OP Malhotra Class 12 Maths Solutions Chapter 12 Maxima and Minima Ex 12(b)

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S Chand Class 12 ICSE Maths Solutions Chapter 12 Maxima and Minima Ex 12(b)

Question 1.
x³ – 9x² + 24x – 12
Solution:
Let y = x³ – 9x² + 24x – 12
∴ \(\frac { dy }{ dx }\) = 3x² – 18x + 24
& \(\frac{d^2 y}{d x^2}\) = 6x – 18
for maxima/minima, \(\frac { dy }{ dx }\) = 0
⇒ 3(x² – 6x – 8) = 0
⇒ (x – 2) (x – 4) = 0
⇒ x = 2, 4
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=2}\) = 12 – 18 = – 6 < 0
∴ x = 2 be a point of maxima & maximum value of y = 8 – 36 + 48 – 12 = 8
& \(\left(\frac{d^2 y}{d x^2}\right)_{x=4}\) = 24 – 18 = 6 > 0
∴ x = 4 be a point of minima & minimum value = 64 – 144 + 96 – 12 = 4

Question 2.
x³ + 12x² – 5
Solution:
Let y = -x³ + 12x² – 5
∴ \(\frac { dy }{ dx }\) = -3x² + 24x; \(\frac{d^2 y}{d x^2}\) = -6x + 24
for maxima/minima, \(\frac { dy }{ dx }\) = 0
⇒ -3x² + 24x = 0 ⇒ x = 0, 8
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=0}\) = 24 > 0
∴ x = 0 be a point of minima & minimum value = 0 + 0 – 5 = -5
& \(\left(\frac{d^2 y}{d x^2}\right)_{x=8}\) = – 48 + 24 = -24 < 0
∴ x = 8 be a point of maxima & minimum value = -512 + 768 – 5 = 251

Question 3.
2x³ – 6x² + 17.
Solution:
Let y = 2x³ – 6x² + 17
∴ \(\frac { dy }{ dx }\) = 6x² – 12x
& \(\frac{d^2 y}{d x^2}\) = 12x – 12
for maxima/minima, \(\frac { dy }{ dx }\) = 0
⇒ x = 0 be a point of maxima & maximum value = 0 – 0 + 17 = 17
& \(\left(\frac{d^2 y}{d x^2}\right)_{x=2}\) = 24 – 12 = 12 > 0
∴ x = 2 be a point of minima & maximum of y = 16 – 24 + 17 = 9

Question 4.
2x³ – 3x² – 12x + 4
Solution:
Let y = 2x³ – 3x² – 12x + 4
∴ \(\frac { dy }{ dx }\) = 6x² – 6x – 12
& ∴ \(\frac{d^2 y}{d x^2}\) = 12x – 6
for maxima/minima, \(\frac { dy }{ dx }\) = 0
⇒ 6 (x² – x – 2) = 0
⇒ (x – 2) (x + 1) = 0
⇒ x = 2, – 1
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=2}\) = 24 – 6 = 18 > 0
∴ x = 2 is a point of minima
∴ minimum value = 16 – 12 – 24 + 4 = – 16
& \(\left(\frac{d^2 y}{d x^2}\right)_{x=-1}\) = – 12 – 6 = – 18 < 0
thus x = – 1 is a point of maxima & maximum value of y = – 2 – 3 + 12 + 4 = 11

Question 5.
(x – 1) (x + 2)².
Solution:
Given f(x) = (x – 1) (x + 2)²
∴ f ‘ (x) = (x – 1) 2(x + 2) + (x + 2)² = (x + 2) [2x – 2 + x + 2] = 3x (x + 2)
For local maxima/minima, f ‘ (x) = 0
⇒ x = 0, -2
∴ f ” (x) = 3[2x + 2]
at x = 0 ; f ”(0) = 6 > 0
∴ x = 0 is a point of local minima and local minimum value = -4
at x = -2 ; f ”(-2) = -6 < 0
∴ x = -2 is a point of local maxima and maximum value = 0.

Question 6.
(x – 1)³(x+1)².
Solution:
Given f(x) = (x – 1)³ (x + 1)²
∴ f ‘ (x) = -3 (x – 1)² (x + 1)² – 2(x – 1)² (x + 1)
= – (x – 1)² (x + 1)[3(x + 1) + 2(x – 1)]
= – (x – 1)² (x + 1)(5x + 1)
The critical points of f (x) are given by f ‘ (x) = 0
⇒ – (x – 1)² (x + 1)(5x + 1) = 0
⇒ x = 1, -1, \(\frac{-1}{5}\)
Now f ‘ (x) = – (x² – 1) (5x² – 4x – 1)
∴ f ” (x) = -[(x² – 1) (10x – 4) + (5x² – 4x – 1) 2x]
When x = -1 :
f ” (-1) = -[0 + (5 + 4 – 1)(- 2)] = 16 > 0
∴ x = -1 is a point of local minima and local minimum value of f(x) = f(- 1) = -(- 8) × 0 = 0

When x = 1 :
f ” (1) = – (o + 5 – 4 – 1) 2 = 0
∴ f’ ”’ (x) = -(60x² – 8x – 12)
∴ f ”’ (1) = – (60 – 8 – 12) = – 40 ≠ 0
Thus x = 1 is a point of neither maxima nor minima.

When x = -1/5 :
f ” (-1/5) = –\(\left[\frac{-24}{25}(-6)-\frac{2}{5}\left(\frac{1}{5}+\frac{4}{5}-1\right)\right]\)
∴ f ” \(\left(\frac{-1}{5}\right)\) = – \(\frac{144}{25}\) < 0
Thus x = – \(\frac{1}{5}\) is a point of local maxima and local max. value of f(x) = f\(\left(-\frac{1}{5}\right)\)
= – \(\left(-\frac{6}{5}\right)^3\) \(\left(\frac{4}{5}\right)^2\) = +\(\frac{3456}{3125}\)

Question 7.
sin 2x – x – \(\frac{\pi}{2}\) ≤ x ≤ \(\frac{\pi}{2}\).
Solution:
Given f(x) = sin 2x – x; – \(\frac{\pi}{2}\) ≤ x ≤ \(\frac{\pi}{2}\)
∴ f ‘ (x) = 2 cos 2x – 1
For critical points f ‘ (x) = 0
⇒ 2 cos 2x – 1 = 0
⇒ cos 2x = \(\frac{1}{2}\) = cos\(\frac{\pi}{3}\)
⇒ 2x = \(\frac{\pi}{3}\), – \(\frac{\pi}{3}\) ⇒ x = \(\frac{\pi}{6}\), – \(\frac{\pi}{6}\)
[∵ cos θ = cos α ⇒ θ = 2nπ ± α ∀ n ∈ |]
[∵ – π/2 ≤ x ≤ π/2 ⇒ – π ≤ 2x ≤ π]
when x slightly < \(\frac{\pi}{6}\) ⇒ 2x < \(\frac{\pi}{3}\) ⇒ cos 2x > \(\frac{1}{2}\)
⇒ 2 cos 2x > 1
⇒ 2 cos 2x – 1 > 0
[∵ f (x) is a decreasing function in x ∈ (0, π/2)]
⇒ f ‘ (x) < 0 when x slightly > π/6 ⇒ 2x > π/3
⇒ cos 2x < 1/2
[∵ cos x is a decreasing function in first quad.]
⇒ 2 cos 2x – 1 < 0 ⇒ f ‘ (x) < 0
∴ f ‘ (x) changes its sign from +ve to -ve as x increase through π/6.
∴ x = π/3 is a point of local maxima and local maximum value = f\(\left(\frac{\pi}{6}\right)\) = \(\frac{\sqrt{3}}{2}\) – \(\frac{\pi}{6}\)
When x slightly < – π/6 ⇒ 2x < – π/3
Thus 2x lies in 4th quadrant and cosine function is increasing in 4th quadrant.
∴ cos 2x < cos (- π/3) ⇒ 2 cos 2x – 1 < 0
⇒ f ‘ (x) < 0 When x slightly > – π/6
⇒ 2x > – \(\frac{\pi}{3}\)
⇒ cos 2x > + \(\frac{1}{2}\)
⇒ 2 cos 2x – 1 > 0 ⇒ f ‘ (x) > 0
Hence, f ‘(x) changes its sign from negative to positive as x increases through -π/6. Thus x = -π/6 is a point of minima
and local min. value = f\(\left(-\frac{\pi}{6}\right)\)
= sin \(\left(-\frac{\pi}{3}\right)\) + \(\frac{\pi}{6}\) = –\(\frac{\sqrt{3}}{2}\) + \(\frac{\pi}{6}\)

Question 8.
sin x +cos x, 0 < x < \(\frac{\pi}{2}\)
Solution:
Given f(x) = sin x + cos x, 0 < x < \(\frac{\pi}{2}\)
∴ f ‘(x) = cos x – sin x
For local maxima or minima, f ‘(x) = 0
⇒ cos x – sin x = 0 ⇒ tan x = 1 = tan\(\frac{\pi}{4}\)
∴ x = \(\frac{\pi}{4}\) ∈ \(\left(0, \frac{\pi}{2}\right)\)
[ ∵ tan x = tan α ⇒ x = nπ + α, n ∈ 1]
∴ f ” (x) = – sin x – cos x
Now at x = \(\frac{\pi}{4}\)
f ” \(\frac{\pi}{4}\) = – \(\frac{1}{\sqrt{2}}\) – \(\frac{1}{\sqrt{2}}\) = √2 < 0
∴ x = \(\frac{\pi}{4}\)is a pt. of maxima
and Max. value = \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) = √2

Question 9.
The function y = a logx+bx²+x has extreme values at x = 1 and x = 2. Find a nad b.
Solution:
Given y = a log x + bx² + x ….(1)
∴ \(\frac{d y}{d x}\) = \(\frac{a}{x}\) + 2bx + 1
since given function (1) has extreme value at x = 1 and x = 2
∴ \(\left(\frac{d y}{d x}\right)_{x=1}\) = 0 = \(\left(\frac{d y}{d x}\right)_{x=2}\)
∴ \(\left(\frac{d y}{d x}\right)_{x=1}\) = 0
⇒ a + 2b + 1 = 0 ⇒ a + 2b = – 1 …(2)
and \(\left(\frac{d y}{d x}\right)_{x=1}\) = 0
⇒ \(\frac{a}{2}\) + 4b + 1 = 0
⇒ a + 8b = – 2 …(3)
eqn. (2) – eqn. (3) gives ;
– 6b = 1 ⇒ b = – 1/6
∴ from (2); a – 1/3 = -1 ⇒ a = -2/3

Question 10.
Show that y = sin³ x cos x has a maximum value at x = \(\frac{\pi}{3}\) and find its value.
Solution:
Let y = sin³ x cos x
∴ \(\frac{d y}{d x}\) = sin³ x(-sin x) + cos x(3 sin² x) cos x = sin²x (3 cos²x – sin²x)
& \(\frac{d^2 y}{d x^2}\) = sin² x(-6 cos x sin x – 2sin x cos x) + (3 cos² x – sin² x) 2 sin x cos x
= sin²x (-4 sin 2x) + (3 cos²x – sin²x) sin 2x
for maxima/minima, \(\frac{d y}{d x}\) = 0
⇒ sin²x ( 3 cos²x – sin²x) = 0
⇒ sin x = 0 or tan²x = (√3)² = tan² π/3
⇒ x = nπ or x = nπ ± π/3
⇒ x = 0, π, 2π, \(\frac{\pi}{3}\), \(\frac{2 \pi}{3}\), \(\frac{4 \pi}{3}\), \(\frac{5 \pi}{3}\) ∈ [0, 2π]

thus x = π/3 be a point of maxima & maximum value of y = sin³\(\frac{\pi}{3}\) cos\(\frac{\pi}{3}\)
= \(\left(\frac{\sqrt{3}}{2}\right)^3\) x \(\frac{1}{2}\) = \(\frac{3 \sqrt{3}}{16}\)

Question 11.
A quadratic function in x has the value 19 , when x = 1 and has a maximum value 20 when x = 2. Find the function.
Solution:
Let the required quadratic function be
f(x) = ax² + bx + c ; a ≠ 0 (*)
∴ f(1) = 19 ⇒ a + b + c = 19 …(i)
Now f ‘(x) = 2ax + b
since function f(x) has maximum value = 0, when x = 2
i.e. f ‘(2) = 0 ⇒ 4 a + b = 0 …(ii)
& f(2) = 20 ⇒ 4a + 2b + c = 20 …(iii)
from (ii) & (iii); we have
⇒ 4a + 2(-4a) + c = 20 …(iv)
⇒ -4 a + c = 20
equation (ii) -equation (i) gives;
3a – c = – 19 …(iv)
on adding (iv) & (v); we have
– a = 1, ⇒ a = -1
∴ from (iv); 4 + c = 20 ⇒ c = 16
∴ from (ii); b = -4a = 4
thus putting the values of a, b & c in equations(*); we have
y = f(x) = -x² + 4x + 16

Question 12.
Find the absolute maximum and absolute minimum value of the following functions in the given interval :
(i) f(x) = x³ in [-2, 2]
(ii) f(x) = (x – 1)² + 3 in [-3, 1]
(iii) y = \(\sqrt{x-4}\) on 4 ≤ x ≤ 29.
(iv) 3 x⁴ – 8x³ + 12x² – 48x + 1 on the interval [1, 4].
(v) f(x) = 2x³ – 24x + 107 in [-3, -1].
(vi) f(x) = x⁵⁰ – x²⁰ in [0, 1]
(vii) x + sin 2x in [0, 2 π].
(viii) f(x) = cos² x + sin x = x ∈[0, π]
Solution:
(i) Given f(x) = x³ in [-2, 2]
∴ f ‘(x) = 3x² i.e. f ‘ (x) = 0
⇒ 3 x² = 0 ⇒ x = 0 ∈[-2, 2]
Now y]_{x = 0} = 0; y]_{x = 2} = 8;
y]_{x = -2} = – 8
∴ absolute max. value = 8;
absolute min value = -8.

(ii) Given f(x) = (x – 1)² + 3 in [-3, 1]
∴ f ‘ (x) = 2(x – 1)
∴ f ‘ (x) = 0 ⇒ 2x – 2 = 0 ⇒ x = 1 ∈[-3, 1]
Now y_{x = 1} = 3; y]_{x = -3} = 16 + 3 = 19
∴ Absolute maximum value = 19 and Absolute minimum value = 3.

(iii) Given y = \(\sqrt{x-4}\); 4 ≤ x ≤ 29
∴ \(\frac{d y}{d x}\) = \(\frac{1}{2 \sqrt{x-4}}\)
since \(\frac{d y}{d x}\) ≠ 0 ∀x∈R
so we find the value of function at x = 4, 29
∴ f(4) = (y)_{x = 4} = \(\sqrt{4-4}\) = 0
f(29) = \(\sqrt{29-4}\) = \(\sqrt{25}\) = 5
∴ absolute maximum value of on [4, 29] = 5 & absolute minimum value of function f on [4, 29] = 0

(iv) Given f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25
∴ f ‘(x) = 12x³ – 24x² + 24x – 48
=12 (x³ – 2x² + 2x – 4)
For local maxima/minima, we have
f ‘ (x) = 0
⇒ 12 [x³ – 2x² + 2x – 4] = 0
⇒ x² (x – 2) + 2 (x – 2) = 0
⇒ (x – 2) (x² + 2) = 0
⇒ x – 2 = 0 or x² + 2 = 0 does not gives real values of x
∴ x = 2
Now let us compute the values of f(x) at stationary points and also at the end points of given interval [0, 3].
f(2) = 3 × 16 – 8 × 8 + 12 × 4 – 96 + 25
= 48 – 64 + 48 – 96 + 25 = – 39
f(0) = 0 – 0 + 0 – 0 + 25 = 25
f(3) = 3 × 81 – 8 × 27 + 12 × 9 – 48 × 3 + 25
= 243 – 216 + 108 – 144 + 25 = 16
Out of these values, maximum value be f(0) = 25
and least value be f(2) = – 39
Hence absolute maximum value = 25 at x = 0
and absolute minimum value = -39 at x = 2.

(v) Given f(x) = 2x³ – 24x + 107
∴ f ‘ (x) = 6x² – 24 = 6(x – 2)(x + 2)
For critical points we have f ‘ (x) = 0
⇒ 6 (x – 2)(x + 2) = 0
⇒ x = 2, -2
if f(x) be defined on [1, 3]
∴ f ‘ (x) = 0 at x =2
Now let us compute the values of f(x) at critical point x = 2 and also at the end point of given interval [1, 3].
∴ f(2) = 16 – 48 + 107 = 123 – 48 = 75
f(1) = 2 – 24 + 107 = 109 – 24 = 85
f(3) = 54 – 72 + 107 = 161 – 72 = 89
out of these values, maximum value of f(x) be f(3) = 89
∴ absolute max. value = 89 at x = 3.
if f(x) be defined in [-3, -1]
Then f ‘ (x) = 0 at x = -2.
Thus we compute the values of f(x) at critical point x = -2 and also at the end points of given interval [-3, -1]
∴ f(-2) = – 16 + 48 + 107 = 139
f(-1) = -2 + 24 + 107 = 129
f(-3) = – 54 + 72 + 107 = 125
Out of these values, maximum value of f(x) be f(-2) = 139.
Hence, absolute maximum value = 139 at x = -2

(vi) Given f(x) = x⁵⁰ – x²⁰ in [0, 1]
∴ f ‘(x) = 50x⁴⁹ – 20x¹⁹
f ” (x) = 50 × 49x⁴⁸ – 20 × 19x¹⁸
for maxima/minima, we put f ‘ (x) = 0
⇒ x¹⁹ (50x²⁰ – 20) = 0
⇒ x = 0, \(\left(\frac{2}{5}\right)^{1 / 30}\)
∴ at x = 0; f ‘ (x) = 0……f⁽¹⁹⁾(0) = 0
∴ f²⁰(x) = 50 × 49 × 48 × …… × x³⁰ – 20 × 19 × 18 ……. × 2 × 1x⁰
∴ f²⁰ = -20 ! < 0 ∴ x = 0 be a point of maxima & miximum value of y = 0

(vii) Let y = x + sin 2x on [0, 2π]
∴ \(\frac{d y}{d x}\) = 1 + 2 cos 2x, Now \(\frac{d y}{d x}\) = 0
⇒ 1 + 2 cos 2x = 0 ⇒ cos 2x = –\(\frac { 1 }{ 2 }\)
∴ cos 2x = – cos\(\left(\frac{\pi}{3}\right)\) = c0s \(\left(\pi-\frac{\pi}{3}\right)\)
⇒ cos x = cos\(\frac{2 \pi}{3}\)
⇒ 2x = 2n π±\(\frac{2 \pi}{3}\) [∵ cos θ = cos α ⇒ θ = 2nπ ± α]

∴ y is absolute maximum at x = 2π
and absolute Max. value = 2π
and y is absolute minimum at x = 0
and absolute minimum value = 0

(viii) Given f (x) = cos³⁰ x + sin x
∴ f ‘ (x) = – sin 2x + cos x
= – 2 sin x cos x + cos x
= cos x(1 – 2 sin x)
For critical points we must have f ‘ (x) = 0
⇒ cos x(1 – 2 sin x) = 0
∴ cos x = 0 or 1 – 2 sin x = 0
⇒ x = π/ 2 or sin x = 1/2
⇒ x = π/6, or x = π/6, π – π/6
⇒ x = π/2, π/6, 5 π/6 ∈[0, π]
Now let us compute the values of f(x) at all stationary (critical points) and also at the end points of given internal [0, π].

Out of these values, maximum value of f(x) be f\(\left(\frac{\pi}{6}\right)\) = \(\frac { 5 }{ 4 }\) = f\(\left(\frac{5 \pi}{6}\right)\)
and minimum value of f(x) be 1.
Hence absolute max. value = \(\frac { 5 }{ 4 }\) at x = \(\frac{\pi}{6}\), \(\frac{5 \pi}{6}\)
and absolute min. value = 1 at x = 0, \(\frac{\pi}{2}\), π

Question 13.
Show that sin x(1 + cos x), x ∈[0, π] is maximum at x = π/3.
Solution:
Let y = sin x(1 + cos x); x ∈[0, π]
∴ \(\frac{d y}{d x}\) = -sin²x + cos x + cos² x
= cos x + cos 2x
for maxima/minima, \(\frac{d y}{d x}\) = 0
⇒ cos 2x + cos x = 0
⇒ 2 cos² x + cos x – 1 = 0
⇒ (cos x + 1)(2 cos x – 1) = 0
⇒ cos x = -1 or cos x = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
⇒ x = 2 nπ ± π or x = 2nπ ± π/3; n ∈ I Also x ∈ [0, π]
Thus, x = π, π/3
We calculate the value of f(x) at x = 0, π/3, π.
When x = 0; f(0) = 0
When x = π/3;
f(π/3) = sin\(\frac{\pi}{3}\)(1 + cos\(\frac{\pi}{3}\))
= \(\frac{\sqrt{3}}{2}\left(1+\frac{1}{2}\right)\) = \(\frac{3 \sqrt{3}}{4}\)
When x = π; f(x) = sin π (1 + cos π) = 0 thus clearly maximum value of f(x) be \(\frac{3 \sqrt{3}}{4}\) at x = π/3.

Question 14.
Find the maximum and minimum values of f(x) = (sin x + \(\frac{1}{2}\) cos 2x) in 0≤x≤\(\frac{\pi}{2}\).
Solution:
Given,
f(x) = sin x + \(\frac{1}{2}\)cos 2x; 0 ≤ x ≤ \(\frac{\pi}{2}\)
∴ f ‘(x) = cos x – sin 2x
Now f ‘(x) = 0 ⇒ cos x – sin 2x = 0
⇒ cos x – 2 sin x cos x = 0
⇒ cos x (1 – 2 sin x) = 0
⇒ cos x = 0 or 1 – 2 sin x= 0
⇒ x = \(\frac{\pi}{2}\) or x = \(\frac{\pi}{6}\) [∵ 0 ≤ x ≤ \(\frac{\pi}{2}\)]
Since x = \(\frac{\pi}{2}\) be the end point so x = π/6 be the only critical point.
Now f\(\left(\frac{\pi}{6}\right)\) = sin\(\frac{\pi}{6}\) + \(\frac{1}{2}\)cos \(\frac{2 \pi}{6}\)
= \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) = \(\frac { 3 }{ 4 }\)
f(0) = sin 0 + \(\frac { 1 }{ 2 }\) cos 0 = \(\frac { 1 }{ 2 }\)
f\(\left(\frac{\pi}{2}\right)\) = sin \(\frac{\pi}{2}\) + \(\frac { 1 }{ 2 }\) cos π = 1 – \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\)
thus maximum value of f(x) be \(\frac { 3 }{ 4 }\) at x = π/6 and minimum value of f(x) be \(\frac { 1 }{ 2 }\) and attains at x = 0, \(\frac{\pi}{2}\).

Question 15.
Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 + 24x – 18x².
Solution:
Given P(x) = 41 + 24x – 18x²
∴ \(\frac{dp}{dx}\)(x) = 24 – 36x
& \(\frac{d^2 \mathrm{P}(x)}{d x^2}\) = – 36
For maxima/minima, \(\frac{d \mathrm{P}(x)}{d x}\) = 0
⇒ 24 – 36x = 0 ⇒ x = \(\frac { 2 }{ 3 }\)
at x = \(\frac { 2 }{ 3 }\); \(\frac{d^2 \mathrm{P}(x)}{d x^2}\) = – 36 < 0
∴ Profit is maximum at x = \(\frac { 2 }{ 3 }\)
Thus maximum profit = P\(\left(\frac{2}{3}\right)\)
= 41 + 24 × \(\frac { 2 }{ 3 }\) – 18 × \(\frac { 4 }{ 9 }\)
= 41 + 16 – 8 = 49

Question 16.
(i) Find the point on the parabola y² = 2x which is closest to the point (1, 4).
(ii) Find the point on the curve y² = 4x which is nearest to the point (2,-8).
(iii) Find the point on the curve x² = 8y which is nearest to the point (2, 4).
Solution:
(i) Let P(x, y) be any point on given curve y² = 2x …(1)
Let A(1, 4) be the given point
Then AP² = (x – 1)² + (y – 4)²
⇒ AP² = \(\left(\frac{y^2}{2}-1\right)^2\) + (y – 4)²
Let T = AP², Then T is minimum/maximum according as AP is minimum/maximum.
∴ T = \(\left(\frac{y^2}{2}-1\right)^2\) + (y – 4)²
∴ \(\frac { dT }{ dy }\) = \(2\left(\frac{y^2}{2}-1\right)\) y + 2(y – 4)
= y³ – 2y + 2y – 8 = y³ – 8
For critical points of T;
We must have \(\frac { dT }{ dy }\) = 0
⇒ y³ – 8 = 0 ⇒ y = 2
∴ \(\frac{d^2 \mathrm{~T}}{d y^2}\) = 3y²
⇒ \(\left(\frac{d^2 \mathrm{~T}}{d y^2}\right)_{y=2}\) = 12 > 0
Thus T is minimum for y = 2
∴ from (1); x = \(\frac{y^2}{2}\) = \(\frac { 4 }{ 2 }\) = 2
Thus the required point on given curve be P (2, 2).
Hence the point P(2, 2) on the given curve is at a minimum distance from given point.

(ii) Let P(x, y) be any point on the curve
y² = 4x …(1)
Let A(2, -8) be the given point.
∴ |AP| = \(\sqrt{(x-2)^2+(y+8)^2}\)
⇒ AP² = (y + 8)²
= \(\left(\frac{y^2}{4}-2\right)^2\) + (y + 8)² [using (1)]
Let S =AP², Then S is maximum or minimum according as AP is maximum or minimum.

For max/minima, \(\frac { dS }{ dy }\) = 0 ⇒ y³ = -64
⇒ y³ = -64 ⇒ y = -4
∴ \(\left(\frac{d^2 \mathrm{~S}}{d y^2}\right)_{y=-4}\) = \(\frac { 3 }{ 4 }\)(16) = 12 > 0
∴ S is minimise for y = -4
and from (1); x = 4
Hence the point (4, -4) on curve y² = 4x is nearest to the given point (2, -8).

(iii) Let P(x, y) be any point on the curve x² = 8y ….(1) and let the given point be A (2, 4).
∴ AP² = (x – 2)² + (y – 4)²
= (x – 2)² + \(\left(\frac{x^2}{8}-4\right)^2\) [using (1)]
Let S = AP², Now S is max ./ m i n$. according as AP is max./min. according as AP is max./min

∴ S is minimize for x = 4 ∴ from (1); y = 2
∴ Required point on given curve be (4, 2).

Question 17.
An enemy jet is flying along the curve y = x² + 2. A soldier is placed at the point (3, 2). Find the nearest distance between the soldier and the jet.
Solution:
Given eqn. of curve be y=x² + 2 ….(i) Let the jet is at point P(x, y) and soldier is placed at given point A(3, 2). So we want to minimise the distance AP so it is conuenient to minimise AP² i.e. z
Let z = AP² = (x – 3)² + (y – 2)² = (x – 3)² + (x²)² [Using eqn. (i)]
⇒ z = (x – 3)² + x⁴;
Differentiate both sides w.r.t. x; we have \(\frac { dz }{ dx }\) + 4x³
& \(\frac{d^2 z}{d x^2}\) = 2 + 12x²
For maxima/minima, \(\frac { dz }{ dx }\) = 0
⇒ 2(x – 3) + 4x³ = 0
⇒ 2x³ + x – 3 = 0
⇒ (x – 1) (2x² + 2x + 3) = 0
⇒ x = 1
Since 2x² + 2x + 3 – 0 does not gives real values of x.
∴ \(\left(\frac{d^2 z}{d x^2}\right)_{x=1}\) = 2 + 12 = 14 > 0
Thus z is minimise at x = 1
∴ from (i) ; y = 1² + 2 = 3
Hence z i.e. AP² is minimise at (1, 3)
i.e. distance AP is minimise at (1, 3)
Thus nearest point be (1, 3) & min distance
= \(\sqrt{(1-3)^2+(3-2)^2}\)
= \(\sqrt{4+1}\) = \(\sqrt{5}\)

Question 18.
Find the coordinates of a point on the parabola y = x² + 7x + 2 which is closest to the straight line y = 3x – 3.
Solution:
Let P(x, y) be any point on given curve y = x² + 7x + 2
and given eqn. of straight line be 3x – y – 3 = 0
Let the distance of point P(x, y) from given line (2) is

Hence S is minimise for x = -2.
∴ from (1); y = 4 – 14 + 2 = -8
Hence the required point on given curve be (-2, -8).
Thus the point (-2,-8) on given parabola is closet to given straight line y = 3x – 3.

Question 19.
A straight line is drawn through a given point P(1, 4). Determine the least value of the sum a the intercepts on the coordinate axes.
Solution:
The equation of line through P(1, 4) is given by
y – 4 = m (x – 1), where m < 0
⇒ mx – y = m – 4

Hence the intercepts on the axes are \(\frac{m-4}{m}\) and – (m – 4).
Let S = sum of the intercepts
= \(\frac{m-4}{m}\) – (m – 4)
⇒ S = 4 – \(\frac{4}{m}\) – m
⇒ \(\frac { dS }{ dm }\) = \(\frac{4}{m^2}\) – 1
For maxima/minima, \(\frac { dS }{ dm }\) = 0

Thus S is minimum for m = -2
Hence least value of S = (S)_m=-2
= 5 – \(\frac{4}{-2}\) + 2 = 5 + 2 + 2 = 9

Question 20.
Find the maximum slope of the curve y = -x³ + 3 x² + 2x – 27.
Solution:
Given eqn. of curve be
y = -x³ + 3x² + 2x – 27 ….(1)
thus slope of curve m = -3x² + 6x + 2
∴ \(\frac{dm}{dx}\) = -6x + 6
To maximise/minimise m, we have \(\frac{dm}{dx}\) = 0
⇒ -6x + 6 = 0 ⇒ x = 1.
∴ \(\frac{d^2 m}{d x^2}\) = -6
⇒ \(\left(\frac{d^2 m}{d x^2}\right)_{x=1}\) = -6 < 0
Hence m is maximum for x = 1
∴ from (1); y = – 1 + 3 + 2 – 27 = – 23
Thus, the slope is maximum at the point (1, -23)
and maximum value of slope = (m)_{(1, -23)} = – 3 + 6 + 2 = 5