RBSE Solutions for Class 12 Maths Chapter 8 Application of Derivatives is part of RBSE Solutions for Class 12 Maths. Here we have given Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives.
Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Ex 8.3
Question 1.
For curve y = x3 – x find slope of tangent at x = 2.
Solution:
Question 2.
For curve y = \(\frac { x-1 }{ x-2 } \), x ≠ 2 find slope of tangent at x = 10.
Solution:
Question 3.
Find all points where slope of tangent to the curve
Solution:
Question 4.
Find all equations of lines which are tangent to the curve y + \(\frac { 2 }{ x-3 } \) = 0 and slope of those is 2.
Solution:
Putting x = 4 in eq. (i)
y = \(\frac { -2 }{ 4-3 } \) = -2
Then, point = (4, – 2)
Again putting, x = 2 in eq. (i)
y = \(\frac { -2 }{ 2-3 } \) = 2
Then, point = (2, 2)
Now, tangent at ponit (4, – 2),
y – (- 2) = 2 (x – 4)
⇒ y + 2 = 2x – 8
⇒ 2x – y – 10 = 0
Again, tangent at point (2, 2),
y – 2 = 2 (x – 2)
⇒ y – 2 = 2x – 4
⇒ 2x – y – 2 = 0
Hence, required equations are 2x – y – 10 = 0 and 2x – y – 2 = 0.
Question 5.
For curve
find those points where tangent is
(i) Parallel to x-axis
(ii) Parallel to y-axis.
Solution:
(i) When tangent is parallel to x-axis then,
Hence, tangents are parallel to x-axis at points (0, ± 5).
(ii) When tangent is parallel to y-axis then,
Hence, tangents are parallel to 7-axis at points (± 2, 0)
Question 6.
For curve x – a sin3 t,y = b cos3 t find equation n of tangent at t = \(\frac { \pi }{ 2 } \)
Solution:
So, at t = \(\frac { \pi }{ 2 } \) or at (a, 0), equation of tangent of given curve
y – 0 = 0 (x – a)
or y = 0
Question 7.
For curves = sin2 x, find equation of normal at
Solution:
Question 8.
Find equation of tangent and normal following curves, at given points.
Solution:
(a) Given curve, y = x2 + 4x + 1 …..(i)
Putting x = 3 in equation (i),
y = (3)2 + 4 (3) + 1
y = 22
So, point of contact = (3, 22)
DifF. (i) w.r.t. x,
⇒ 10y – 220 = – x + 3
⇒ x + 10y = 223
So, equation of normal x + 10y = 223
(b) Given curve, y2 = 4ax …..(i)
Putting at x = a in equation (i)
y2 = 4a (a)
⇒ y2 = 4a2
⇒ y = ± 2a
Hence, points of contact are (a, 2a) and (a, – 2a).
DifF. (i) w.r.t. x,
Hence, at (a, 2a) equation of tangent is x – y + a = 0 and equation of normal is x + y – 3a = 0
⇒ m2 (my – 2d) = – (m2x – a)
⇒m3y – 2am2 = – m2x + a
⇒ m2x + m3 y = a (2m2 + 1)
Hence, equation of normal
m2x + m3y = a (2m2 + 1)
⇒ ay tan θ – ab tan2 θ = bx sec θ – ab sec2 θ
⇒ bx sec θ – ay tan θ = ab(sec2 θ – tan2 θ)
Equation of tangent
y – 2at = 1 (x – at2)
At t = 1
y – 2a = x – a
⇒ x – y + a = 0
Hence, equation of tangent x – y + a = 0
Again, equation of normal
y – 2at = – 1 (x – a t2)
At t = 1
y – 2a = – (x – a)
⇒ x + y – 3a = 0
Hence, equation of normal x + y – 3a = 0
(g) Given,
x = θ + sin θ, y – 1 – cos θ
Diff. w.r.t. θ,
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