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# Class 12 NCERT Solutions- Mathematics Part I – Chapter 6 Application of Derivatives -Exercise 6.3 | Set 1

### Question 1. Find the slope of the tangent to the curve y = 3x4 â€“ 4x at x = 4.

Solution:

Given curve: y = 3x4 – 4x

On differentiating w.r.t x, we get

dy/dx = 12x3 – 4

Now, we find the slope of the tangent to the given curve at x = 4 is

= dy/dx = 12(4)3 – 4 = 764

Hence, the slope is 764

### Question 2. Find the slope of the tangent to the curve , x â‰  2 at x = 10.

Solution:

Given curve:

On differentiating w.r.t x, we get

Now, we find the slope of the tangent to the given curve at x = 10 is

Hence, the slope is -1/64

### Question 3. Find the slope of the tangent to curve y = x3 â€“ x + 1 at the point whose x-coordinate is 2.

Solution:

Given curve: y = x3 – x + 1

On differentiating w.r.t x, we get

Now, we find the slope of the tangent to the given curve at x = 2 is

Hence, the slope is 11

### Question 4. Find the slope of the tangent to the curve y = x3 â€“3x + 2 at the point whose x-coordinate is 3.

Solution:

Given curve: y = x3 – 3x + 2

On differentiating w.r.t x, we get

dy/dx = 3x2 – 3

Now, we find the slope of the tangent to the given curve at x = 3 is

dy/dx = 3(3)2 – 3 = 24

Hence, the slope is 24

### Question 5. Find the slope of the normal to the curve x = acos3Î¸, y = asin3Î¸ at Î¸ = Ï€/4.

Solution:

Given curve: x = acos3Î¸ = f(Î¸)

y = asin3Î¸ = g(Î¸)

To find slope of the normal of the curve at Î¸ = Ï€/4

Now, slope of the normal is

-(1)

= a.3.cos2 Î¸.(-sin Î¸)

= a.3sin2 Î¸.cos Î¸

-(using eq(1))

Now, we find the slope of the tangent to the given curve at Î¸ = Ï€/4 is

The slope of normal of the parametric curve

Hence, the slope is 1

### Question 6. Find the slope of the normal to the curve x = 1 – asinÎ¸, y = bcos2Î¸ at Î¸ = Ï€/2.

Solution:

Given curve: x = 1 – a sinÎ¸

y = b cos2Î¸

Now, slope of normal is

-(1)

-(using eq(1))

Now, we find the slope of the tangent to the given curve at Î¸ = Ï€/2 is

Hence, the slope is -a/2b.

### Question 7. Find points at which the tangent to the curve y = x3 â€“ 3x2 â€“ 9x + 7 is parallel to the x-axis.

Solution:

Given curve: y = x3 â€“ 3x2 â€“ 9x + 7

On differentiating w.r.t x, we get

dy/dx = 3x2 – 6x – 9

For tangent to be parallel to x-axis, slope is 0. So dy/dx = 0.

3x2 – 6x – 9 = 0

3(x2 – 2x – 3) = 0

3(x2 + x – 3x – 3) = 0

3(x(x + 1) – 3(x + 1)) = 0

3(x + 1)(x – 3) = 0

x = -1 or x = 3

For x = -1, y = (-1)3 – 3(-1)2 – 9(-1) + 7

x = -1, y = -1 – 3 + 9 + 7 = 12

Hence, the first point is (-1, 12)

### Question 8. Find a point on the curve y = (x â€“ 2)2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).

Solution:

Given curve: y = (x – 2)2

On differentiating w.r.t x, we get

dy/dx = 2(x – 2)                 -(1)

Given that, the tangent is parallel to the chord joining the points (2, 0) & (4, 4)

Slope of the chord =

Now equality dy/dx = slope of chord

2(x – 2) = 2

x – 2 = 1

x = 3

y = (x – 2)2

y = (3 – 2)2 = 1

Hence, the point on the curve y = (x – 2)2 is (3, 1)

### Question 9. Find the point on the curve y = x3 â€“ 11x + 5 at which the tangent is y = x â€“ 11.

Solution:

Given curve: y = x3 – 11x + 5

Given tangent: y = x – 11

From the given tangent, we can find out the slope comparing y = x – 11 with y = mx + c, we get

Slope(m) = 1

Now y = x3 – 11x + 5

dy/dx = 3x2 – 11                  -(1)

dy/dx = slope = 1

So, from eq(1), we get

3x2 – 11 = 1

3x2 = 12

x2 = 4

x = Â±2

If x = +2, y = 23 – 11(2) + 5 = -9

If x = -2, y = (-2)3 – 11(-2) + 5 = 19

The points must lie on the tangent as well.

Only (2,-9) is satisfying the tangent equation.

So the point on the curve whose tangent is y = x – 11 is (2,-9).

### Question 10. Find the equation of all lines having slope â€“1 that are tangents to the curve y = [Tex]  [/Tex], x â‰  1.

Solution:

Given curve: y =

-(1)

Now given slope = -1 & we know that dy/dx = slope, so

dy/dx = -1                  -(1)

From 1 & 2, we get

-1 =

(x – 1)2 = 1

x = 1 Â± 1

x1 = 2 & x2 = 0

Now corresponding to these x1 & x2 we need to find out y1 & y

The points are (2, 1) & (0, -1)

Now equations slope is -1

Using point slope form the first tangent equation is

(y – y1) = m(x – x1)

y – 1 = -1(x – 2)

= x + y = 3

Using point slope from the second tangent equation is

(y – y2) = m(x – x2)

y – (-1) = -1(x – 0)

= x + y + 1 = 0

### Question 11. Find the equation of all lines having slope 2 which are tangents to the curve , x â‰  3.

Solution:

Given curve: y = 1/(x – 3)

dy/dx = -1/(x – 3)2 = slope               -(dy/dx is slope)

Now given slope is 2, so

dy/dx = -1/(x – 3)2 = 2

(x – 3)2 = -1/2              -(1) (not possible)

Now because there is no real value of x which can satisfy 1, therefore no such tangent exists on the curve y = 1/x – 3 whose is 2.

### Question 12. Find the equations of lines having slope 0 which are tangent to the curve .

Solution:

Given curve,

On differentiating w.r.t x, we get

-(chain rule)

Given slope = 0 = dy/dx

So,

x – 1 = 0

x = 1

For x = 1, y =

So the equation of tangent from point slope from is

y – y1 = m(x – x1)

= 0(x – 1)

2y – 1 = 0

### (i) Parallel to x-axis            (ii)Parallel to y-axis

Solution:

Given curve:

-(1)

(i) If tangent is parallel to x-axis then it means slope is 0 or dy/dx = 0

On differentiating both sides of equations (1) we get,

Now slope = 0, so

= x1 = 0

For x1 = 0,

y12 = 16

y1 = Â±4

The coordinates are (0, 4) & (0, -4)

(ii) Now, if tangent is parallel to y-axis to the dy/dx or slope is not defined or dy/dx = 0

On differentiating equations(1) with respect to y, we get

y2 = 0

For y2 = 0,

x22 = 9

x2 = Â±3

Hence, the coordinates are (3, 0) & (-3, 0)