Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 4 Applications of Derivatives Ex 4.1 Questions and Answers.
Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1
Solution & Step-by-Step Answer:
y = 3x2 – x + 1 ∴ (3x2 – x + 1) = 3 × 2x – 1 + 0 = 6x – 1 ∴ = 6(1) – 1 = 5 = slope of the tangent at (1, 3). ∴ the equation of the tangent at (1, 3) is y – 3 = 5(x – 1) ∴ y – 3 = 5x – 5 ∴ 5x – y – 2 = 0. The slope of the normal at (1, 3) = ∴ the equation of the normal at (1, 3) is y – 3 = (x – 1) ∴ 5y – 15 = -x + 1 ∴ x + 5y – 16 = 0 Hence, the equations of the tangent and normal are 5x – y – 2 = 0 and x + 5y – 16 = 0 respectively.
(ii) 2x2+ 3y2= 5 at (1, 1)
Solution:
2x2+ 3y2= 5
Differentiating both sides w.r.t. x, we get
= slope of the tangent at (1, 1)
∴ the equation of the tangent at (1, 1) is
y – 1 = (x – 1)
∴ 3y – 3 = -2x + 2
∴ 2x + 3y – 5 = 0.
The slope of normal at (1, 1) =
∴ the equation of the normal at (1, 1) is
y – 1 = (x – 1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0
Hence, the equations of the tangent and normal are 2x + 3y – 5 = 0 and 3x – 2y – 1 = 0 respectively.
(iii) x2+ y2+ xy = 3 at (1, 1)
Solution:
x2+ y2+ xy = 3
Differentiating both sides w.r.t. x, we get
= slope of the tangent at (1, 1)
the equation of the tangent at (1, 1) is
y – 1= -1(x – 1)
∴ y – 1 = -x + 1
∴ x + y = 2
The slope of the normal at (1, 1) =
=
= 1
∴ the equation of the normal at (1, 1) is y – 1 = 1(x – 1)
∴ y – 1 = x – 1
∴ x – y = 0
Hence, the equations of tangent and normal are x + y = 2 and x – y = 0 respectively.
Solution & Step-by-Step Answer:
Let P(x1, y1) be the point on the curve y = x2 + 5 where the tangent is parallel to the line 4x – y + 1 = 0. Differentiating y = x2 + 5 w.r.t. x, we get (x2 + 5) = 2x + 0 = 2x = slope of the tangent at (x1, y1) Let m1 = 2x1 The slope of the line 4x – y + 1 = 0 is m2 = = 4 Since, the tangent at P(x1, y1) is parallel to the line 4x – y + 1 = 0, m1 = m2 ∴ 2x1 = 4 ∴ x1 = 2 Since, (x1, y1) lies on the curve y = x2 + 5, y1 = + 5 ∴ y1 = (2)2 + 5 = 9 ……[x1 = 2] ∴ the coordinates of the point are (2, 9) and the slope of the tangent = m1 = m2 = 4. ∴ the equation of the tangent at (2, 9) is y – 9 = 4(x – 2) ∴ y – 9 = 4x – 8 ∴ 4x – y + 1 = 0 Slope of the normal = ∴ the equation of the normal at (2, 9) is y – 9 = (x – 2) ∴ 4y – 36 = -x + 2 ∴ x + 4y – 38 = 0 Hence, the equations of tangent and normal are 4x – y + 1 = 0 and x + 4y – 38 = 0 respectively.
Solution & Step-by-Step Answer:
Let P(x1, y1) be the point on the curve y = 3x2 – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0. Differentiating y = 3x2 – 3x – 5 w.r.t. x, we get (3x2 – 3x – 5) = 3 × 2x – 3 × 1 – 0 = 6x – 3 ∴ = slope of the tangent at (x1, y1) Let m1 = 6x1 – 3 The slope of the line 3x – y + 1 = 0 m2 = = 3 Since, the tangent at P(x1, y1) is parallel to the line 3x – y + 1 = 0, m1 = m2 ∴ 6x1 – 3 = 3 ∴ 6x1 = 6 ∴ x1 = 1 Since, (x1, y1) lies on the curve y = 3x2 – 3x – 5, , where x1 = 1 = 3(1)2 – 3(1) – 5 = 3 – 3 – 5 = -5 ∴ the coordinates of the point are (1, -5) and the slope of the tangent = m1 = m2 = 3. ∴ the equation of the tangent at (1, -5) is y – (-5) = 3(x – 1) ∴ y + 5 = 3x – 3 ∴ 3x – y – 8 = 0 Slope of the normal = ∴ the equation of the normal at (1, -5) is y – (-5) = (x – 1) ∴ 3y + 15 = -x + 1 ∴ x + 3y + 14 = 0 Hence, the equations of tangent and normal are 3x – y – 8 = 0 and x + 3y + 14 = 0 respectively.