OP Malhotra Class 12 Maths Solutions Chapter 11 Applications of Derivatives Ex 11(a)

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S Chand Class 12 ICSE Maths Solutions Chapter 11 Applications of Derivatives Ex 11(a)

Question 1.
Find die slopes of the following curves :
(i) y = \(\frac { 4 }{ x }\) at the point (2, 2)
(ii) y = 2x² – 1 at x = 1
(iii) y = 2x – x² at x = 1
(iv) y = x² – sin x at x = 0
(v) f (x) = 9 sin x + sin 3x at x = \(\frac { π }{ 3 }\)
Solution:
(i) Given equation of curve be y = \(\frac { 4 }{ x }\); Differentiate both sides w.r.t. x, we get
\(\frac { dy }{ dx }\) = – \(\frac { 4 }{ x² }\)
∴ slope of tangent to given curve at (2, 2)
= \(\left(\frac{d y}{d x}\right)_{(2,2)}=\frac{-4}{22}\) = – 1

(ii) Given equation of curve be y = 2x – x²
differentiate both sides w.r.t. x, we get
\(\frac { dy }{ dx }\) = 4x
∴ slope of tangent ro given curve at
x = 1 = \(\left(\frac{d y}{d x}\right)_{x=1}\) = 4 x 1 = 4

(iii) Given equation of curve be y = 2x – x²
Differentiate both sides w.r.t. x; we have \(\frac { dy }{ dx }\) = 2 – 2x
Thus slope of tangent to given curve at x = 1
= \(\left(\frac{d y}{d x}\right)_{x=1}\) = 2 – 2 = 0

(iv) Given equation of curve be y = x² – sin x
differentiate both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = 2x – cos x
∴ slope of tangent to given curve at x = 0
= \(\left(\frac{d y}{d x}\right)_{x=0}\) = 2 x 0 – 1 = – 1

(v) Given equation of curve be f(x) = 9 sin x + sin 3x
Differentiate both sides w.r.t. x; we have
f'(x) = 9 cos x + 3 cos 3x
Thus slope of tangent to given curve at x = π/3
= (f'(x))_x=π/3 = 9cos\(\frac { π }{ 3 }\) + 3cosπ
= \(\frac { 9 }{ 2 }\) – 3 = \(\frac { 3 }{ 2 }\)

Question 2.
Find the slope of the normal to the curve y = 3x² at the point whose x-coordinate is 2.
Solution:
The equation curve be y = 3x² …(i)
when x = 2
∴ from (f); y = 3 x 2² = 12
so the point on given curve (i) be (2, 12)
Diffrentiate both sides of equation (i) w.r.t.

Question 3.
Find the equation of the tangent and normal to the given curves at the points given:
(i) y = 2x² – 3x – 1 at the point (1, 2)
(ii) x = cos t, y = sin t, at t = \(\frac { π }{ 4 }\)
(iii) y = x² + 4x + 1 at the point whose abscissa is 3
(iv) y² = \(\frac{x^3}{4-x}\) at (2, – 2)
(v) \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 at (x₀, y₀)
Solution:
(i) Given equation of curve be y = 2x² – 3x – 1
y = 2x² – 3x – 1
∴ \(\frac { dy }{ dx }\) = 4x – 3
∴ slope of tangent to given curve at (1, 2) = \(\left(\frac{d y}{d x}\right)_{(1,2)}\) = 4 – 3 = 1
& slope of normal to given curve at (1, 2)
= \(\left(\frac{\frac{-1}{d y}}{d x}\right)_{(1,2)}\) = – \(\frac { 1 }{ 1 }\) = – 1
Thus equation of tangent to given curve at point (1, 2) is given by y – 2 = 1 (x – 1)
⇒ x – y + 1 = 0 & the equation of normal to given curve at point (1, 2) be given by y – 2 = – 1 (x – 1)
⇒ x + y – 3 = 0.

(ii) Given equation of curve be x = cos t … (i)
& y = sin t …(ii)
Differential equation (i) & (ii) w.r.t. t; we have

Thus slope of tangent to given curve at t = π/4

Thus, the equation of tangent to given curve at t = π/4 is given by
y – \(\frac{1}{\sqrt{2}}=-1\left(x-\frac{1}{\sqrt{2}}\right)\)
[∵ coordinates of point are \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) ]
⇒ x + y – \(\sqrt{2}\) = 0
The equation of normal to given curve
at \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) is given by
y – \(\frac{1}{\sqrt{2}}=1\left(x-\frac{1}{\sqrt{2}}\right)\) i.e. y = x.

(iii) Given eqn. of curve be
y = x² + 4x + 1 … (1)
at x = 3 ; y = 3² + 12 + 1 = 9 + 13 = 22
Thus the point of contact be (3, 22).
Diff. (1) both sides w.r.t. x ; we get

(iv) Given eqn. of curve be,
y² = \(\frac{x^3}{4-x}\) … (1)
Diff. both sides w.r.t. x ; we get

(v) given eqn. of curve be,

The required equation of tangent at (x₀, y₀) is given by

The required eqn. of Normal at (x₀, y₀) is given by

Question 4.
Find the equation of the tangent and noun al to the parabola y² = 4ax at (at², 2at).
Solution:
Give equation of parabola be y² = 4ax … (1)
Differentiate both sides of equation (i) w.r.t, x;

Thus equation of tangent to given curve at (at², 2at) be given by
y – 2at = \(\frac { 1 }{ t }\) (a – at²)
ty – 2at² = x – at²
⇒ x – ty + at² = 0.
The equation of normal to given curve at (at², 2at) be given by
y – 2at = – t(x – at²)
⇒ tx + y – 2 at – at³ = 0

Question 5.
Find the equations to the tangant to the curve y = (x² – 1) (x – 2) at the points where the curve cuts the x-axis.
Solution:
Given equation of curve be y = (x² – 1) (x – 2) … (i)
it meets x-axis i.e. y = 0
∴ from (f); we have 0 = (x² – 1) (x – 2)
⇒ x = 2, ± 1
∴ points on given curve (i) are (2, 0) & 0, 0); (- 1, 0)
Differentiate both sides of equation (i) w.r.t. x; we have
\(\frac{d y}{d x}=\frac{d}{d x}\left(x^3-2 x^2-x+2\right)=3 x^2-4 x-1\)
Thus slope of tangent to given curve at (2, 0)
= \(\left(\frac{d y}{d x}\right)_{(2,0)}\) = 12 – 8 – 1 = 3
and corresponding equation of tangent to given curve at (2, 0) be given by y – 0 = 3 (x – 2) ⇒ 3x – y – 6 = 0 the slope of tangent to given curve at (1,0)
= \(\left(\frac{d y}{d x}\right)_{(1,0)}\)
= 3 – 4 – 1 = – 2
& corresponding tangent at point (1, 0) be given by
y – 0 = – 2(x – 1) ⇒ 2x + y – 2 = 0
∴ Slope of tangent to given curve at
(-1, 0) = \(\left(\frac{d y}{d x}\right)_{(-1,0)}\) = 3 + 4 – 1 = 6
The eqn of tangent to given curve at (-1, 0) be given by y – 0 = 6 (x + 1)
⇒ 6x – y + 6 = 0.

Question 6.
Find the point on the curve y = 2x² – 6x – 4 at which the tangent is parallel to the x-axis.
Solution:
Given equation of curve
y = 2x² – 6x – 4 … (i)
Let the point on curve (i) be (x₁, y₁) differentiate equation (i) w.r.t. x; we have
\(\frac { dy }{ dx }\) = 4x – 6
∴ slope of tangent to given curve (i) at (x₁, y₁) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\) = 4x₁ – 6 since the tangent is parallel to x-axis.
∴ slope the tangent given curve at (x₁, y₁) = 0

Thus required point on given curve be \(\left(\frac{3}{2}, \frac{-17}{2}\right)\).

Question 7.
Find the points on the curve
y = 12x – x³ at which the slope is zero.
Solution:
Given equation of curve be y = 12x – x³ … (i)
let die point on given curve be (x₁, y₁)
∴ y₁ = 12x – x₁³ … (ii)
differentiate equation (i) both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = 12 – 3x²
∴ slope of tangent to given curve (i) at
(x₁, y₁) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= 12 – 3x₁²
It is given that slope of tangent to given curve at (x₁, y₁) be 0.
∴ 12 – 3x₁² = 0 ⇒ x₁² = 4 ⇒ x₁ = ± 2
when x₁ = 2
∴ from (ii); y₁ = 24 – 2³ = 24 – 8 = 16
when x₁ = – 2
∴ from (ii); y₁ = – 24 – (- 2)³
= – 24 + 8 = – 16
hence the required points on given curve are (2, 16) and (- 2, – 16)

Question 8.
Find the points on the curve x² + y² = 25, the tangents at which are (i) parallel to the x-axis, (ii) parallel to the y-axis.
Solution:
Given equation of curve be x² + y² = 25 … (i)
let the point on given curve (i) be (x₁, y₁)
∴ x₁² + y₁² = 25 … (ii)
differentiate both sides of equation (i) w.r.t.x; we have

Thus required points on curve are (0, ± 5).

(ii) Since the tangent to given curve is parallel to y-axis

Hence the required points on given curve are (± 5, 0)

Question 9.
Find the point on the curve y² = 4x, the tangent at which is parallel to the straight line y = 2x + 4.
Solution:
Equation of given curve be y² = 4x … (i)
let (x₁, y₁) be any point on given curve (i)
∴ y₁² = 4x₁ … (ii)
Differentiate (i) w.r.t. x; we have
2y\(\frac { dy }{ dx }\) = 4 ⇒ \(\frac { dy }{ dx }\) = \(\frac { 2 }{ y }\)
∴ slope of tangent to given curve at (x₁, y₁)
= \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=\frac{2}{y_1}\)
given equation of straight line 2x – y + 4 = 0 … (iii)
∴ slope of line = \(\frac { -2 }{ -1 }\) = 2
since it is given that tangent is parallel to line (iii)
Thus, \(\frac { dy }{ dx }\) = 2 ⇒ y₁ = 1
∴ from (i); 4x₁ = 1 ⇒ x₁ = \(\frac { 1 }{ 4 }\)
Hence, the required point on given curve be (\(\frac { 1 }{ 4 }\), 2)

Question 10.
Find the equation of the tangent line to y = 2x² + 1 which is parallel to the line 4x – y + 3 = 0.
Solution:
Given equation of curve be y = 2x² + 7 … (i)
Let P(x₁, y₁) be any point on curve (i)
∴ y₁ = x₁² + 7 … (ii)
Differentiate equation (i) w.r.t. x; we have
\(\frac { dy }{ dx }\) = 4x
∴ Slope of tangent to curve (1) at (x₁, y₁)
= (\(\frac { dy }{ dx }\)) (x₁, y₁) = 4x₁
Also given eqn. of straight line be 4x – y + 3 = 0
∴ slope of line (3) = \(\frac { -4 }{ -1 }\) = 4
Since it is given that tangent is parallel to line (3).
∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\) = 4 ⇒ 4x₁ = 4
∴ from (2) ; y₁ = 2 x 1² + 7 = 9
Thus, required point on given curve be (1, 9)
Hence the required eqn. of tangent to given curve at (1, 9) be given by
y – 9 = 4(x – 1)
⇒ 4x – y + 5 = 0

Question 11.
Find the equation of the normal line to y = x³ + 2x + 6 which is parallel to the line 14y + x + 4 = 0.
Solution:
Given curve be y = x³ + 2x – 6 … (1)
∴ slope of tangent = \(\frac { dy }{ dx }\) = 3x² + 2
Then slope of Normal = – \(\frac{1}{3 x^2+2}\)
But the normal is || to line x + 14y + 4 = 0,
whose slope is – \(\frac { 1 }{ 14 }\)
Thus both slopes must be equal.
∴ – \(\frac{1}{3 x^2+2}=\frac{-1}{14}\) ⇒ 3x² + 2 = 14
⇒ x² = 4 ⇒ x = ± 2
when x = 2 ; from (1), we have
y = 8 + 4 + 6 = 18
when x = – 2 ; from (1), we get
y = – 8 – 4 + 6 = – 6
Hence the points of contact are (2, 18) and (-2, -6).
∴ eqn. of Normal at (2, 18) is given by
y – 18 = – \(\frac { 1 }{ 14 }\)(x – 2) ⇒ x + 14y – 254 = 0
∴ eqn. of Normal at (- 2, – 6) is given by
y + 6 = – \(\frac { 1 }{ 14 }\) (x + 2) ⇒ x + 14y + 86 = 0

Question 12.
Find a point on the curve y = (x – 3)² where the tangent is parallel to the chord joining (3, 0) and (4, 1).
Solution:
Given eqn. of curve be
y = (x – 3)² … (1)
Let the required point on given curve (1) be (x₁, y₁)
∴ \(\frac { dy }{ dx }\) = 2(x – 3)
Thus, slope of tangent to given curve at
\(\left(x_1, y_1\right)=\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=2\left(x_1-3\right)\)
Now slope of chord joining (3, 0) and (4, 1)
= \(\frac{y_2-y_1}{x_2-x_1}=\frac{1-0}{4-3}\) = 1
Since it is given that tangent is parallel to the given chord so their slopes must be euqal.
∴ 2(x₁ – 3) = 1 ⇒ x₁ = \(\frac { 7 }{ 2 }\)
Also point (x₁, y₁) lies on given curve
∴ y₁ = (x₁ – 3)²
⇒ y₁ = (\(\frac { 7 }{ 2 }\) – 3)² = \(\frac { 1 }{ 4 }\)
Thus the required point on given curve be (\(\frac { 7 }{ 2 }\), \(\frac { 1 }{ 4 }\)).

Question 13.
On the curve y = x + \(\frac { 1 }{ x }\), find the points at which the tangents to the curve are parallel to the x-axis.
Solution:
Given eqn. of curve be
y = x + \(\frac { 1 }{ x }\) … (1)
Let the point on curve (1) be (x₁, y₁)
∴ y₁ = x₁ + \(\frac{1}{x_1}\)
Diff. eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}=1-\frac{1}{x^2}\)
∴ slope of tangent to given curve (1) at (x₁, y₁)
= \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=1-\frac{1}{x_1^2}\)
Since the tangent is parallel to x-axis
∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1, y\right)}=0 \Rightarrow 1-\frac{1}{x_1^2}\) = 0
⇒ x₁² = 1 ⇒ x₁ = ± 1
When x₁ = 1 ∴ from (2) ; we have y₁ = 1 + 1 = 2
When x₁ – 1 ∴ from (2) ; we have y₁ = – 1 – 1 = – 2
Thus, the required point on given curve are (1, 2) and (- 1, – 2).

Question 14.
Find the equation of the tangent to the curve y = cot² x – 2 cot x + 2 at x = \(\frac { π }{ 4 }\).
Solution:
Given eqn. of given curve be
y = cot² x – 2cot x + 2 …(1)
at x = \(\frac { π }{ 4 }\) ∴ from (1) ; we have
y = cot² \(\frac { π }{ 4 }\) – 2 cot \(\frac { π }{ 4 }\) + 2 = 1 – 2 + 2 = 1
Hence the required point on given curve be (\(\frac { π }{ 4 }\), 1)
Diff. eqn. (1) both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = 2cotx (-cosec²x) + 2 cosec²x
at x = \(\frac { π }{ 4 }\);
\(\frac { dy }{ dx }\) = – 2 cot \(\frac { π }{ 4 }\) cosec² \(\frac { π }{ 4 }\) + 2cosec² \(\frac { π }{ 4 }\)
= – 2 x 1 x (\(\sqrt{2}\))² + 2(\(\sqrt{2}\))² = 0
Thus eqn. of tangent to given curve at
(\(\frac { π }{ 4 }\), 1) is given by y – 1 = 0 (x – \(\frac { π }{ 4 }\))
y = 1

Question 15.
The equation of tangent at (2, 3) on the curve y² = ax³ + b is y = 4x – 5. Find the values of a and b.
Solution:
Given eqn. of curve be y² = ax³ + b …(1)
Diff. both sides of eqn. (1) w.r.t. x ; we get
2y \(\frac{d y}{d x}=3 a x^2 \Rightarrow \frac{d y}{d x}=\frac{3 a x^2}{2 y}\)
∴ slope of tangent at (2, 3) = \(\left(\frac{d y}{d x}\right)_{(2,3)}\)
= \(\frac { 3a×4 }{ 6 }\) = 2a
also eqn. of tangent to given curve be y = 4x – 5
∴ slope of given tangent = \(\frac { -4 }{ -1 }\) = 4
Now \(\left(\frac{d y}{d x}\right)_{(2,3)}\) = 4
⇒ 2a = 4 ⇒ a = 2
Also the given point (2, 3) lies in eqn. (1)
∴ 9 = 8a + b … (2)
⇒ 9 = 16 + b
⇒ b = – 7 [∵ a = 2]
Hence a = 2 and b = – 7