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RD Sharma Solutions Class 12 Mathematics Chapter 15 MCQ

RD Sharma Solutions Class 12 Mathematics Chapter 15 MCQ

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 01:20 PM IST

The RD Sharma solutions are the most prescribed solution books by the CBSE schools to their students. Students would not face challenges in doing their mathematics sums when they possess this set of books. Especially when it comes to chapter 15 MCQs, students tend to confuse the options. The RD Sharma Class 12th MCQ will lend a helping hand in those circumstances.

This Story also Contains
  1. RD Sharma Class 12 Solutions Chapter 15 Tangents and Normals - Other Exercise
  2. Tangents and Normals Excercise: MCQ
  3. RD Sharma Chapter wise Solutions

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 15 Tangents and Normals - Other Exercise

Chapter 15 - Tangents and Normals Ex 15.1

Chapter 15 - Tangents and Normals Ex 15.2

Chapter 15 - Tangents and Normals Ex 15.3

Chapter 15 -Tangents and Normals Ex-FBQ

Chapter 15 -Tangents and Normals Ex-VSA

Tangents and Normals Excercise: MCQ

Tangents and Normals Exercise Multiple Choice Questions Question 1

Answer:
(c)x+y=0
Hint:
Use differentiation
Given:
y=sinx
Solution:
y=sinx
dydx=cosx
dydx|(0,0)=cos0=1
Equation of normal: slop of normal
=1dydx=1
Equation of normal at (0,0)
y0=1(x0)
y=x
y+x=0
x+y=0 Is the requried
Equation of normals to curve y=sinx at (0,0)

Tangents and Normals Exercise Multiple Choice Questions Question 2

Answer:
(d)2x=π
Hint:
Use differentiation
Given:
y=x+sinxcosx
Solution:
y=x+sinxcosx=x+sin(2x)2y=1+cos(2x) Substituting x=π2y(π2)=1+cos(π)=0
Therefore, slope of tangent is 0. this means that the normal is a vertical line of the x=c
Since the point has x-coordinateπ2 , the equation of the line is
x=π2
2x=π is requried

Tangents and Normals Exercise Multiple Choice Questions Question 3

Answer:
(a)x2y=2
Hint:
Use slope of tangent m=y2y1x2x1
Given:
y=x(2x)
Solution:

T he given curve is y=2xx2
i.e. x22x+y=0
Now the equation of the tangent at the point (x1,x2)=(2,0)
Or
xx1(x+x1)+12(y+y1)=02x+(1)(x+2)+12(y+0)=02x4+y=0y=2x+4
So, the slope of the tangent is (2)
Therefore the slope of the normal=12
So, the equation of the normal at the point (2,0)is
y0=12(x2)
2y=x2
I.e. x2y=2 is required

Tangents and Normals Exercise Multiple Choice Questions Question 4

Answer:
(b)(14,12)
Hint:
Use slope of tangent m=tanθ
Given:
y2=x
Solution:
dydx Slope of the tangent at that point of the curve
Given curve is x=y2
Point (a2,a)
ddx(y2x)=0dydx=12y=12a=tan(45)a=12,a2=14
Point (14,12) is requried.


Tangents and Normals Exercise Multiple Choice Questions Question 6

Answer:
(b)(1,0)
Hint:
Use differentiation
Given:
y=x23x+2
y=x
Solution:
Differentiate both the side w.r.t x we get
dydx=1
Let (x,y)be the required point
It is given that point lies on the curve
y=x123x1+2y=x23x+2
Differentiating both the sides w.r.t x, we get
dydx=2x3
Slope of the tangent=(dydx)(x1,y1)=2x13
The tangent is perpendicular to the line
Slope of the tangent =-1/slope of the line=11=1
Now,
2x13=12x1=2x1=1y1=x123x1+2y1=(1)23(1)+2y1=13+2y1=(x1,y1)=(1,0)

Tangents and Normals Exercise Multiple Choice Questions Question 7

Answer:
(b)(14,12)
Hint:
Use slop of the tangent m=tanθ
Given:
The curve y2=x
Solution:
dydx=Slope of the tangent at that point of the curve
Given curve is x=y2
Keep y=athen x=y2
Point (a2,a)
dydx(y2x)=0dydx=12y=12a=tan(45)a=12,a2=14
Point(14,12)

Tangents and Normals Exercise Multiple Choice Question Question 9

Answer:
(b)tan134

Hint:

Use slop of the tangent m=tanθ
Given:
The curve y2=x and x2=y2at (1,1)
Solution:

Slope of the tangent to curve
y=x2At (1,1)is m1=2
Slope of the tangent to the curve
x=y2At (1,1)is m2=12
Angle between tangents to the curve is tanθ=34
Or θ=tan134


Tangents and Normals Exercise Multiple Choice Question Question 10

Answer:
 (c) x+3y±8=0
Hint:
Use differentiation
Given:
3x2y2=8Is parallel tox+3y=8
Solution:
Given the equation of the line 3x2y2=8 now differentiating both sides with respect to x, we getd(3x2y2)dx=d(8)dx

Now applying the sum rule of differentiation an differentiation of constant =0,so we get

d(3x2)dxd(y2)dx=03d(x2)dxd(y2)dx=03×2(x21)×d(x)dx2(y21)×d(y)dx=06x2y×dydx=06x=2y×dydxdydx=6x2y=3xy

So, this is the slope of the given curve. We know the slope of the normal to the curve is

=1dydx

=13xy=(y3x) (1)

Now the given equation of the line x+3y=8

3y=8x

Differentiating w.r.t x we get

d(3y)dx=d(8x)dx3dydx=1dydx=13
So, the slope of the line is 13
Now, as the normal to the curve is parallel to this line, hence the slope of the line should be equal to the slope of the normal to the given curve,
(y3x)=133y=3xy=x

On substituting this value of the given equation of the curve, we get

3x2y2=83x2(x)2=82x2=8x2=4x=±2

When x=2 the equation of the curve becomes,

3x2y2=8

3(2)2y2=8

3(4)y2=8


128=y2

y2=4

y=±2

When x=-2, the equation of the curve becomes,

3x2y2=83(2)2y2=83(4)y2=8128=y2y2=4y=±2
So, the points at which normal is parallel to the given line are (±2,±2)
And required equation of the normal to the curve at (±2,±2) is
y(±2)=13[x(±2)]3(y(±2))=(x(±2))3y(±6)=x+(±2)x+3y(±6)(±2)=0x+3y+(±8)=0
Hence the equation of normal to the curve
3x2y2=8 Which is parallel to the line x+3y=8 is x+3y+(±8)=0


Tangents and Normals Exercise Multiple Choice Questions Question 15

Answer:
(c)6
Hint:
Use slope of the tangent m=y2y1x2x1
Given:
ay+x2=7 And x3=y
Solution:
Slope of the tangent at (1,1) to ay+x2=7 is m1=2a
Slope of the tangent at (1,1) to x3=y is m2=3
Curve cuts through orthogonally m1×m2=3
m1×m2=2a3=1a=6

Tangents and Normals Exercise Multiple Choice Question Question 16

Answer:
(b)b=1,c=1
Hint:
Use y=mx+c
Given:
y=x Touches the curve y=x2+bx+cat the point (1,1)
Solution:
y=x
Clearly ,m=1
y=x2+bx+c (1)

Differentiate w.r.t, x1we get

dydx=2x+b
Putting the point (1,1)and m=1 in above curve
1=2(1)+b
b=1
Putting the point(1,1) in (1)
1=(1)2+b(1)+c1=11+c [Since b=1]c=1b=1,c=1

Tangents and Normals Exercise Multiple Choice Questions Question 17

Answer:
(b) Is the correct option
Hint:
Substituting x=1in x=3t21
Given:
x=3t2+1,y=t31
Solution:
x=3t2+1, (1)

Differentiate w.r.t t

dxdt=6t And, y=t31 Differentiating w.r.t t dydx=3t2

 Substituting x=1 in (1) 1=3t2+13t2=0t=0dydx=dydtdxdt=3t26t=t2[dydx]t=0=0 slope =0

Tangents and Normals Exercise Multiple Choice Question Question 18

Answer:
(c) Is the correct option
Hint:
Use slope(1)×slope(2)=1
Given:
y=2ex,y=bex
Solution:
y=aex (1)
Differentiating w.r.t x, we get
dydx=aex
And y=bex (2)
Since equation (1) and equation (2) intersect orthogonally
 slope (1)×slope(2)=1aex×bex=1ab=1

Tangents and Normals Exercise Multiple Choice Questions Question 19

Answer:
(c) Is the correct option
Hint:
Equation of normal yy1=m(xx1)
Given:
x=acos3θ,y=asin3θ
Solution:
x=acos3θ (1)
Differentiating w.r.t θ
dxdθ=a3cos2θ(sinθ)=3acos2θsinθdxdθ|θπ4=3a×12×12=3a22y=asin3θ
Differentiating w.r.t θ
dydθ=a×3sin2θcosθdydθ|θπ4=3a×12×12=3a22
Slope of normal of the given curve M(N)=dxdy=(dxdθdydθ)=(3α223α22)=1
Equation of normal at θ=π4
ya(2)3=1(x(a2)3)y=x

Tangents and Normals Exercise 15 point 1 Question 20

Answer:
(a) 12Is the correct option
Hint:
slope(1)×slope(2)=1
Given:
y=2ex And y=aex
Solution:
y=2ex (1)
Differentiating w.r.t x, we get
dydx=2ex
Again y=aex (2)
Differentiating w.r.t x we get
dydx=aex
Since (1) and (2) intersect orthogonally
 slope (1)×slope(2)=1aex×aex=12a=1a=12

Tangents and Normals Exercise Multiple Choice Question , Question 21

Answer : (b) is the correct option
Hint : tanθ=|m111m1|
Given : y=6xx2
Solution :
y=6xx2 (1)
Differentiating (1) w.r.t x we get
dydx=62x=m1
And x+y=2 (2)
Since
tanθ=|m1+11m1|tanπ4=|m1+11m1|1=m1+11m1
m1+1=±(1m1)
Taking
m1+1=1m12m1=0m1=062x=0x=3
Putting the values of x in (1) we get y=9.
Point is (3,9)

Tangents and Normals Exercise Multiple Choice Question , Question 22

Answer : (c) is the correct option.
Hint :
tanθ=|m1m21m1m2|
Given : y2=4ax And x2=4ay
Solution :
y2=4ax(1)x2=4ay(2)
Differentiating (1) w.r.t x we get
2ydydx=4a|dydx|(0,0)=4a2x=0=m1
And
2x=4adydx|dydx|(0,0)=|4a2x|=0=m2
tanθ=|m1m21+m1m2|==|001+00|=θ=π2

Tangents and Normals Exercise Multiple Choice Question , Question 23

Answer : (c) is the correct option
Hint :
tanθ=|m1m21+m1m2|
Given : y=2sin2x,y=cos2x
Solution :
y=2sin2x (1)
Differentiating (1) w.r.t x, we get
dydx=4sinxcosx[dydx]x=π6=4sinπ6cosπ6=4×12×32=3=m1
And y=cos2x (2)
Differentiating (2) w.r.t y, we get
dydx=2sin2x[dydx]x=π6=2sin(2×π6)=2×sinπ3=2×32=3
tanθ=|m1m213|=|3+32|=|232|=+3
tanθ=+3θ=π3

Tangents and Normals Exercise Multiple Choice Question , Question 24

Answer : (c) is the correct option
Hint : Obtain the slope
Given : y=2x7+3x+5
Solution :
y=2x7+3x+5 (1)
Differentiating (1) w.r.t x
dydx=14x6+3
Here the slope cannot be zero. Hence it make an acute angle with x-axis

Tangents and Normals Exercise Multiple Choice Question , Question 25

Answer : (a) is the correct option
Hint : Slope of the mormal is 1 or -1
Given : 9y2=x3
Solution :
9y2=x3 (1)
Differentiating (1) w.r.t x, we get
18ydydx=3x2dydx=3x218y=x26y
 Slope of normal =dxdy=6yx2=1 or 1 Case 1, m=1
6yx2=16y=x26y1=x12y1=x126
P(x,x26) Putting P(x,x26)ln(1)
9(x26)2=x39×x436=x3x3(x4)=0
x=0,4
Putting the values of x in (1) we get y=83
Similarly in case (2) if y=±83
The point (4,±83)

Tangents and Normals Exercise Multiple Choice Question , Question 26

Answer : (b) is the correct option
Hint :  Putting x=2 and y=1 in (1) and (2)
Solution :
x=t2+3t8 (1)
y=2t22t5 (2)
dxdt=2t+3dydt=4t2dydx=dydtdxdt=4t22t+3 (3)
Putting x=2 in (1)
2=t2+3t8t2+3t10=0t2+5t2t10=0
t(t+5)2(t+5)=0(t+5)(t2)=0t=5,2
Putting y=1 in (2)
1=2t22t52t22t4=02t2t2=0t22t+t2=0
t(t2)+1(t2)=0(t2)(t+1)=0t=2,1
Putiing t=2 in (3)
dydt=4(2)22(2)+3=824+3=67
The slope of tangent is 67


Tangents and Normals Exercise Multiple Choice Question , Question 27

Answer : (a) is the correct option
Hint :
Use c=am
Given : The line y=mx+1 is the tangent of the curve y2=4x
Solution :
y2=4x=4×1×x (1)
From equation (1) we equate with y2=4ax so that we get the value of c=1, the y intercept of the parabola
c=am, Where m is slope
From (1) , the value of a is 1
1=1mm=1
The slope of the curve is 1
Hence the slope of the tangent is 1

Tangents and Normals Exercise Multiple Choice Question , Question 28

Answer : (b) is the correct option
Hint :
Equation of normal yy1xx1=m(N)
Given : 2y+x2=3
Solution :
2y+x2=3 (1)
Differentiating equation (1) w.r.t y
2+2xdxdy=0dxdy=1x
dxdy=1x[dydx](1,1)=11=1=m(N)
Equation of the normal at the point (1,1) is y1x1=m(N)
y1x1=1y1=x1xy=0

Tangents and Normals Exercise Multiple Choice Question , Question 29

Answer : (a) is the correct option
Hint :
Slope yy1xx1
Given : x2=4y
Solution :
x2=4y (1)
Differentiating (1) w.r.t x
4dydx=2xdydx=x2dxdy=2x1
Slope of normal =2x1
Since the normal passes through (1,2)
 Slope =y12x112x1=x242x12[ From (1)]
2x1=x1284(x12)x138x1=8x18x1=2
Putting the value of x1=2 in (1) , was y=1
Point of contact (2,1)
Equation of normal
y1=23(x2)y1=(x2)x+y=3

Tangents and Normals Exercise Multiple Choice Question , Question 30

Answer : (a) is the correct option
Hint : Equation of the normal, yy1=N(xx1)
Given : 3y=6x5x3
Solution :
Let (x1,y1) be the point on the curve 3y=6x5x3 at which normal passes through origin
3y=6x5x3 ....(1)
Differentiate (1) w.r.t x, we get
3dydx=615x2dydx=25x2
Slope of the tangent at P is 25x12
Slope of normal at P is 15x122
Equation of normal at P is
yy1=15x122(xx1)
Since it passes through the origin
y=x5x122 .....(2)
On solving eqn 1 and 2
We get x=1 and y=13
As we know, abscissa is the x-coordinate
So, option (a) satisfies above equation i.e., 1

Tangents and Normals Exercise Multiple Choice Question, Question 31

Answer : (b) is the correct option
Hint : Multiply the slope of both curve
Given : x33xy2+2=0 And 3x2yy3=2
Solution :
Given x33xy2+2=0 (1)
Differentiating (1) we get
3x23y23x(2y)dydx=03(x2y2)=6xym1m1=x2y22xy
And 3x2yy32=0 (2)
Differentiating (2), we get
6xy3x2dydx3y2dydx=06xy=6m2(y2x2)m2=2xyy2x2
m1×m2=x2y2xy×2xyy2x2=1
Equation (1) and (2) at right angle

Tangents and Normals Exercise Multiple Choice Question, Question 32

Answer : (d) is the correct option
Hint :
 Put t=π4 in dydx
Given : x=etcost,y=etsint
Solution :
x=etcost (1)
Differentiating (1) w.r.t t, we get
dxdy=et(costsint)y=etsint (2)
Differentiating (2) w.r.t t, we get
dydt=et(cost+sint)dydx=dydtdxdt=cost+sintcostsint
[dydx](0,0)=tan1()=π2

Tangents and Normals Exercise Multiple Choice Question, Question 33

Answer : (b) is the correct option
Hint : Equation of tangent yy1=m(xx1)
Given : y=e2x
Solution :
y=e2x (1)
Differentiating (1) w.r.t x, we get
dydx=2e2xdydx|x=0=2e0=2
Equation of the tangent at the point (0, 1) and having the slope 2 is
y1=2(x0)y1=2xy=2x+1
Since it (2) meets in x-axis
y=0
Putting y=0 in (2)
0=2x+1x=12
The tangent meets at the point P(12,0)

Tangents and Normals Exercise Multiple Choice Question, Question 34

Answer : (a) is the correct option
Hint :
Equation of tangent yy1=m(xx1)
Given :
y(1+x2)=2x
Solution :
y(1+x2)=2x (1)
Since it crosses x -axis
2x=0x=2
The point of contact (2,0)
Differentiating (1) w.r.t x
y(2x)+(1+x2)dydx=1 At (2,0)0(2×2)+(1+22)dydx=1dydx=15
Equation of tangent at (1,2)
y0=15(x2)5y=x+2x+5y=2

Tangents and Normals Exercise Multiple Choice Question, Question 35

Answer : (d) is the correct option
Hint :  Parallel to x-axis y=0
Given : y=x312x+18
Solution :
y=x312x+18 (1)
Differentiating (1) w.r.t x, we get
dydx=3x212
Since it's parallel to x-axis
dydx=03x212=0x2=4x=±2
When x=2
y=(2)312(2)+18=824+18=2
When x=2
y=(2)312(2)+18=8+24+18=34
The points are (2,2) and (2,34)

Tangents and Normals Exercise Multiple Choice Question, Question 36

Answer : (a) is the correct option
Hint :  Put (0,0) in dydx
Given :
y=x15
Solution :
y=x15 (1)
Differentiating (1) w.r.t x, we get
dydx=15x151=15x45(dydx)(0,0)=
y=x15 Has vertical tangent

Chapter 15, Tangents and Normals, is where many students lose marks due to a lack of clarity. This chapter consists of three exercises, ex 15.1 to ex 15.3. RD Sharma Class 12 Solutions Tangents and Normals MCQ contains questions from concepts like an equation to the normal curve, the point of the curve when the tangent is perpendicular, the angle between curves, the slope of the tangent, and the angle of intersection. There are around 36 MCQ questions present in this chapter. The RD Sharma Class 12 Chapter 15 MCQ will lend a helping hand.

Experts in the teaching field provide the given solutions in the RD Sharma Class 12th MCQ book. It follows the NCERT pattern making it easier for the CBSE school students to utilize. The sums are solved in shortcut methods that make the students understand how the answer has arrived. Hence, they can learn the concept easily.

If you are confused between the options given in the MCQ, try answering it and check with the Class 12 RD Sharma Chapter 15 MCQ solutions book. You can use it while doing homework, preparing assignments, and even while preparing for exams. With a bit of regular practice in MCQ questions, you will become familiar with it. Hence, you would not lose marks in MCQs during the exams, which eventually makes you cross the benchmark score.

The best part is that the RD Sharma Class 12 Solutions Tangents and Normals can be downloaded from the Career360 website for free of cost. Many students have benefitted by using these solution books to prepare for their examinations. The solutions are given in shortcut methods to find the answers for the MCQs, and it saves a lot of time during exams.

There are chances of questions being taken from the RD Sharma Class 12th MCQ book’s practice questions. Hence, RD Sharma books help a lot when a student prepares for their exam with it. So, download the RD Sharma Class 12 Solutions Chapter 15 MCQ from the Career360 website and start preparing for your exams. You can very well score good marks when you have the best guide to prepare for the challenging mathematics exams.

RD Sharma Chapter wise Solutions

Frequently Asked Questions (FAQs)

1. Which solution book can I use to clarify the doubts in the mathematics chapter 15 MCQ?

The best solution book that clears all your doubts in the mathematics chapter 15 multiple choice questions is the RD Sharma Class 12th MCQ book

2. Where can I find the RD Sharma solution books for free?

You can find the RD Sharma solutions book on the Career360 website for free of cost. It is accessible as well as can be downloaded for later use.

3. Why is the RD Sharma Class 12th MCQ book the most prescribed one for the students?

Most of the schools recommend the RD Sharma Class 12 Chapter 15 FBQ solution books due to the high quality of their answers. As experts provide these answers, the students can understand and can trust the accuracy of it.  

4. Is it enough if I can solve the MCQs given in the RD Sharma books?

The RD Sharma has every question and answer as given in the Class 12 textbook.  Furthermore, it also contains extra sums for practice. Hence, the students can become familiar with the concept if they use this book.

5. Can everyone access the RD Sharma solution books given on the Career 360 website?

The students utilize this resource material to clarify their doubts and the teachers use it to prepare question papers in most cases. Hence, the RD Sharma books can be accessible by anyone.

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