In this chapter, we provide NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 pdf, free NCERT solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 book pdf download.

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

∴ = 12x^{3} – 4

∴Req. slope =

= 12 x 4^{3} – 4 = 764.

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

Find the slope of the tangent to the curve at x = 10.**Solution:**

The curve is

**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

= 3x² – 1

∴slope of tangent =

= 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

= 3x² – 3

∴slope of tangent =

= 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 = .**Solution:**

**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 = **Solution:**

**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; = 3 (x – 3) (x + 1)

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

or

⇒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

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

Slope of AB = = 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

⇒ = 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 , x≠1**Solution:**

Here

⇒

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

Find the equation of ail lines having slope 2 which are tangents to the curve , x≠3.**Solution:**

Here

∵ slope of tangent = 2

Which is not possible as (x – 3)² > 0

Thus, no tangent to 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 **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 at which the tangents are

(a) parallel to x-axis

(b) parallel to y-axis**Solution:**

The equation of the curve is …(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 = **Solution:**

Putting x = 0, 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)

= 2x – 2 = 2(x – 1)

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

⇒ Slope of tangent = = 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 = 21 x²

Now m_{1} = slope at x = 2 is = 21 x 2² = 84

and m_{2} = slope at x = -2 is = 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

=>

Tangent is parallel to x-axis, if 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}

**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 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 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

**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)

(c) -3

(d) **Solution:**

(d) ∵ y = 2x² + 3sinx

∴ at

x = 0,

∴ slope = 3

⇒ slope of normal is =

**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,

∴

Slope of the given line y = x + 1 is 1 ∴

y = 2 Putting y= 2 in y² = 4x 2² = 4x

⇒ x = 1

∴ Point of contact is (1,2)

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