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.
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Also Read - RD Sharma Solution for Class 9 to 12 Maths
Chapter 15 - Tangents and Normals Ex 15.1
Chapter 15 - Tangents and Normals Ex 15.2
Chapter 15 - Tangents and Normals Ex 15.3
Tangents and Normals Exercise Multiple Choice Questions Question 1
Answer:Tangents and Normals Exercise Multiple Choice Questions Question 2
Answer:Tangents and Normals Exercise Multiple Choice Questions Question 3
Answer:T he given curve is $y=2x-x^{2}$
i.e. $x^{2}-2x+y=0$
Now the equation of the tangent at the point $\left ( x_{1},x_{2} \right )=\left ( 2,0 \right )$
Or
$\begin{aligned} &x x_{1}-\left(x+x_{1}\right)+\frac{1}{2}\left(y+y_{1}\right)=0 \\ &2 x+(-1)(x+2)+\frac{1}{2}(y+0)=0 \\ &2 x-4+y=0 \\ &y=-2 x+4 \end{aligned}$
So, the slope of the tangent is $\left ( -2 \right )$
Therefore the slope of the normal=$\frac{1}{2}$
So, the equation of the normal at the point $\left ( 2,0 \right )$is
$y-0=\frac{1}{2}\left ( x-2 \right )$
$2y=x-2$
I.e. $x-2y=2$ is required
Tangents and Normals Exercise Multiple Choice Questions Question 4
Answer:Tangents and Normals Exercise Multiple Choice Questions Question 6
Answer:Tangents and Normals Exercise Multiple Choice Questions Question 7
Answer:Tangents and Normals Exercise Multiple Choice Question Question 9
Answer:Hint:
Use slop of the tangent $m=\tan \theta$
Given:
The curve $y^{2}=x$ and $x^{2}=y^{2}$at $\left ( 1,1 \right )$
Solution:
Slope of the tangent to curve
$y=x^{2}$At $\left ( 1,1 \right )$is $m_{1}=2$
Slope of the tangent to the curve
$x=y^{2}$At $\left ( 1,1 \right )$is $m_{2}=\frac{1}{2}$
Angle between tangents to the curve is $\tan \theta =\frac{3}{4}$
Or $\theta =tan^{-1}\frac{3}{4}$
Tangents and Normals Exercise Multiple Choice Question Question 10
Answer:Now applying the sum rule of differentiation an differentiation of constant =0,so we get
$\begin{aligned} &\frac{d\left(3 x^{2}\right)}{d x}-\frac{d\left(y^{2}\right)}{d x}=0 \\ &3 \frac{d\left(x^{2}\right)}{d x}-\frac{d\left(y^{2}\right)}{d x}=0 \\ &3 \times 2\left(x^{2-1}\right) \times \frac{d(x)}{d x}-2\left(y^{2-1}\right) \times \frac{d(y)}{d x}=0 \\ &6 x-2 y \times \frac{d y}{d x}=0 \\ &6 x=2 y \times \frac{d y}{d x} \\ &\frac{d y}{d x}=\frac{6 x}{2 y}=\frac{3 x}{y} \end{aligned}$
So, this is the slope of the given curve. We know the slope of the normal to the curve is
$=-\frac{1}{\frac{dy}{dx}}$
$=-\frac{1}{\frac{3x}{y}}=\left ( -\frac{y}{3x} \right )$ (1)
Now the given equation of the line $x+3y=8$
$3y=8-x$
Differentiating w.r.t x we get
$\begin{aligned} &\frac{d(3 y)}{d x}=\frac{d(8-x)}{d x} \\ &3 \frac{d y}{d x}=-1 \\ &\frac{d y}{d x}=-\frac{1}{3} \end{aligned}$
So, the slope of the line is $-\frac{1}{3}$
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,
$\begin{aligned} &\therefore\left(-\frac{y}{3 x}\right)=-\frac{1}{3} \\ &3 y=3 x \\ &y=x \end{aligned}$
On substituting this value of the given equation of the curve, we get
$\begin{aligned} &3 x^{2}-y^{2}=8 \\ &3 x^{2}-(x)^{2}=8 \\ &2 x^{2}=8 \\ &x^{2}=4 \\ &x=\pm 2 \end{aligned}$
When x=2 the equation of the curve becomes,
$3x^{2}-y^{2}=8$
$3\left ( 2 \right )^{2}-y^{2}=8$
$3\left ( 4 \right )-y^{2}=8$
$12-8=y^{2}$
$y^{2}=4$
$y=\pm 2$
When x=-2, the equation of the curve becomes,
$\begin{aligned} &3 x^{2}-y^{2}=8 \\ &3(-2)^{2}-y^{2}=8 \\ &3(4)-y^{2}=8 \\ &12-8=y^{2} \\ &y^{2}=4 \\ &y=\pm 2 \end{aligned}$
So, the points at which normal is parallel to the given line are $\left ( \pm 2,\pm 2 \right )$
And required equation of the normal to the curve at $\left ( \pm 2,\pm 2 \right )$ is
$\begin{aligned} &y-(\pm 2)=-\frac{1}{3}[x-(\pm 2)] \\ &3(y-(\pm 2))=-(x-(\pm 2)) \\ &3 y-(\pm 6)=-x+(\pm 2) \\ &x+3 y-(\pm 6)-(\pm 2)=0 \\ &x+3 y+(\pm 8)=0 \end{aligned}$
Hence the equation of normal to the curve
$3x^{2}-y^{2}=8$ Which is parallel to the line $x+3y=8$ is $x+3y+\left ( \pm 8 \right )=0$
Tangents and Normals Exercise Multiple Choice Questions Question 15
Answer:Tangents and Normals Exercise Multiple Choice Question Question 16
Answer:Differentiate w.r.t, $x_{1}$we get
$\frac{dy}{dx}=2x+b$
Putting the point $\left ( 1,1 \right )$and m=1 in above curve
$1=2\left ( 1 \right )+b$
$b=-1$
Putting the point$\left ( 1,1 \right )$ in (1)
$\begin{aligned} &1=(1)^{2}+b(1)+c \\ &1=1-1+c \quad \text { [Since } b=-1] \\ &c=1 \\ &\therefore b=-1, c=1 \end{aligned}$
Tangents and Normals Exercise Multiple Choice Questions Question 17
Answer:Differentiate w.r.t t
$\begin{aligned} &\frac{d x}{d t}=6 t\\ &\text { And, } y=t^{3}-1\\ &\text { Differentiating w.r.t t }\\ &\frac{d y}{d x}=3 t^{2} \end{aligned}$
$\begin{aligned} &\text { Substituting } x=1 \text { in (1) }\\ &1=3 t^{2}+1\\ &3 t^{2}=0\\ &t=0\\ &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{3 t^{2}}{6 t}=\frac{t}{2}\\ &\left[\frac{d y}{d x}\right]_{t=0}=0\\ &\therefore \text { slope }=0 \end{aligned}$
Tangents and Normals Exercise Multiple Choice Question Question 18
Answer:Tangents and Normals Exercise Multiple Choice Questions Question 19
Answer:Tangents and Normals Exercise 15 point 1 Question 20
Answer:Tangents and Normals Exercise Multiple Choice Question , Question 21
Answer : (b) is the correct optionTangents and Normals Exercise Multiple Choice Question , Question 22
Answer : (c) is the correct option.Tangents and Normals Exercise Multiple Choice Question , Question 23
Answer : (c) is the correct optionTangents and Normals Exercise Multiple Choice Question , Question 24
Answer : (c) is the correct optionTangents and Normals Exercise Multiple Choice Question , Question 25
Answer : (a) is the correct optionTangents and Normals Exercise Multiple Choice Question , Question 26
Answer : (b) is the correct optionTangents and Normals Exercise Multiple Choice Question , Question 27
Answer : (a) is the correct optionTangents and Normals Exercise Multiple Choice Question , Question 28
Answer : (b) is the correct optionTangents and Normals Exercise Multiple Choice Question , Question 29
Answer : (a) is the correct optionTangents and Normals Exercise Multiple Choice Question , Question 30
Answer : (a) is the correct optionTangents and Normals Exercise Multiple Choice Question, Question 31
Answer : (b) is the correct optionTangents and Normals Exercise Multiple Choice Question, Question 32
Answer : (d) is the correct optionTangents and Normals Exercise Multiple Choice Question, Question 33
Answer : (b) is the correct optionTangents and Normals Exercise Multiple Choice Question, Question 34
Answer : (a) is the correct optionTangents and Normals Exercise Multiple Choice Question, Question 35
Answer : (d) is the correct optionTangents and Normals Exercise Multiple Choice Question, Question 36
Answer : (a) is the correct optionChapter 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.
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