya^2}$
$y-y_1=m(x-x_1)\\ y-y_0=\frac{x_0b^2}{y_0a^2}(x-x_0)\\ yy_0a^2-y_0^2a^2 = xx_0b^2-x_0^2b^2\\ xx_0b^2 - yy_0a^2 = x_0^2b^2-y_0^2a^2$
Now, divide both sides by $a^2b^2$
$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = \left ( \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} \right )$
$=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left ( \because \frac{x_0^2}{a^2}-\frac{y_0^2}{b^2 } = 1\right )$
$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$
Hence, the equation of tangent is

$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$
We know that
$Slope \ of \ normal= \frac{-1}{slope \ of \ tangent } = -\frac{y_0a^2}{x_0b^2}$
equation of normal at the point $(x_0 , y_0 )$ with slope $-\frac{y_0a^2}{x_0b^2}$
$y-y_1=m(x-x_1)\\ y-y_0=-\frac{y_0a^2}{x_0b^2}(x-x_0)\\ \frac{y-y_0}{y_0a^2} + \frac{x-x_0}{x_0b^2} = 0$

Question:25 Find the equation of the tangent to the curve $y = \sqrt{3x-2}$ which is parallel to the line $4x - 2y + 5 = 0 .$

Answer:

Parellel to line $4x - 2y + 5 = 0 \Rightarrow y = 2x + \frac{5}{2}$ means the slope of tangent and slope of line is equal
We know that the equation of line is
y = mx + c
on comparing with the given equation we get the slope of line m = 2 and c = 5/2
Now, we know that the slope of the tangent at a given point to given curve is given by $\frac{dy}{dx}$
Given the equation of curve is
$y = \sqrt{3x-2}$
$\frac{dy}{dx} = \frac{1}{2}.\frac{3}{\sqrt{3x-2}}=\frac{3}{2\sqrt{3x-2}}$
$\frac{3}{2\sqrt{3x-2}} = 2\\ 3^2 = (4\sqrt{3x-2})^2\\ 9 = 16(3x-2)\\ 3x-2=\frac{9}{16}\\ 3x = \frac{9}{16} +2\\ 3x= \frac{41}{16}\\ x = \frac{41}{48}$
Now, when

$x = \frac{41}{48}$ , $y = \sqrt{3x-2} \Rightarrow y = \sqrt{3\times\frac{41}{48}-2 } = \sqrt{\frac{41}{16}-2}=\sqrt\frac{9}{16 } = \pm \frac{3}{4}$

but y cannot be -ve so we take only positive value
Hence, the coordinates are

$\left ( \frac{41}{48},\frac{3}{4} \right )$
Now, equation of tangent paasing through

$\left ( \frac{41}{48},\frac{3}{4} \right )$ and with slope m = 2 is
$y - y_1=m(x-x_1)\\ y-\frac{3}{4}=2(x-\frac{41}{48})\\ 48y-36=2(48x-41)\\ 48x-24y=41-18\\ 48x-24y=23$
Hence, equation of tangent paasing through $\left ( \frac{41}{48},\frac{3}{4} \right )$ and with slope m = 2 is 48x - 24y = 23

Question:26 The slope of the normal to the curve $y = 2x ^2 + 3 \sin x \: \: at \: \: x = 0$ is
(A) 3 (B) 1/3 (C) –3 (D) -1/3

Answer:

Equation of the given curve is
$y = 2x ^2 + 3 \sin x$
Slope of tangent = $\frac{dy}{dx} = 4x +3 \cos x$
at x = 0
$\frac{dy}{dx} = 4(0) +3 \cos 0= 0 + 3$
$\frac{dy}{dx}= 3$
Now, we know that
$Slope \ of \ normal = \frac{-1}{\ Slope \ of \ tangent} = \frac{-1}{3}$
Hence, (D) is the correct option

Question:27 The line $y = x+1$ is a tangent to the curve $y^2 = 4 x$ at the point
(A) (1, 2) (B) (2, 1) (C) (1, – 2) (D) (– 1, 2)

Answer:

The slope of the given line $y = x+1$ is 1
given curve equation is
$y^2 = 4 x$
If the line is tangent to the given curve than the slope of the tangent is equal to the slope of the curve
The slope of tangent = $2y\frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}$
$\frac{dy}{dx} = \frac{2}{y} = 1\\ y = 2$
Now, when y = 2, $x = \frac{y^2}{4} = \frac{2^2}{4} = \frac{4}{4} = 1$
Hence, the coordinates are (1,2)

Hence, (A) is the correct answer

Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3

Solving exercise 6.3 Class 12 Maths will be beneficial for both the CBSE board exam and JEE Main exam.
One question from NCERT solutions for Class 12 Maths chapter 6 exercise 6.3 can be expected for the CBSE Class 12 Maths board paper.
The applications of tangents will be used in Class 12 Physics and Chemistry also.

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Key Features Of NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 6

Comprehensive Coverage: The solutions encompass all the topics covered in ex 6.3 class 12, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 maths ex 6.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 ex 6.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 6.3 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
Inclusive Approach: Solutions for ex 6.3 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
Relevance to Curriculum: The solutions for class 12 maths ex 6.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What is the equation of the tangent to a curve y=f(x) at (x0,y0)?

y-y0=f’(x0)(x-x0)

2. Give the equation of normal to the tangent y-y0=f’(x0)(x-x0)

(y-y0)f’(x0)+(x-x0)=0

3. Equation of a tangent with slope=0 st (x0, y0) is

The slope =0. Therefore the equation of tangent is y=y0

4. The slope of a tangent is infinity, then what is its equation at (x0,y0)?

The equation of tangent with infinite slope is x=x0

5. Normal is ……………...to the tangent

Normal is perpendicular to the tangent

6. What is the relation between the slope of a tangent and the normal to the tangent?

The slope of normal is the negative of the inverse of the slope of the tangent.

7. Give the number of questions discussed in Exercise 6.3 Class 12 Maths

27 questions are answered in the NCERT Solutions for Class 12 Maths chapter 6 exercise 6.3. For more questions refer to NCERT exemplar. Following NCERT syllabus will be useful for CBSE board exams.

8. How many solved examples are given in the topic tangents and normals?

There are 7 solved examples explained before the NCERT book Class 12 Maths chapter 6 exercise 6.3.

Also Read

NCERT Class 12 Maths Notes - Download Chapter wise Free PDFs

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NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 10 - Application of Derivatives

NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.3

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