NCERT Solutions For Class 12 Maths Chapter 6 Application Of Derivatives

Q1 (Medium)

Prove that the following functions do not have maxima or minima:
(i) f(x) = e^x    (ii) g(x) = log x    (iii) h(x) = x³ + x² + x + 1

Solution:
(i) f'(x)=e^x>0 for all x ⇒ strictly increasing ⇒ no maxima/minima.
(ii) g'(x)=1/x > 0 for x>0 ⇒ strictly increasing on its domain ⇒ no maxima/minima.
(iii) h'(x)=3x²+2x+1; discriminant = -8 < 0 and leading coeff >0 ⇒ h'(x)>0 ∀x ⇒ strictly increasing ⇒ no maxima/minima.

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Q2 (Easy)

If the radius of a sphere is measured as 7 m with an error of 0.02 m, find the approximate error in calculating its volume.

Solution:
V=(4/3)πr³ ⇒ dV=4πr²dr
r=7, dr=0.02 ⇒ dV=4π·49·0.02=3.92π ≈ 12.32 m³

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Q3 (Easy)

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 ⇒ dy/dx=3x²
3x²=x³ ⇒ x²(3-x)=0 ⇒ x=0,3
Points: (0,0), (3,27)

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Q4 (Easy)

Show that the tangents to the curve y = 7x³ + 11 at x = 2 and x = −2 are parallel.

Solution:
dy/dx=21x²
At x=2: slope=84; at x=-2: slope=84 ⇒ slopes equal ⇒ tangents parallel.

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Q5 (Easy)

Find the equation of the tangent line to y = x² − 2x + 7 which is:
(a) parallel to 2x – y + 9 = 0
(b) perpendicular to 5y − 15x = 13

Solution:
y=x²-2x+7 ⇒ y'=2x-2.

(a) slope=2 ⇒ 2a-2=2 ⇒ a=2 ⇒ point (2,7) ⇒ tangent: y=2x+3

(b) 5y−15x=13 ⇒ slope=3 ⇒ perpendicular slope = -1/3
2a-2=-1/3 ⇒ a=5/6, y=217/36 ⇒ tangent: y = (-1/3)x + 37/6

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Q6 (Medium)

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) same curve at (1,3)
(iii) y = x³ at (1,1)
(iv) y = x² at (0,0)
(v) x = cos t, y = sin t at t = π/4

Solution (answers):
(i) y' = 4x³-18x²+26x-10 ⇒ at x=0, m=-10
Tangent: y=-10x+5, Normal: y-5=(1/10)x

(ii) at x=1, m=2 ⇒ Tangent: y=2x+1, Normal: y-3=-(1/2)(x-1)

(iii) y'=3x² ⇒ m=3 ⇒ Tangent: y=3x-2, Normal: y-1=-(1/3)(x-1)

(iv) y'=2x ⇒ m=0 ⇒ Tangent: y=0, Normal: x=0

(v) dy/dx = -cot t ⇒ at t=π/4, m=-1; point (√2/2, √2/2)
Tangent: x+y=√2, Normal: y=x

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Q7 (Easy)

Find the equation of all lines having slope 2 which are tangents to the curve y = 1/(x − 3), x ≠ 3.

Solution:
y' = -1/(x-3)² (always negative) ⇒ slope 2 possible nahi.
No such tangent exists.

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Q8 (Easy)

Find the equation of all lines having slope −1 that are tangents to the curve y = 1/(x − 1), x ≠ 1.

Solution:
y' = -1/(x-1)². Set = -1 ⇒ (x-1)²=1 ⇒ x=0,2
At x=0, y=-1 ⇒ y=-x-1
At x=2, y=1 ⇒ y=-x+3

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Q9 (Medium)

Find the point on the curve y = x³ − 11x + 5 at which the tangent is y = x − 11.

Solution:
y' = 3x²−11; tangent slope=1 ⇒ 3x²−11=1 ⇒ x=±2
Check line: x=2 gives y=-9 and line gives -9 ✓ ; x=-2 fails ✗
Point: (2, −9)

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Q10 (Easy)

Find the slope of the normal to the curve x = 1 − a sin θ, y = b cos² θ at θ = π/2.

Solution:
dx/dθ = -a cosθ, dy/dθ = -2b cosθ sinθ ⇒ dy/dx = (2b/a) sinθ
At θ=π/2 ⇒ slope tangent = 2b/a ⇒ slope normal = -a/(2b)

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Q11 (Easy)

Find the slope of the normal to the curve x = a cos³ θ, y = a sin³ θ at θ = π/4.

Solution:
dy/dx = -tanθ ⇒ at θ=π/4, slope tangent = -1 ⇒ slope normal = 1

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Q12 (Easy)

Find the slope of the tangent to y = x³ − 3x + 2 at x = 3.

Solution:
y' = 3x² − 3 ⇒ at x=3 ⇒ 27−3= 24

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Q13 (Easy)

Find the slope of the tangent to y = x³ − x + 1 at x = 2.

Solution:
y' = 3x² − 1 ⇒ at x=2 ⇒ 12−1= 11

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Q14 (Medium)

Find the least value of a such that f(x)=x²+ax+1 is strictly increasing on (1,2).

Solution:
f'(x)=2x+a. On (1,2) minimum near x=1 ⇒ require 2+a ≥ 0 for least a.
a = -2

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Q15 (Medium)

Show that y = log(1+x) − (2x)/(2+x), x > −1, is increasing throughout its domain.

Solution:
y' = 1/(1+x) − 4/(2+x)²
Combine: y' = x²/[(1+x)(2+x)^{2</sup] ≥ 0 for x>-1 ⇒ increasing.}

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Q16 (Medium)

Find intervals where f(x)=2x³−3x²−36x+7 is strictly increasing/decreasing.

Solution:
f'(x)=6(x-3)(x+2)
Increasing: (-∞,-2) ∪ (3,∞)
Decreasing: (-2,3)

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Q17 (Medium)

Find intervals where f(x)=2x²−3x is strictly increasing/decreasing.

Solution:
f'(x)=4x-3
Increasing: (3/4,∞)
Decreasing: (-∞,3/4)

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Q18 (Easy)

Show that f(x)=e^2x is strictly increasing on R.

Solution:
f'(x)=2e^2x > 0 for all x ⇒ strictly increasing on R.

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