NCERT Solutions for Maths Class 12 Exercise 6.3

NCERT Solutions for Maths Class 12 Exercise 6.3


        NCERT Solutions for Maths Class 12 Exercise 6.3

 

Hello Students! Welcome to maths-formula.com. In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 6.3.

 

You can download the PDF of NCERT Books Maths Chapter 6 for your easy reference while studying NCERT Solutions for Maths Class 12 Exercise 6.3.

 

Class 12th is a very crucial stage of your student’s life, since you take all important decisions about your career on this stage. Mathematics plays a vital role to take decision for your career because if you are good in mathematics, you can choose engineering and technology field as your career.

 

NCERT Solutions for Maths Class 12 Exercise 6.3 helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.

 

All the schools affiliated with CBSE, follow the NCERT books for all subjects. You can check your syllabus from NCERT Syllabus for Mathematics Class 12.

 

NCERT Solutions for Maths Class 12 Exercise 6.3 are prepared by the experienced teachers of CBSE board. If you are preparing for JEE Mains and NEET level exams, then it will definitely make your foundation strong.

 

If you want to recall All Maths Formulas for Class 12, you can find it by clicking this link.

If you want to recall All Maths Formulas for Class 11, you can find it by clicking this link.


NCERT Solutions for Maths Class 12 Exercise 6.1

NCERT Solutions for Maths Class 12 Exercise 6.2

NCERT Solutions for Maths Class 12 Exercise 6.4

NCERT Solutions for Maths Class 12 Exercise 6.5

 

NCERT Solutions for Maths Class 12 Exercise 6.3

 

Maths Class 12 Ex 6.3 Question 1.
Find the slope of the tangent to the curve y = 3x4 – 4x at x = 4.

Solution:
The given curve is y = 3x4 – 4x
dy/dx = 12x3 – 4
Slope of the tangent to the curve = (dy/dx)x = 4
                                                                = 12 × 43 – 4 = 764

Maths Class 12 Ex 6.3 Question 2.
Find the slope of the tangent to the curve  at x = 10.


Solution:
The given curve is 



Maths Class 12 Ex 6.3 Question 3.
Find the slope of the tangent to curve y = x3 – x + 1 at the point whose x-coordinate is 2.

Solution:
The given curve is y = x3 – x + 1
dy/dx = 3x² – 1
Slope of the tangent to the curve = (dy/dx)x = 2
                                                                = 3 × 2² – 1
                                                                = 11

Maths Class 12 Ex 6.3 Question 4.
Find the slope of the tangent to the curve y = x3 – 3x + 2 at the point whose x-coordinate is 3.

Solution:
The given curve is y = x3 – 3x + 2
dy/dx = 3x² – 3
Slope of the tangent to the curve = (dy/dx)x = 3
                                                                = 3 × 3² – 3
                                                                = 24

Maths Class 12 Ex 6.3 Question 5.
Find the slope of the normal to the curve x = a cos3 θ, y = a sin3 θ at θ = π/4.

Solution:

The given curve is x = a cos3 θ, y = a sin3 θ


Maths Class 12 Ex 6.3 Question 6.

Find the slope of the normal to the curve x = 1 – a sin θ, y = b cos² θ at θ = π/4.

Solution:

The given curve is x = 1 – a sin θ, y = b cos² θ


Maths Class 12 Ex 6.3 Question 7.

Find points at which the tangent to the curve y = x3 – 3x2 – 9x + 7 is parallel to the x-axis.

Solution:
Differentiating w.r.t. x; dy/dx = 3(x – 3) (x + 1)
Tangent is parallel to x-axis if the slope of tangent = 0
or dy/dx = 0
3(x + 3)(x + 1) = 0
x = –1, 3
When x = –1, y = 12 and when x = 3, y = –20
Hence, the tangent to the given curve are parallel to x-axis at the points (–1, 12), (3, –20)

Maths Class 12 Ex 6.3 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).

Solution:
The given equation of the curve is y = (x – 2)²
Differentiating w.r.t x, we get dy/dx = 2(x – 2)
The points A and B are (2, 0) and (4, 4), respectively.

Slope of AB = (y2 – y1)/(x2 – x1) = (4 – 0)/(4 – 2) = 4/2 = 2    … (i)
Slope of the tangent = 2(x – 2)    … (ii)
From eq. (i) and (ii), 2(x – 2) = 2


x – 2 = 1 or x = 3
when x = 3, y = (3 – 2)² = 1
The tangent is parallel to the chord AB at (3, 1).

Maths Class 12 Ex 6.3 Question 9.
Find the point on the curve y = x3 – 11x + 5 at which the tangent is y = x – 11.

Solution:
The given curve is y = x3 – 11x + 5
⇒ dy/dx = 3x² – 11
The slope of tangent line y = x – 11 is 1.
3x² – 11 = 1
3x² = 12
x² = 4, x = ±2
When x = 2, y = –9 and when x = –2, y = 19
But (–2, 19) does not lie on the curve.
y = x – 11 is the tangent at (2, –9).

Maths Class 12 Ex 6.3 Question 10.
Find the equation of all lines having slope –1 that are tangents to the curve y = 1/(x – 1), x ≠ 1.

Solution:
The given curve is y = 1/(x – 1), x ≠ 1

The point where the tangents to the given curve have the slope -1 are (2, 1) and (0, -1).

Equation of the tangent at (2, 1) is

Y – 1 = -1(x – 2) or x + y = 3

Equation of the tangent at (0, -1) is

Y + 1 = -1(x – 0) or x + y + 1 = 0

Thus, the required lines are x + y = 3 and x + y + 1 = 0.

 

Maths Class 12 Ex 6.3 Question 11.
Find the equation of all lines having slope 2 which are tangents to the curve y = 1/ (x – 3), x ≠ 3.

Solution:
The given curve is y = 1/(x – 3), x ≠ 3
dy/dx = 1/(x – 3)2
Since slope of the tangent = 2
1/(x – 3)2 = 2 or (x – 3)2 = –1/2  

Which is not possible as (x – 3)² > 0
Thus, no tangent to y = 1/(x – 3) has slope 2.

Maths Class 12 Ex 6.3 Question 12.
Find the equations of all lines having slope 0 which are tangent to the curve y = 1/(x2 – 2x + 3).

Solution:
Let the tangent at the point (x1, y1) to the curve


Maths Class 12 Ex 6.3 Question 13.
Find points on the curve x2/9 + y2/16 = 1 at which the tangents are
(a) parallel to x-axis
(b) parallel to y-axis

Solution:
The equation of the curve is 

Therefore, the tangents are parallel to y-axis at (3, 0) and (–3, 0).

Maths Class 12 Ex 6.3 Question 14.
Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)
(ii) y = x4 – 6x3 + 13x2 – 10x + 5 at (1, 3)
(iii) y = x3 at (1, 1)
(iv) y = x2 at (0, 0)
(v) x = cos t, y = sin t at t = π/4 

Solution:
The given curve is y = x4 – 6x3 + 13x2 – 10x + 5

dy/dx = 4x3 – 18x2 + 26x – 10
Putting x = 0, dy/dx at (0, 5) = –10


Maths Class 12 Ex 6.3 Question 15.
Find the equation of the tangent line to the curve y = x2 – 2x + 7 which is
(a) parallel to the line 2x – y + 9 = 0
(b) perpendicular to the line 5y – 15x = 13.

Solution:
The given curve is y = x² – 2x + 7       … (i)
dy/dx = 2x – 2 = 2(x – 1)
(a) Slope of the line 2x – y + 9 = 0 is 2
Slope of tangent = dy/dx = 2(x – 1) = 2


Maths Class 12 Ex 6.3 Question 16.
Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = – 2 are parallel.

Solution:
The given curve is y = 7x3 + 11
dy/dx = 21x²
Now m1 = slope at x = 2 is (dy/dx)x = 2 = 21 × 2² = 84
And m2 = slope at x = -2 is (dy/dx)x = -2 = 21 × (-2)² = 84
Here, m1 = m2 

Thus, the tangents to the given curve at the points where x = 2 and x = –2 are parallel.

Maths Class 12 Ex 6.3 Question 17.
Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinate of the point.

Solution:
Let P(x1, y1) be the required point.
The given curve is y = x3      … (i)


Maths Class 12 Ex 6.3 Question 18.
For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.

Solution:
Let (x1, y1) be the required point on the given curve y = 4x3 – 2x5,

then y1 = 4x13 – 2x15        … (i)


Maths Class 12 Ex 6.3 Question 19.
Find the points on the curve x2 + y2 – 2x – 3 = 0 at which the tangents are parallel to the x-axis.

Solution:
The given curve is x2 + y2 – 2x – 3 = 0
dy/dx = (1 – x)/y
Tangent is parallel to x-axis, if dy/dx = 0, i.e. if 1 – x = 0
x = 1
Putting x = 1 in (i), we get y = ±2
Hence, the required points are (1, 2), (1, –2).

Maths Class 12 Ex 6.3 Question 20.
Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3.

Solution:
The given curve is ay2 = x3


Maths Class 12 Ex 6.3 Question 21.

Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Solution:
Let the required normal be drawn at the point (x1, y1).
The equation of the given curve is y = x3 + 2x + 6   … (i)


Maths Class 12 Ex 6.3 Question 22.

Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at², 2at).

Solution:

Maths Class 12 Ex 6.3 Question 23.

Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1.

Solution:
The given curves are x = y²     … (i)
And xy = k        … (ii)


Maths Class 12 Ex 6.3 Question 24.

Find the equations of the tangent and normal to the hyperbola x2/a2 – y2/b2 = 1 at the point (x0 ,y0).

Solution:


Maths Class 12 Ex 6.3 Question 25.

Find the equation of the tangent to the curve  which is parallel to the line 4x – 2y + 5 = 0.

Solution:

Let the point of contact of the tangent line parallel to the given line be P(x1, y1).

The equation of the curve is 


Choose the correct answer in Exercises 26 and 27.

Maths Class 12 Ex 6.3 Question 26.

The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is
(A) 3
(B) 1/3  
(C) 3
(D) –1/3

Solution:
(D) The given curve is y = 2x² + 3 sin x
∴ dy/dx = 4x + 3 cos x 

At x = 0, dy/dx = 3
Slope of the tangent = 3
Slope of the normal is –1/3.

Hence, the correct option is (D).

Maths Class 12 Ex 6.3 Question 27.

The line y = x + 1 is a tangent to the curve y² = 4x at the point
(A) (1, 2)
(B) (2, 1)
(C) (1, 2)
(D) (1, 2)  

Solution:
(A) The given curve is y² = 4x
 dy/dx = 4/2y = 2/y
Slope of the given line y = x + 1 is 1

∴ 2/y = 1 ⇒ y = 2

Putting y = 2 in y² = 4x, we get 2² = 4x x = 1
Point of contact is (1, 2).

Hence, the correct option is (A).

 NCERT Solutions for Maths Class 12 Exercise 6.1

NCERT Solutions for Maths Class 12 Exercise 6.2

NCERT Solutions for Maths Class 12 Exercise 6.4

NCERT Solutions for Maths Class 12 Exercise 6.5

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