NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3.

- Application of Derivatives Class 12 Ex 6.1
- Application of Derivatives Class 12 Ex 6.2
- Application of Derivatives Class 12 Ex 6.4
- Application of Derivatives Class 12 Ex 6.5

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Application of Derivatives |

Exercise |
Ex 6.3 |

Number of Questions Solved |
27 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3

**Ex 6.3 Class 12 Maths Question 1.**

Find the slope of the tangent to the curve y = 3x^{4} – 4x at x = 4.

**Solution:**

The curve is y = 3x^{4} – 4x

∴\(\frac { dy }{ dx }\) = 12x^{3} – 4

∴Req. slope = \({ \left( \frac { dy }{ dx } \right) }_{ x=4 }\)

= 12 x 4^{3} – 4 = 764.

**Ex 6.3 Class 12 Maths Question 2.**

Find the slope of the tangent to the curve \(y=\frac { x-1 }{ x-2 } ,x\neq 2\) at x = 10.

**Solution:**

The curve is \(y=\frac { x-1 }{ x-2 } \)

**Ex 6.3 Class 12 Maths Question 3.**

Find the slope of the tangent to curve y = x^{3} – x + 1 at the point whose x-coordinate is 2.

**Solution:**

The curve is y = x^{3} – x + 1

\(\frac { dy }{ dx }\) = 3x² – 1

∴slope of tangent = \({ \left( \frac { dy }{ dx } \right) }_{ x=2 }\)

= 3 x 2² – 1

= 11

**Ex 6.3 Class 12 Maths Question 4.**

Find the slope of the tangent to the curve y = x^{3} – 3x + 2 at the point whose x-coordinate is 3.

**Solution:**

The curve is y = x^{3} – 3x + 2

\(\frac { dy }{ dx }\) = 3x² – 3

∴slope of tangent = \({ \left( \frac { dy }{ dx } \right) }_{ x=3 }\)

= 3 x 3² – 3

= 24

**Ex 6.3 Class 12 Maths Question 5.**

Find the slope of the normal to the curve x = a cos^{3} θ, y = a sin^{3} θ at θ = \(\frac { \pi }{ 4 } \) .

**Solution:**

\(\frac { dx }{ d\theta } =-3a\quad { cos }^{ 2 }\theta sin\theta ,\frac { dy }{ d\theta } =3a\quad { sin }^{ 2 }\theta cos\theta \)

**Ex 6.3 Class 12 Maths Question 6.**

Find the slope of the normal to the curve x = 1 – a sin θ, y = b cos² θ at θ = \(\frac { \pi }{ 2 } \)

**Solution:**

\(\frac { dx }{ d\theta } =-a\quad cos\theta \quad \& \quad \frac { dy }{ d\theta } =2b\quad cos\theta (-sin\theta )\)

**Ex 6.3 Class 12 Maths Question 7.**

Find points at which the tangent to the curve y = x^{3} – 3x^{2} – 9x + 7 is parallel to the x-axis.

**Solution:**

Differentiating w.r.t. x; \(\frac { dy }{ dx }\) = 3 (x – 3) (x + 1)

Tangent is parallel to x-axis if the slope of tangent = 0

or \(\frac { dy }{ dx }=0\)

⇒3(x + 3)(x + 1) = 0

⇒x = -1, 3

when x = -1, y = 12 & When x = 3, y = – 20

Hence the tangent to the given curve are parallel to x-axis at the points (-1, -12), (3, -20)

**Ex 6.3 Class 12 Maths 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 equation of the curve is y = (x – 2)²

Differentiating w.r.t x

\(\frac { dy }{ dx }=2(x-2)\)

The point A and B are (2,0) and (4,4) respectively.

Slope of AB = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 4-0 }{ 4-2 } =\frac { 4 }{ 2 } \) = 2 …(i)

Slope of the tangent = 2 (x – 2) ….(ii)

from (i) & (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)

**Ex 6.3 Class 12 Maths Question 9.**

Find the point on the curve y = x^{3} – 11x + 5 at which the tangent is y = x – 11.

**Solution:**

Here, y = x^{3} – 11x + 5

⇒ \(\frac { dy }{ dx }\) = 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 & when x = -2,y = -13

But (-2, -13) does not lie on the curve

∴ y = x – 11 is the tangent at (2, -9)

**Ex 6.3 Class 12 Maths Question 10.**

Find the equation of all lines having slope -1 that are tangents to the curve \(y=\frac { 1 }{ x-1 }\), x≠1

**Solution:**

Here

\(y=\frac { 1 }{ x-1 }\)

⇒ \(\frac { dy }{ dx } =\frac { -1 }{ { (x-1) }^{ 2 } } \)

**Ex 6.3 Class 12 Maths Question 11.**

Find the equation of ail lines having slope 2 which are tangents to the curve \(y=\frac { 1 }{ x-3 }\), x≠3.

**Solution:**

Here

\(y=\frac { 1 }{ x-3 }\)

\(\frac { dy }{ dx } ={ (-1)(x-3) }^{ -2 }=\frac { -1 }{ { (x-3) }^{ 2 } } \)

∵ slope of tangent = 2

\(\frac { -1 }{ { (x-3) }^{ 2 } } =2\Rightarrow { (x-3) }^{ 2 }=-\frac { 1 }{ 2 } \)

Which is not possible as (x – 3)² > 0

Thus, no tangent to \(y=\frac { 1 }{ x-3 }\) has slope 2.

**Ex 6.3 Class 12 Maths Question 12.**

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

**Solution:**

Let the tangent at the point (x_{1}, y_{1}) to the curve

**Ex 6.3 Class 12 Maths Question 13.**

Find points on the curve \(\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 16 } =1\) at which the tangents are

(a) parallel to x-axis

(b) parallel to y-axis

**Solution:**

The equation of the curve is \(\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 16 } =1\)…(i)

**Ex 6.3 Class 12 Maths Question 14.**

Find the equations of the tangent and normal to the given curves at the indicated points:

(i) y = x^{4} – 6x^{3} + 13x^{2} – 10x + 5 at (0,5)

(ii) y = x^{4} – 6x^{3} + 13x^{2} – 10x + 5 at (1,3)

(iii) y = x^{3} at (1, 1)

(iv) y = x^{2} at (0,0)

(v) x = cos t, y = sin t at t = \(\frac { \pi }{ 4 } \)

**Solution:**

\(\frac { dy }{ dx } ={ 4x }^{ 3 }-18{ x }^{ 2 }+26x-10\)

Putting x = 0, \(\frac { dy }{ dx } \) at (0,5) = – 10

**Ex 6.3 Class 12 Maths Question 15.**

Find the equation of the tangent line to the curve y = x^{2} – 2x + 7 which is

(a) parallel to the line 2x – y + 9 = 0

(b) perpendicular to the line 5y – 15x = 13.

**Solution:**

Equation of the curve is y = x² – 2x + 7 …(i)

\(\frac { dy }{ dx } \) = 2x – 2 = 2(x – 1)

(a) Slope of the line 2x – y + 9 = 0 is 2

⇒ Slope of tangent = \(\frac { dy }{ dx } \) = 2(x – 1) = 2

**Ex 6.3 Class 12 Maths Question 16.**

Show that the tangents to the curve y = 7x^{3} + 11 at the points where x = 2 and x = – 2 are parallel.

**Solution:**

Here, y = 7x^{3} + 11

=> x \(\frac { dy }{ dx }\) = 21 x²

Now m_{1} = slope at x = 2 is \({ \left( \frac { dy }{ dx } \right) }_{ x=2 }\) = 21 x 2² = 84

and m_{2} = slope at x = -2 is \({ \left( \frac { dy }{ dx } \right) }_{ x=-2 }\) = 21 x (-2)² = 84

Hence, m_{1} = m_{2} Thus, the tangents to the given curve at the points where x = 2 and x = – 2 are parallel

**Ex 6.3 Class 12 Maths 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

**Solution:**

Let P (x_{1}, y_{1}) be the required point.

The given curve is: y = x^{3}

**Ex 6.3 Class 12 Maths Question 18.**

For the curve y = 4x^{3} – 2x^{5}, find all the points at which the tangent passes through the origin.

**Solution:**

Let (x_{1}, y_{1}) be the required point on the given curve y = 4x^{3} – 2x^{5}, then y_{1} = 4x_{1}^{3} – 2x_{1}^{5} …(i)

**Ex 6.3 Class 12 Maths 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.

**Solution:**

Here, x^{2} + y^{2} – 2x – 3 = 0

=> \(\frac { dy }{ dx } =\frac { 1-x }{ y } \)

Tangent is parallel to x-axis, if \(\frac { dy }{ dx }=0\) i.e.

if 1 – x = 0

⇒ x = 1

Putting x = 1 in (i)

⇒ y = ±2

Hence, the required points are (1,2), (1, -2) i.e. (1, ±2).

**Ex 6.3 Class 12 Maths Question 20.**

Find the equation of the normal at the point (am^{2}, am^{3}) for the curve ay^{2} = x^{3}.

**Solution:**

Here, ay^{2} = x^{3}

\(2ay\frac { dy }{ dx } ={ 3x }^{ 2 }\Rightarrow \frac { dy }{ dx } =\frac { { 3x }^{ 2 } }{ 2ay } \)

**Ex 6.3 Class 12 Maths Question 21.**

Find the equation of the normal’s to the curve y = x^{3} + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

**Solution:**

Let the required normal be drawn at the point (x_{1}, y_{1})

The equation of the given curve is y = x^{3} + 2x + 6 …(i)

**Ex 6.3 Class 12 Maths Question 22.**

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

**Solution:**

**Ex 6.3 Class 12 Maths 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)

**Ex 6.3 Class 12 Maths 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}).

**Solution:**

**Ex 6.3 Class 12 Maths 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.

**Solution:**

Let the point of contact of the tangent line parallel to the given line be P (x_{1}, y_{1}) The equation of the curve is \(y=\sqrt { 3x-2 } \)

**Choose the correct answer in Exercises 26 and 27.**

**Ex 6.3 Class 12 Maths Question 26.**

The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is

(a) 3

(b) \(\frac { 1 }{ 3 }\)

(c) -3

(d) \(-\frac { 1 }{ 3 }\)

**Solution:**

(d) ∵ y = 2x² + 3sinx

∴\(\frac { dy }{ dx }=4x+3cosx\) at

x = 0,\(\frac { dy }{ dx }=3\)

∴ slope = 3

⇒ slope of normal is = \(\frac { 1 }{ 3 }\)

**Ex 6.3 Class 12 Maths 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 curve is y² = 4x,

∴ \(\frac { dy }{ dx } =\frac { 4 }{ 2y } =\frac { 2 }{ y } \)

Slope of the given line y = x + 1 is 1 ∴ \(\frac { 2 }{ y }=1\)

y = 2 Putting y= 2 in y² = 4x 2² = 4x

⇒ x = 1

∴ Point of contact is (1,2)

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