RD Sharma Solutions Class 12 Mathematics Chapter 3 FBQ

RD Sharma Solutions Class 12 Mathematics Chapter 15 FBQ

Updated on Jan 21, 2022 12:59 PM IST

The CBSE schools recommend the student use the RD Sharma solution books to prepare for their tests and exams. The students might face numerous doubts while finding the answers for the FBQs. The concept of Tangents and Normals make it even harder to find accurate answers. As the FBQ section is where students tend to lose marks, extra practice is required to avoid it. This is where the RD Sharma Class 12th FBQ comes to help.

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RD Sharma Class 12 Solutions Chapter 15 Tangents and Normals - Other Exercise
Tangents and Normals Excercise:FBQ
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-MCQ

Chapter 15 -Tangents and Normals Ex-VSA

Tangents and Normals Excercise:FBQ

Tangents and Normals exercise Fill in the blanks question 1

Answer:
Equation of normal,

y = - x

Hint:
Use equation of normal formula,

(y - y_{1}) = \frac{- 1}{\frac{d y}{d x}} (x - x_{1})

Given:
Here given the curve ,

y = \tan x

To find:
We have to find the equation of normal to the curve at

(0, 0)

Solution:
Here given,

y = \tan x

Differentiating both side with respect to

x

, we get

\begin{aligned} \Rightarrow \frac{d y}{d x} = \sec^{2} x \\ \Rightarrow {(\frac{d y}{d x})}_{(0, 0)} = \sec^{2} (0) = 1 \\ \Rightarrow {(\frac{d y}{d x})}_{(0.0)} = 1 \end{aligned}

Slope of normal at

(0, 0) = \frac{- 1}{\frac{d y}{d x}}

= - 1

Now, equation of normal to the curve

y = \tan x

(0, 0)

(y - y_{1}) = \frac{- 1}{\frac{d y}{d x}} (x - x_{1})

\Rightarrow

(y - 0) = (- 1) x

\Rightarrow

y = - x

Hence, this is the required equation of normal.

Tangents and Normals exercise Fill in the blanks question 2

Answer:

\pm 10

Hint:
Since the given curve touches

x -

axis. i.e.

y = 0

Given:

y = x^{2} + a x + 25

… (i)
To find:
We have to find the value of

a

for which the given curve touches

x -

axis.
Solution:
We know that,
Since the given curve,

y = x^{2} + a x + 25

touches the

x -

axis

\Rightarrow \frac{d y}{d x} = 0

Differentiating equation (i) with respect to

x

, we get

\Rightarrow 2 x + a = 0

\begin{array}{ll} \Rightarrow & \frac{d y}{d x} = 2 x + a \end{array}

\Rightarrow

x = \frac{- a}{2}

Putting the value of

x = \frac{- a}{2}

in equation (i), we get
Since

y = 0

\Rightarrow \frac{a^{2}}{4} + a (\frac{- a}{2}) + 25 = 0

\Rightarrow \frac{a^{2}}{4} - \frac{a^{2}}{2} + 25 = 0

\Rightarrow a = \pm 10

Hence, the required value of

a

are

\pm 10

Tangents and Normals exercise Fill in the blanks question 3

Answer:

(2, 16) (- 2, - 16)

Hint:
Gradient is zero. i.e.

\frac{d y}{d x} = 0

Given:
Given curve,

y = 12 x - x^{3}

To find:
We have to find the points on the given curve at which the gradient is zero.
Solution:
Here

y = 12 x - x^{3}

… (i)
Given that gradient is zero
i.e.

\frac{d y}{d x} = 0

Differentiating equation (i), we get

\Rightarrow \frac{d y}{d x} = 12 - 3 x^{2}

Since

\frac{d y}{d x} = 0

\begin{array}{lrl} \Rightarrow & 12 - 3 x^{2} = 0 \\ \Rightarrow & - 3 x^{2} = - 12 \\ \Rightarrow & x^{2} = 4 \\ \Rightarrow & x = \pm 2 \end{array}

Putting the value

x = \pm 2

in equation (i), we get

\begin{aligned} y = 12 (2) - (2)^{3} or y = 12 (- 2) - (- 2)^{3} \\ y = 16 or y = - 16 \end{aligned}

Hence the points are

(2, 16)

and

(- 2, - 16)

Tangents and Normals Exercise Fill in the blanks Question 4

Answer:

(e^{- 2}, - 2 e^{- 2})

Hint:

Slope of normal $\frac{- 1}{\frac{d y}{d x}}$
Slope of line is $m$ . i.e. $y = m x + c$

Given:
Given curve,

y = x \log_{e} x

and the line

2 x - 2 y = 3

To find:
We have to find the co-ordinates of a point on the given curve at which the normal is parallel to the line

2 x - 2 y = 3

Solution:
We have,

y = x \log_{e} x

…(i)
Differentiating both side with respect to

x

, we get

\Rightarrow \frac{d y}{d x} = x \frac{d (\log x)}{d x} + \log x \frac{d (x)}{d x}

[∵ \frac{d (u v)}{d x} = u \frac{d v}{d x} + \frac{d u}{d x} v]

\Rightarrow \frac{d y}{d x} = x \cdot \frac{1}{x} + \log x (1)

[\begin{array}{l} ∵ \frac{d (\log x)}{d x} = \frac{1}{x} \\ ∵ \frac{d (x^{n})}{d x} = n x^{n - 1} \end{array}]

\Rightarrow \frac{d y}{d x} = 1 + \log x

The Slope of normal

= \frac{- 1}{\frac{d y}{d x}}

= \frac{- 1}{1 + \log x}

The line,

2 x - 2 y = 3

\begin{aligned} \Rightarrow 2 y = 2 x - 3 \\ \Rightarrow y = x - \frac{3}{2} \end{aligned}

Comparing this equation with the formula

y = m x + c

, we can get the slope

m

.
Here the slope of line

2 x - 2 y = 3

1

As the slope of parallel lines are equal,
Therefore,

\frac{- 1}{1 + \log x} = 1

\begin{array}{ll} \Rightarrow & \log x = - 2 \\ \Rightarrow & x = e^{- 2} \end{array}

Putting the value of

x

in equation (i)

\Rightarrow y = - 2 e^{- 2}

Therefore the coordinate of the point is

(e^{- 2}, - 2 e^{- 2})

Tangents and Normals Exercise Fill in the blanks Question 5

Answer:

(6, 7)

Hint:
First we find slope of curve then equate with given slope

\frac{2}{5}

Given:
Given curve,

y = 2 + \sqrt{4 x + 1}

and tangent has slope

\frac{2}{5}

To find:
We have to find the coordinates of the point on the given curve where tangent has slope

\frac{2}{5}

.
Solution:
We have,

y = 2 + \sqrt{4 x + 1}

… (i)
Differentiating equation (i) on both side with respect to

x

, we get

\begin{aligned} \Rightarrow y = 2 + (4 x + 1)^{\frac{1}{2}} \\ \Rightarrow \frac{d y}{d x} = 0 + \frac{1}{2} (4 x + 1)^{\frac{1}{2} - 1} \frac{d (4 x + 1)}{d x} \end{aligned}

[∵ \frac{d (x^{n})}{d x} = n x^{n - 1}]

\begin{aligned} \Rightarrow \frac{d y}{d x} = \frac{1}{2} (4 x + 1)^{\frac{- 1}{2}} (4) \\ \Rightarrow \frac{d y}{d x} = \frac{2}{\sqrt{4 x + 1}} \end{aligned}

And tangent has slope

\frac{2}{5}

\begin{aligned} \Rightarrow \frac{2}{\sqrt{4 x + 1}} = \frac{2}{5} \\ \Rightarrow \frac{10}{2} = \sqrt{4 x + 1} \end{aligned}

Squaring on both sides,

\begin{array}{ll} \Rightarrow & (5)^{2} = 4 x + 1 \\ \Rightarrow & 25 = 4 x + 1 \\ \Rightarrow & x = 6 \end{array}

Substituting the value of

x = 6

in equation (i), we get

\Rightarrow y = 2 + \sqrt{4 (6) + 1}

\begin{array}{ll} \Rightarrow & y = 2 + \sqrt{25} \\ \Rightarrow & y = 2 + 5 = 7 \end{array}

Hence the required coordinate is

(6, 7)

Tangents and Normals Exercise Fill in the blanks Question 6

Answer:
Slope of tangent =

0

Hint:
First put

x = 1

in equation

x = 3 t^{2} + 1

, then find

t

and

\frac{d y}{d x}

.
Given:
Given curve,

x = 3 t^{2} + 1

and

y = t^{3} - 1

To find:
We have to find the slope of tangent to the given curve at

x = 1

Solution:
Given,

x = 3 t^{2} + 1

… (i)

y = t^{3} - 1

… (ii)
Put

x = 1

in equation (i), we get

\begin{aligned} x = 3 t^{2} + 1 \\ \Rightarrow & 1 = 3 t^{2} + 1 \\ \Rightarrow & 3 t^{2} = 0 \\ \Rightarrow & t = 0 \end{aligned}

Differentiating equation (i) and (ii) with respect to

x

, we get

\begin{aligned} \Rightarrow \frac{d x}{d t} = 6 t and \frac{d y}{d t} = 3 t^{2} \\ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}

= \frac{3 t^{2}}{6 t} = \frac{t}{2}

Thus we get the slope of tangent

{(\frac{d y}{d x})}_{t = 0} = \frac{0}{2} = 0

Tangents and Normals Exercise Fill in the blanks Question 7

Answer:

\frac{π}{2}

Hint:
First we need to find their point of intersection. So solve these given equations.
Given:
Given curves,

y = x^{2}

and

x = y^{2}

To find:
We have to find the angle of intersection of the given curves at

(0, 0)

Solution:
Here

y = x^{2}

… (i)

x = y^{2}

… (ii)
From equation (i) and (ii), we get

\begin{array}{ll} \Rightarrow & x^{4} = x \\ \Rightarrow & x^{3} (x - 1) = 0 \end{array}

Which gives

x = 0, x = 1

Therefore points of intersection of the curves are

(0, 0)

and

(1, 1)

On differentiating equation (i) and (ii), we get

\Rightarrow \frac{d y}{d x} = 2 x

Similarly,

2 y \frac{d y}{d x} = 1

\Rightarrow \frac{d y}{d x} = \frac{1}{2 y}

To find the angle of intersection at

(0, 0)

So,

m_{1}

slope of tangent to curve

y = x^{2}

(0, 0)

= 2 x = 0

So tangent is parallel to

x -

axis.

m_{2}

= slope of tangent to curve

x = y^{2}

(0, 0)

= \frac{1}{2 y}

which is not defined.
So tangent is parallel to

y -

axis.
Now, one tangent is parallel to

x -

axis and other is parallel to

y -

axis.
Hence angle between tangents is right angle. i.e.

\frac{π}{2}

Tangents and Normals Exercise Fill in the blanks Question 8

Answer:

\frac{- b}{a}

Hint:
Use

(0, y_{1})

is the point where tangent to the curve crosses

y -

axis, then proceed next to find slope of tangent.
Given:
Given curve,

y = b e^{\frac{- x}{a}}

To find:
We have to find the slope of tangent where the curves crosses

y -

axis.
Solution:
Given equation of curve is,

y = b e^{\frac{- x}{a}}

… (i)
Let

p (0, y_{1})

be the point where tangent to curve crosses

y -

axis.

\Rightarrow y_{1} = b

So the point

p

p (0, b)

Differentiating equation (i), we get

\Rightarrow \frac{d y}{d x} = b e^{\frac{- x}{a}} = {\frac{b e}{a}}^{- \frac{x}{a}} = \frac{- y}{a}

Hence slope of tangent at

p (0, b)

\frac{- b}{a}

Tangents and Normals Exercise Fill in the blanks Question 9

Answer:
Point

= (\frac{- 1}{2}, 0)

Hint:

Using equation of tangent is $(y - y_{1}) = \frac{d y}{d x} (x - x_{1})$
Tangent line meets the $x -$ axis. i.e. $y = 0$

Given:
Given equation of curve,

y = e^{2 x}

To find:
We have to find the point where curve at

(0, 1)

cut

x -

axis.
Solution:
The given equation of curve,

y = e^{2 x}

… (i)
Differentiating equation (i), we get

\begin{aligned} \Rightarrow \frac{d y}{d x} = 2 \cdot e^{2 x} \\ \Rightarrow {(\frac{d y}{d x})}_{(0.1)} = 2 \cdot e^{2 (0)} = 2 \end{aligned}

Slope of tangent to the curve =2
So equation of tangent is

(y - 1) = 2 (x - 0)

\Rightarrow y = 2 x + 1

Above tangent line cuts the

x -

axis, where

y = 0

∴ x = \frac{- 1}{2}

Hence the required point is

(\frac{- 1}{2}, 0)

Tangents and Normals Excercise Fill in the blanks Question 10

Answer:
Slope of normal

= - 6

Hint:
Slope of normal

= \frac{- 1}{\frac{d y}{d x}}

Given:
Given equation of curve,

y^{3} - x y - 8 = 0

To find:
We have to find the slope of normal to the given curve at the point

(0, 2)

Solution:
Given curve,

y^{3} - x y - 8 = 0

… (i)
On differentiating with respect to

x

, we get

\Rightarrow 3 y^{2} \frac{d y}{d x} - x \frac{d y}{d x} - y = 0

[∵ \frac{d (x^{n})}{d x} = n x^{n - 1}]

\begin{aligned} \Rightarrow & 3 y^{2} \frac{d y}{d x} - x \frac{d y}{d x} - y = 0 \\ \Rightarrow & (3 y^{2} - x) \frac{d y}{d x} = y \\ \Rightarrow & \frac{d y}{d x} = \frac{y}{3 y^{2} - x} \end{aligned}

\Rightarrow {(\frac{d y}{d x})}_{(0.2)} = \frac{2}{3 (2)^{2} - 0} = \frac{1}{6}

Hence the slope of the normal is

\frac{- 1}{\frac{d y}{d x}} = - 6

Tangents and Normals Excercise Fill in the blanks Question 11

Answer:

a + b = 10

Hint:
First find the equation of normal then comparing with

a x - b y + b = 0

Given:
The given equation of curve,

y^{2} = 5 x - 1

To find:
We have to find the value of

a + b

Solution:
We have,

y^{2} = 5 x - 1

… (i)
Differentiating equation (i) with respect to

x

, we get

\Rightarrow 2 y \frac{d y}{d x} = 5

[∵ \frac{d (x^{n})}{d x} = n x^{n - 1}]

\begin{aligned} \Rightarrow \frac{d y}{d x} = \frac{5}{2 y} \\ ∴ {(\frac{d y}{d x})}_{(1, - 2)} = \frac{- 5}{4} \end{aligned}

Then the equation of normal at point

(1, - 2)

\begin{array}{ll} (y - (- 2)) = \frac{- 1}{\frac{- 5}{4}} (x - 1) \\ \Rightarrow & 5 (y + 2) = 4 (x - 1) \\ \Rightarrow & 4 x - 5 y - 14 = 0 \end{array}

… (ii)
As the normal is of the form

a x - 5 y + b = 0

… (iii)
Comparing equation (ii) and (iii), we get

a = 4

and

b = - 14

Hence

a + b = - 10

Tangents and Normals Excercise Fill in the blanks Question 12

Answer:

\frac{a}{b}

Should be negative. i.e.

(- \infty, 0)

Hint:
First find the slope of normal to the given curve then compare with slope of

a x + b y + c

Given:
Given the equation of curve,

x y = 1

To find:
We have to find the set of

\frac{a}{b}

if the line

a x + b y + c = 0

is normal to the given curve.
Solution:
Given,

\begin{aligned} x y = 1 \\ \Rightarrow & y = \frac{1}{x} \end{aligned}

Differentiating with respect to

x

, we get

\Rightarrow \frac{d y}{d x} = \frac{- 1}{x^{2}}

[∵ \frac{d (x^{n})}{d x} = n x^{n - 1}]

Thus slope of normal

= \frac{- 1}{\frac{d y}{d x}} = x^{2}

Which is always positive and it is given

a x + b y + c = 0

is normal
Slope =

\frac{- a}{b}

\Rightarrow \frac{- a}{b} > 0

\Rightarrow \frac{a}{b} < 0

a

and

b

are of opposite sign.
Hence

a < 0, b > 0

\frac{a}{b}

should be negative. i.e

(- \infty, 0)

Tangents and Normals Excercise Fill in the blanks Question 13

Answer:

f^{'} (3) = 1

Hint:
First find the slope of given curve at point

(3, 4)

, then compare with

\frac{3 π}{4}

Given:
Given curve,

y = f (x)

To find:
We have to find

f^{'} (3)

, if the normal to the given curve at

(3, 4)

makes an angle

\frac{3 π}{4}

with positive x-axis.
Solution:
Here we have,

y = f (x)

On differentiating with respect to

x

, we get

\Rightarrow \frac{d y}{d x} = f^{'} (x)

Slope of tangent at

(3, 4) = \frac{d y}{d x} = f^{'} (x)_{(3, 4)}

Therefore, slope of normal

\begin{aligned} = \frac{- 1}{f^{'} (x)_{(3.4)}} \\ = \frac{- 1}{f^{'} (3)} \end{aligned}

But

\frac{- 1}{f^{'} (3)} = \tan (\frac{3 π}{4})

[Given]

\begin{aligned} \Rightarrow \frac{- 1}{f^{'} (3)} = \tan (\frac{π}{2} + \frac{π}{4}) \\ \Rightarrow \frac{- 1}{f^{'} (3)} = - 1 \end{aligned}

Hence

f^{'} (3) = 1

Tangents and Normals Exercise Fill in the blanks Question 14

Answer:
Equation of tangent is

y = 3

Hint:
Tangent is parallel to X-axis so slope becomes 0
Given:
Given curve

y = x + \frac{4}{x^{2}}

To find:
We have to find the equation of the tangent to the given curve that is parallel to X-axis
Solution:
Given tangent is parallel to X-axis so slope is 0

\Rightarrow \frac{d y}{d x} = 0

… (i)
Here

y = x + \frac{4}{x^{2}}

… (ii)
On differentiate both side with respect to

x

we get

\frac{d y}{d x} = 1 - \frac{8}{x^{3}}

Substituting the value of

\frac{d y}{d x}

in equation (i), we get

\begin{aligned} 1 - \frac{8}{x^{3}} = 0 \\ x^{3} = 8 \\ x = 2 \end{aligned}

∵

From (ii) we get

y = 2 + \frac{4}{4} = 3

Hence equation of tangent is

y = 3

[

∵

equation of tangent

y - 3 = 0 (x - 2)

]

Tangents and Normals Exercise Fill in the blanks Question 15

Answer:
Point:

(1, 0)

Hint:

First we will find the slope of line $x - y = 0$ then evaluate with the slope of curve
If tangent is perpendicular to the line then $m_{1} \cdot m_{2} = - 1$

Given:
Given curve

y = x^{2} - 3 x + 2

where the tangent is perpendicular to the line

x - y = 0

To find:
We have to find the co-ordinate of the point on the given curve.
Solution:
We have,

\begin{aligned} x - y = 0 \\ \Rightarrow & y = x \end{aligned}

… (i)
Comparing equation (i) with equation

y = m x + c

\Rightarrow m_{1} = 1

Again, if the line is perpendicular to the tangent then

\begin{aligned} \Rightarrow m_{1} \cdot m_{2} = - 1 \\ \Rightarrow 1 \cdot m_{2} = - 1 \end{aligned}

[∵ m_{1} = 1]

\Rightarrow m_{2} = - 1

Given curve,

y = x^{2} - 3 x + 2

… (ii)
On differentiating both side with respect to

x

, we get

\Rightarrow \frac{d y}{d x} = 2 x - 3

[∵ \frac{d (x^{n})}{d x} = n x^{n - 1}]

Here slope of tangent is

- 1

\begin{aligned} \Rightarrow 2 x - 3 = - 1 \\ \Rightarrow 2 x = 2 \\ \Rightarrow x = 1 \end{aligned}

Substituting the value of

x

in equation (ii), we get

\begin{aligned} y = 1 - 3 + 2 \\ y = 0 \end{aligned}

Hence the point is

(x, y) = (1, 0)

Tangents and Normals Exercise Fill in the blanks Question 16

Answer:
Points are

(0, 0)

and

(3, 27)

Hint:
If the slope of tangent is equal to ordinate of point that means

\frac{d y}{d x} = y

Given:
The slope of tangent to curve

y = x^{3}

at a point is equal to ordinate of point.
To find:
We have to find the point
Solution:
We have,

y = x^{3}

… (i)
On differentiating with respect to

x

, we get

\frac{d y}{d x} = 3 x^{2}

[∵ \frac{d (x^{n})}{d x} = n x^{n - 1}]

Now, we know that the given slope of tangent to the given curve at a point is equal to ordinate of point

3 x^{2} = y

… (ii)
Putting value of

y

in equation (i), we get

\Rightarrow

3 x^{2} = x^{3}

\Rightarrow

x = 3, 0

and

y = 0, 27

[From equation (i)]
Thus the two points are

(0, 0)

and

(3, 27)

Tangents and Normals Excercise Fill in the blanks Question 17

Answer:
Equation of normal,

y = 3 x - 5

Hint:
Use equation of normal,

(y - y_{1}) = - \frac{d x}{d y} (x - x_{1})

Given:
Given curve,

x^{2} + y^{2} - 2 x + 4 y - 5 = 0

To find:
We have to find the slope of normal to the given curve at point

(2, 1)

Solution:
Given equation is

x^{2} + y^{2} - 2 x + 4 y - 5 = 0

On differentiating both side with respect to

x

, we get

\Rightarrow 2 x + 2 y \frac{d y}{d x} - 2 + 4 \frac{d y}{d x} = 0

[∵ \frac{d (x^{n})}{d x} = n x^{n - 1}]

\Rightarrow (y + 2) \frac{d y}{d x} = 1 - x

[Taking common

2

from each term]

\begin{aligned} \Rightarrow \frac{d y}{d x} = \frac{1 - x}{y + 2} \\ \Rightarrow {(\frac{d y}{d x})}_{(2, 1)} = \frac{1 - 2}{1 + 2} = \frac{- 1}{3} \end{aligned}

\Rightarrow - {(\frac{d y}{d x})}_{(2, 1)} = \frac{1}{3}

\Rightarrow - {(\frac{d x}{d y})}_{(2, 1)} = 3

Now, equation of normal is

(y - y_{1}) = - \frac{d x}{d y} (x - x_{1})

Substituting these value, we get
i.e.

y_{1} = 1, x_{1} = 2, \frac{- d x}{d y} = 3

\begin{aligned} (y - 1) = 3 (x - 2) \\ \Rightarrow & y - 1 = 3 x - 6 \\ \Rightarrow & y = 3 x - 5 \end{aligned}

Hence,

y = 3 x - 5

is required equation of normal.

Tangents and Normals Excercise Fill in the blanks Question 18

Answer:

(\frac{1}{4}, \frac{1}{2})

Hint:
First find the slope of curve

y^{2} = x

, then compare with slope

45^{\circ}

with x-axis.
i.e. slope

\tan \frac{π}{4} = 1

Given:
Given curve,

y^{2} = x

, the tangent at which makes an angle of

45^{\circ}

with x-axis.
To find:
We have to find the point on the given curve.
Solution:
We have

y^{2} = x

… (i)
Differentiate both side with respect to

x

, we get

\Rightarrow

2 y \frac{d y}{d x} = 1

[∵ \frac{d (x^{n})}{d x} = n x^{n - 1}]

\Rightarrow

\frac{d y}{d x} = \frac{1}{2 y}

… (ii)
Also tangent makes an angle of

45^{\circ} (\frac{π}{4})

with x-axis

\Rightarrow

\frac{d y}{d x} = \tan (\frac{π}{4})

[∵ \tan (\frac{π}{4}) = 1]

\Rightarrow

\frac{d y}{d x} = 1

… (iii)
From equation (ii) and (iii), we get

\begin{aligned} \Rightarrow \frac{1}{2 y} = 1 \\ \Rightarrow y = \frac{1}{2} \end{aligned}

Put value of

y

in equation

y^{2} = x

\Rightarrow

x = \frac{1}{4}

Hence required point is

(\frac{1}{4}, \frac{1}{2})

Tangents and Normals Excercise Fill in the blanks Question 19

Answer:

(3, 34)

Hint:
When both curves touch then slope of both curves should be same.
Given:
Given curves,

y = 4 x^{2} + 2 x - 8

and

y = x^{3} - x + 13

touch each other.
To find:
We have to find the point where the given curves touch each.
Solution:
The curves,

y = 4 x^{2} + 2 x - 8

and

y = x^{3} - x + 13

For first curve say

{(\frac{d y}{d x})}_{1}

∴ {(\frac{d y}{d x})}_{1} = 8 x + 2

For second curve

{(\frac{d y}{d x})}_{2} = 3 x^{2} - 1

When both curves touch the slope both curves should be same

\begin{aligned} ∴ 8 x + 2 = 3 x^{2} - 1 \\ \Rightarrow 3 x^{2} - 8 x - 3 = 0 \end{aligned}

Solving the quadratic equation, we get

\begin{array}{ll} \Rightarrow & (3 x + 1) (x - 3) = 0 \end{array}

\Rightarrow x = \frac{- 1}{3}, x = 3

Now consider,

x = 3

For first curve,

y (3) = 4 {(3)}^{2} + 2 (3) - 8 = 34

For second curve,

y (3) = {(3)}^{3} - 3 + 10 = 34

…(here there should be 10 inplace of 13)
Thus at

(3, 34)

both curves touch
Now consider,

x = \frac{- 1}{3}

For first curve,

y (\frac{- 1}{3}) = 4 {(\frac{- 1}{3})}^{2} + 2 (\frac{- 1}{3}) - 5 = \frac{- 47}{9}

For second curve,

y (\frac{- 1}{3}) = {(\frac{- 1}{3})}^{3} - (\frac{- 1}{3}) + 13 = \frac{8}{27} + 13

Thus at

x = \frac{- 1}{3}

both curves do not meet.
But their tangent are parallel
Hence the only point where both curves touch is

(3, 34)

Tangents and Normals Excercise Fill in the blanks Question 20

Answer:
$y = 0$ is the required equation of normal
Hint:
Equation of normal,
$(y - y_{1}) = - \frac{d x}{d y} (x - x_{1})$
Given:
Given curve,
$y^{2} = 8 x$
To find:
We have to find the equation of normal to the given curve at the origin.
Solution:
We have,
$y^{2} = 8 x$
On differentiating both side with respect to $x$ , we get
$\Rightarrow 2 y \frac{d y}{d x} = 8$ $[∵ \frac{d (x^{n})}{d x} = n x^{n - 1}]$
$\begin{aligned} \Rightarrow \frac{d y}{d x} = \frac{8}{2 y} \\ \Rightarrow \frac{d y}{d x} = \frac{4}{y} \\ \Rightarrow - (\frac{d x}{d y}) = \frac{- y}{4} \end{aligned}$
At origin i.e. $(0, 0)$
Here the equation of normal,
$\begin{aligned} (y - y_{1}) = - {(\frac{d x}{d y})}_{(0.0)} (x - x_{1}) \\ \Rightarrow & (y - 0) = 0 (x - 0) \\ \Rightarrow & y = 0 \end{aligned}$
Hence the required equation of normal is $y = 0$ .

Chapter 15, Tangents and Normals, has three exercises, ex 15.1, ex15.2, and ex 15.3. RD Sharma Class 12 Solutions Tangents and Normals FBQ include the equation of the normal to the curve, coordinates of a point, slope of a tangent, angle of intersection curves, and slope of normal to the curve. There are around 20 FBQ questions asked in the mathematics textbook. Students can utilize the RD Sharma Class 12 Chapter 15 FBQ book to find and verify their answers.

Experts provide accurate answers for these FBQs in the RD Sharma solution books with a broad knowledge of their respective domains. The NCERT pattern is strictly followed for the CBSE board students to adapt it quickly. Apart from the solutions for the textbook, the RD Sharma Class 12th FBQ also contains additional practice questions for the students to work out. This makes them increase their speed in finding answers during the examinations.

You will not find FBQs challenging to solve once you commence your practice with the Class 12 RD Sharma Chapter 15 FBQ resource material. The questions are solved in simple methods using shortcuts to find the answers quickly. This saves time during examinations. Moreover, once the students achieve full marks in FBQs, they can easily cross their benchmark scores.

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Class 12 RD Sharma Chapter 15 FBQ Solution are widely used, and hence questions for the public exams are likely to be taken from this book. Preparing with the RD Sharma Class 12th FBQ book will prepare the students for their public exams. Many students of the previous set have been benefitted from using the RD Sharma Class 12 Solutions Chapter 15 FBQ book. Download your own set of RD Sharma books and start preparing for your exams from today.

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1. Which is the prescribed book for the students to clarify their doubts regarding the mathematics chapter 15 FBQ?

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The RD Sharma books provide solutions for the exercise questions, MCQs, FBQs, and VSA portions.

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RD Sharma Solutions Class 12 Mathematics Chapter 15 FBQ

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