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

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## 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 = 3x4 – 4x at x = 4.
Solution:
The curve is y = 3x4 – 4x $frac { dy }{ dx }$ = 12x3 – 4
∴Req. slope = ${ left( frac { dy }{ dx } right) }_{ x=4 }$
= 12 x 43 – 4 = 764.

Ex 6.3 Class 12 Maths Question 2.
Find the slope of the tangent to the curve $y=frac { x-1 }{ x-2 } ,xneq 2$ 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 = x3 – x + 1 at the point whose x-coordinate is 2.
Solution:
The curve is y = x3 – 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 = x3 – 3x + 2 at the point whose x-coordinate is 3.
Solution:
The curve is y = x3 – 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 cos3 θ, y = a sin3 θ at θ = $frac { pi }{ 4 }$ .
Solution: $frac { dx }{ dtheta } =-3aquad { cos }^{ 2 }theta sintheta ,frac { dy }{ dtheta } =3aquad { sin }^{ 2 }theta costheta$ 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: $frac { dx }{ dtheta } =-aquad costheta quad & quad frac { dy }{ dtheta } =2bquad costheta (-sintheta )$ Ex 6.3 Class 12 Maths Question 7.
Find points at which the tangent to the curve y = x3 – 3x2 – 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 = x3 – 11x + 5 at which the tangent is y = x – 11.
Solution:
Here, y = x3 – 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 $frac { -1 }{ { (x-3) }^{ 2 } } =2Rightarrow { (x-3) }^{ 2 }=-frac { 1 }{ 2 }$
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 (x1, y1) 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 = x4 – 6x3 + 13x2 – 10x + 5 at (0,5)
(ii) y = x4 – 6x3 + 13x2 – 10x + 5 at (1,3)
(iii) y = x3 at (1, 1)
(iv) y = x2 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 = x2 – 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 = 7x3 + 11 at the points where x = 2 and x = – 2 are parallel.
Solution:
Here, y = 7x3 + 11
=> x $frac { dy }{ dx }$ = 21 x²
Now m1 = slope at x = 2 is ${ left( frac { dy }{ dx } right) }_{ x=2 }$ = 21 x 2² = 84
and m2 = slope at x = -2 is ${ left( frac { dy }{ dx } right) }_{ x=-2 }$ = 21 x (-2)² = 84
Hence, m1 = m2 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 = x3 at which the slope of the tangent is equal to the y-coordinate of the point
Solution:
Let P (x1, y1) be the required point.
The given curve is: y = x3 Ex 6.3 Class 12 Maths Question 18.
For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.
Solution:
Let (x1, y1) be the required point on the given curve y = 4x3 – 2x5, then y1 = 4x13 – 2x15 …(i)  Ex 6.3 Class 12 Maths Question 19.
Find the points on the curve x2 + y2 – 2x – 3 = 0 at which the tangents are parallel to the x-axis.
Solution:
Here, x2 + y2 – 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 (am2, am3) for the curve ay2 = x3.
Solution:
Here, ay2 = x3 $2ayfrac { dy }{ dx } ={ 3x }^{ 2 }Rightarrow frac { dy }{ dx } =frac { { 3x }^{ 2 } }{ 2ay }$ Ex 6.3 Class 12 Maths Question 21.
Find the equation of the normal’s to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
Solution:
Let the required normal be drawn at the point (x1, y1)
The equation of the given curve is y = x3 + 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 (x0 ,y0).
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 (x1, y1) 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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