These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.3
Question 1.
Find the slope of the tangent to the curve y = 3x4 – 4x at x = 4.
Solution:
y = 3x4 – 4x
Differentiating w.r.t. x,
∴ \(\frac { dy }{ dx }\) = 12x3 – 4
∴ Req. slope = \({ \left( \frac { dy }{ dx } \right) }_{ x=4 }\)
= 12(43) – 4 = 764.
Question 2.
Find the slope of the tangent to the curve y = \(\frac { x-1 }{ x-2 }\), x ≠ 2 at x = 10.
Solution:
y = \(\frac { x-1 }{ x-2 }\), x ≠ 2
Differentiating w.r.t. x,
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
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 3x² – 1
∴ slope of tangent = \(\frac { dy }{ dx }\)
= 3x² – 1
= 3 x 2² – 1
= 11
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 = \(\frac { dy }{ dx }\)
= 3x² – 3
= 3 x 3² – 3
= 24
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:
y = a sin3θ and x = a cos3θ
Differentiating w.r.t. x,
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:
x = 1 – a sin θ and y = b cos² θ
Differentiating w.r.t. x,
Question 7.
Find points at which the tangent to the curve y = x3 – 3x2 – 9x + 7 is parallel to the x-axis.
Solution:
y = x3 – 3x2 – 9x + 7
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)
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)
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)
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:
y = \(\frac { 1 }{ x-1 }\), x ≠ 1
Differentiating w.r.t x
\(\frac { dy }{ dx }\) = \(\frac{-1}{(x-1)^{2}}\)
Since the tangent have slope – 1,
\(\frac { dy }{ dx }\) = – 1 ⇒ \(\frac{-1}{(x-1)^{2}}\) = – 1
⇒ (x – 1)² = 1 ⇒ x – 1 = ± 1
⇒ x = 2 or x = 0
When x = 0, y = \(\frac { 1 }{ 0 – 1 }\) = – 1
When x = 2, y = \(\frac { 1 }{ 2 – 1 }\) = 1
∴ The required points are (0, – 1) and (2, 1)
Equation of the tangent at (0, – 1) is
y – 1 = – 1 (x – 0) ⇒ x + y + 1 = 0
Equation of the tangent at (2, 1) is
y – 1 = – 1(x – 2) ⇒ x + y – 3 = 0
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 = \(\frac { 1 }{ x-3 }\)
Hence there is no tangent to the curve with slope 2.
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:
y = \(\frac { 1 }{ { x }^{ 2 }-2x+3 }\)
Differentiating w.r.t x
∴ \(\frac { dy }{ dx }\) = \(\frac{-(2 x-2)}{\left(x^{2}-2 x+3\right)^{2}}\)
Since the slope of the tangent is zero,
\(\frac { dy }{ dx }\) = 0
Question 13.
Find points on the curve \(\frac { { x }^{ 2 } }{ 9 } + \frac { { y }^{ 2 } }{ 16 } \) = 1 at which the tangents are
i. parallel to x-axis
ii. parallel to y-axis
Solution:
The equation of the curve is \(\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 16 }\) = 1 … (i)
Differentiating both sides w.r.t x
\(\frac { 2 }{ 9 }\)x + \(\frac { 2y }{ 16 }\)\(\frac { dy }{ dx }\) = 0
∴ \(\frac { dy }{ dx }\) = \(\frac { -16x }{ 9y }\)
i. Since the tangent is parallel to x axis,
\(\frac { dy }{ dx }\) = 0 ⇒ \(\frac{-16x}{9y}\)
Hence x = 0
When x = 0, we get \(\frac{0}{9}\) + \(\frac { { y }^{ 2 } }{ 16 }\) = 1
⇒ y² = 16 ⇒ y = ±4
∴ (0,4) and (0, – 4) are the points at which the tangents are parallel to the x axis
ii. Since the tangent is parallel to y axis, its slope is not defined, then the normal is parallel to x-axis whose slope is zero.
∴ (3, 0) and (- 3, 0) are the points at which the tangents are parallel to they axis.
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:
(i) y = x4 – 6x3 + 13x2 – 10x + 5
Differentiating w.r.t x
Equation of the tangent at (0, 5) is
(y- 5) = – 10(x- 0)
i.e., 10x+y-5 = 0
Equation of the normal at (0, 5) is
(y – 5) = – 10(x – 0)
i.e., x – 10y + 50 = 0
(ii) y = x4 – 6x3 + 13x2 – 10x + 5
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 4x3 – 18x2 + 26x – 10
\(\frac { dy }{ dx }\)(1, 3) = 4(1) – 18(1) + 26(1) – 10 = 2
At (1, 3), the slope of the tangent = 2
At (1, 3), the slope of the normal = \(\frac { – 1 }{ 2 }\)
Equation of the tangent at (1, 3) is y – 3 = 2( x – 1)
i.e., y = 2x + 1
Equation of the normal at (1, 3) is
y – 3 = y(x – 1)
i.e., x + 2y – 7 = 0
(iii) y = x3
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 3x2
At(1, 1), \(\frac { dy }{ dx }\) = 3 x 1² = 3
At (1, 1), the slope of the tangent = 3
Equation of the tangent at(1, 1) is
y – 1 = 3(x – 1)
i.e., y = 3x – 2
At (1, 1), the slope o fthe normal = \(\frac { -1 }{ 3 }\)
Equation of the normal at (1, 1) is
y – 1 = \(\frac { -1 }{ 3 }\) (x – 1)
i.e., x + 3y – 4 = 0
(iv) y = x2
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 2x
[ \(\frac { dy }{ dx }\) ](0, 0) = 0
At (0, 0), the slope of the tangent = 0
Equation of the tangent at (0, 0) is
y – 0 = 0(x – 0)
i.e., y = 0 or the x axis
Since the tangent is the x axis, the normal is parallel to y axis.
∴ The equation of the normal at (0, 0) is x = 0, since the normal passes through (0,0)
(v) y = sin t and x = cos t
Differentiating w.r.t. t, we get \(\frac { dy }{ dt }\) = cos t
Question 15.
Find the equation of the tangent line to the curve y = x2 – 2x + 7 which is
(i) parallel to the line 2x – y + 9 = 0
(ii) perpendicular to the line 5y – 15x = 13.
Solution:
y = x2 – 2x + 7
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 2x – 2
\(\frac { dy }{ dx }\) = 2(x – 1) … (1)
(i) Slope of the line 2x – y + 9 = 0 is 2
Since the tangent to the curve is parallel to the line 2x – y + 9 = 0, we get
\(\frac { dy }{ dx }\) = 2
From (1), we get 2(x – 1) = 2 ⇒ x = 2
When x = 2, y = 2² – 2 x 2 + 7 = 7
∴ At (2, 7) the tangent is parallel to 2x – y + 9 = 0
The equation of the tangent to the given curve at (2, 7) is y – 7 = 2(x – 2)
i.e., 2x – y + 3 = 0 or y – 2x – 3 = 0
Slope of the line – 15x = 13 is 3
(ii) Since the tangent to the curve is perpendicular to the line 5y – 15x = 13,
\(\frac { dy }{ dx }\) = \(\frac { -1 }{ 3 }\)
From (1), we get 2(x – 1) = \(\frac { -1 }{ 3 }\)
The tangent is perpendicular to the line 5y – 15x = 13
The equation of the tangent to the given
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
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:
y = x³
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 3x²
The slope of the tangent = 3x²
Since slope of the tangent = y coordinate,
3x² = y or 3x² = x³
= 3x² – x³ = 0 = x²(3 – x) = 0
⇒ x = 3 or x = 0
When x = 0, y = (0)³ = O
When x = 3, y= (3)³ = 27
∴ (0, 0) and (3, 27) are the points at which the slope of the tangent is equal to the y coordinate of the point.
Question 18.
For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.
Solution:
Let, (h, k) be the point at which the tangent passes through the origin.
y = 4x³ – 2x5
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 12x² – 10x4
At (h, k) the slope of the tangent = \(\frac { dy }{ dx }\)(h,k) = 12h² – 10h4
Equation of the tangent at (h, k) is
y – k = (12h² – 10h4) (x – h)
Since the tangent passes through the origin,
(0 – k) = (12h² – 10h4) (0 – h)
⇒ k = 12h³ – 10h5 … (1)
(h, k) is a point on the curve y = 4x³ – 2x5
∴ k = 4h³ – 2h5 … (2)
From (1) & (2) we get
12h³ – 10h5 = 4h³ – 2h5
⇒ 8h³ – 8h5 = 0
⇒ 8h³ (1 – h²) = 0
⇒ 8h³ (1 – h) (1 +h) = 0
⇒ h = 0 or h= 1 or h = – 1
When h = 0, k = 0
When h = 1, k = 2
When h = – 1, k = – 2
∴ (0, 0), (1, 2) and (- 1, – 2) are the points at which the tangent passes through the origin.
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).
Question 20.
Find the equation of the normal at the point (am², am³ ) for the curve ay² = x³ .
Solution:
ay² = x³
Differentiating w.r.t. x,
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:
y = x3 + 2x + 6
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 3x² + 2
Slope of the normal = \(\frac{-1}{3 x^{2}+2}\)
Slope of the line x + 14y + 4 = 0 is \(\frac { -1 }{ 14 }\)
Since the normal is parallel to x + 14y + 4 = 0
\(\frac{-1}{3 x^{2}+2}\) = \(\frac { -1 }{ 14 }\) ⇒ 3x² + 2 = 14
⇒ 3x² = 12 ⇒ x² = 4 ⇒ x = ± 2
When x = 2, y = 18 and when x = – 2, y = – 6
∴ At (2, 18) and (- 2, – 6), the normals are parallel to x + 14y + 4 = 0.
Slope of the normal = \(\frac { -1 }{ 14 }\)
Equation of the normal at (2, 18) is
y – 8 = \(\frac { -1 }{ 14 }\)(x – 2)
i.e., x + 14y – 254 = 0
Equation of the normal at ( -2, -6) is
y + 6 = \(\frac { -1 }{ 14 }\) (x + 2) i.e., x + 14y + 86 = 0
Question 22.
Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at²,2at).
Solution:
y² = 4ax
Differentiating w.r.t. x,
2y\(\frac { dy }{ dx }\) = 4a
Equation of the tangent at (at², 2at) is
y – 2at = \(\frac { 1 }{ t }\)(x – at²)
yt – 2at² = x – at²
x – yt + at² = 0 or ty = x + at²
Slope of the normal at (at², 2at) is – t.
Equation of the normal at (at², 2at) is
y – 2at = – t(x – at²)
xt + y = at³ + 2at
y = – tx + 2at + at³
Question 23.
Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1.
Solution:
The curve are
x = y² … (1)
xy = k … (2)
Let P be the point of intersection
Substituting x = y² in (2),
Since the two curves cut at right angles at P, the product of their slopes at P is – 1.
Taking the cubes we get k² = \(\frac { 1 }{ 8 }\) or 8k² = I
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:
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:
y = \(\sqrt { 3x – 2 } \)
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = \(\frac{3}{2 \sqrt{3 x-2}}\)
Slope of the tangent = \(\frac{3}{2 \sqrt{3 x-2}}\)
Slope of the line 4x – 2y + 5 = 0 is 2
Since the tangent is parallel to the line dy
4x – 2y + 5 = 0, we get \(\frac { dy }{ dx }\) = 2
∴ At(\(\frac { 41 }{ 48 }\), \(\frac { 3 }{ 4 }\)), the tangent is parallel to the line 4x – 2y + 5 = 0
Equation of the tangent at (\(\frac { 41 }{ 48 }\), \(\frac { 3 }{ 4 }\)) is
y – \(\frac { 3 }{ 4 }\) = 2(x – \(\frac { 41 }{ 48 }\)
24y – 18 = 48x – 41
i.e., 48x – 24y = 23
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 = \(\frac { 1 }{ 3 }\)
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
∴ The Point of contact is (1, 2)