Applications Of Derivatives Class 12 OP Malhotra Exe-11A ISC Maths Solutions Ch-11 Solutions. In this article you would learn about geometrical interpretation of dy / dx. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Applications Of Derivatives Class 12 OP Malhotra Exe-11A ISC Maths Solutions Ch-11

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-11	Applications Of Derivatives
Writer	OP Malhotra
Exe-11(A)	geometrical interpretation of dy/dx

geometrical interpretation of dy / dx

Applications Of Derivatives Class 12 OP Malhotra Exe-11A ISC Maths Solutions Ch-11 Solutions.

Que-1: Find die slopes of the following curves :
(i) y = 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 = π/3

Sol: (i) Given equation of curve be y = 4/x ; Differentiate both sides w.r.t. x, we get
dy/dx = – 4x²
∴ slope of tangent to given curve at (2, 2)
= (dy/dx)_(2,2) = -4/2² = -1

(ii) Given equation of curve be y = 2x – x²
differentiate both sides w.wr.t. x, e get
dy/dx = 4x
∴ slope of tangent ro given curve at
x = 1 = (dy/dx) _x=1 = 4 x 1 = 4

(iii) Given equation of curve be y = 2x – x²
Differentiate both sides w.r.t. x; we have dy/dx = 2 – 2x
Thus slope of tangent to given curve at x = 1
= (dy/dx) _x=1 = 2 – 2 = 0

(iv) Given equation of curve be y = x² – sin x
differentiate both sides w.r.t. x; we have
dy/dx = 2x – cos x
∴ slope of tangent to given curve at x = 0
= (dy/dx) _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 = 9cosπ/3 + 3cosπ
= 9/2 – 3 = 3/2

Que-2: Find the slope of the normal to the curve y = 3x² at the point whose x-coordinate is 2.

Sol: 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.

Que-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 = π/4
(iii) y = x² + 4x + 1 at the point whose abscissa is 3
(iv) y² = x³/4-x at (2, – 2)
(v) x²/a² – y²/b² = 1 at (x₀, y₀)

Sol: (i) Given equation of curve be y = 2x² – 3x – 1
y = 2x² – 3x – 1
∴ dy/dx = 4x – 3
∴ slope of tangent to given curve at (1, 2) = (dy/dx)_(1,2) = 4 – 3 = 1
& slope of normal to given curve at (1, 2)
= (-1/dy/dx)_(1,2) = -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 – 1/√2 = -1(x – 1/√2)
[∵ coordinates of point are (1/√2,1/√2) ]
⇒ x + y – √2 = 0
The equation of normal to given curve
at (1/√2,1/√2) is given by
y – √2 = 1(x -1/√2) 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 = x³/4-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

Que-4: Find the equation of the tangent and noun al to the parabola y² = 4ax at (at², 2at).

Sol: 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 = 1t (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

Que-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.

Sol: 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
dy/dx = d/dx(x³-2x²-x+2) = 3x²-4x+1
Thus slope of tangent to given curve at (2, 0)
= (dy/dx)_(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)
= (dy/dx)_(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) = (dy/dx)_(-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.

Que-6: Find the point on the curve y = 2x² – 6x – 4 at which the tangent is parallel to the x-axis.

Sol: 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
dy/dx = 4x – 6
∴ slope of tangent to given curve (i) at (x₁, y₁) = (dy/dx)_{(x1, y1)} = 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 (3/2,-17/2).

Que-7: Find the points on the curve
y = 12x – x³ at which the slope is zero.

Sol: 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
dy/dx = 12 – 3x²
∴ slope of tangent to given curve (i) at
(x₁, y₁) = (dy/dx)_{(x1, y1)}
= 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)

Que-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.

Sol: 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)

Que-9: Find the point on the curve y² = 4x, the tangent at which is parallel to the straight line y = 2x + 4.

Sol: 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
2 dy/dx = 4 ⇒ dy/dx = 2/y
∴ slope of tangent to given curve at (x₁, y₁)
= (dy/dx)_{(x1, y1)} = 2/y1
given equation of straight line 2x – y + 4 = 0 … (iii)
∴ slope of line = -2/-1 = 2
since it is given that tangent is parallel to line (iii)
Thus, dy/dx = 2 ⇒ y₁ = 1
∴ from (i); 4x₁ = 1 ⇒ x₁ = 1/4
Hence, the required point on given curve be (1/4,2)

Que-10: Find the equation of the tangent line to y = 2x² + 1 which is parallel to the line 4x – y + 3 = 0.

Sol: 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
dy/dx = 4x
∴ Slope of tangent to curve (1) at (x₁, y₁)
= (dy/dx)_{(x1, y1)} = 4x₁
Also given eqn. of straight line be 4x – y + 3 = 0
∴ slope of line (3) = -4/-1 = 4
Since it is given that tangent is parallel to line (3).
∴ (dy/dx)_{(x1, y1)} = 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

Que-11: Find the equation of the normal line to y = x³ + 2x + 6 which is parallel to the line 14y + x + 4 = 0.

Sol: Given curve be y = x³ + 2x – 6 … (1)
∴ slope of tangent = dy/dx = 3x² + 2
Then slope of Normal = – 1/3x²+2
But the normal is || to line x + 14y + 4 = 0,
whose slope is – 1/14
Thus both slopes must be equal.
∴ – 1/3x²+2 = -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 = –1/14(x – 2) ⇒ x + 14y – 254 = 0
∴ eqn. of Normal at (- 2, – 6) is given by
y + 6 = – 1/14(x + 2) ⇒ x + 14y + 86 = 0

Que-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).

Sol: Given eqn. of curve be
y = (x – 3)² … (1)
Let the required point on given curve (1) be (x₁, y₁)
∴ dy/dx = 2(x – 3)
Thus, slope of tangent to given curve at
(x1,y1) = (dy/dx)_{(x1, y1)} = 2(x1-3)
Now slope of chord joining (3, 0) and (4, 1)
= y2-y1/x2-x1 1-0/4-3 = 1
Since it is given that tangent is parallel to the given chord so their slopes must be equal.
∴ 2(x₁ – 3) = 1 ⇒ x₁ = 7/2
Also point (x₁, y₁) lies on given curve
∴ y₁ = (x₁ – 3)²
⇒ y₁ = (7/2 – 3)² = 1/4
Thus the required point on given curve be (7/2,1/4).

Que-13: On the curve y = x +1x, find the points at which the tangents to the curve are parallel to the x-axis.

Sol: Given eqn. of curve be
y = x + 1/x… (1)
Let the point on curve (1) be (x₁, y₁)
∴ y₁ = x₁ + 1/x1
Diff. eqn. (1) w.r.t. x ; we have
dy/dx = 1-1/x²
∴ slope of tangent to given curve (1) at (x₁, y₁)
= (dy/dx)_{(x1, y1)} = 1-1/x₁²
Since the tangent is parallel to x-axis
∴ (dy/dx)_{(x1, y1)} = 0 ⇒ 1-1/x₁² = 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).

Que-14: Find the equation of the tangent to the curve y = cot² x – 2 cot x + 2 at x = π/4.

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

Que-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.

Sol: Given eqn. of curve be y² = ax³ + b …(1)
Diff. both sides of eqn. (1) w.r.t. x ; we get
2y dy/dx = 3ax² ⇒ dy/dx = 3ax²/y
∴ slope of tangent at (2, 3) = (dy/dx)_(2,3)
=3a×4/6 = 2a
also eqn. of tangent to given curve be y = 4x – 5
∴ slope of given tangent = -4/-1 = 4
Now (dy/dx)_(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

–: End Applications Of Derivatives Class 12 OP Malhotra Exe-11A ISC Math Ch-11 Solution :–

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