Practicing OP Malhotra Maths Class 12 Solutions Chapter 12 Maxima and Minima Ex 12(a) is the ultimate need for students who intend to score good marks in examinations.
S Chand Class 12 ICSE Maths Solutions Chapter 12 Maxima and Minima Ex 12(a)
Question 1.
Find the turning values of the following functions, distinguishing in each case whether the value is a maximum, minimum, or inflexional:
(i) 4x3 + 19x2 – 14x + 3
(ii) 2x3 + 3x2 – 12x + 7
(iii) 3x4 + 8x3+ 6x2
(iv) x3 – 2x2 – 4x – 1
Solution:
(i) Let y = 4x3 + 19x2 – 14x + 3
∴ \(\frac { dy }{ dx }\) = (12x2 + 38x – 14)
∴ \(\frac{d^2 y}{d x^2}\) = 24x + 38
for maxima/minima, \(\frac { dy }{ dx }\) = 0
⇒ 12x2 + 38x – 14 = 0
⇒ \(x=\frac{-19 \pm \sqrt{361+168}}{12}\)
= \(\frac{-19 \pm \sqrt{529}}{12}\)
⇒ x = \(\frac{-19 \pm 23}{12}\) = \(\frac { 4 }{ 12 }\), \(\frac { -42 }{ 12 }\)
i.e. x = \(\frac { 1 }{ 3 }\), \(\frac { -7 }{ 2 }\)
when x = \(\frac { 1 }{ 3 }\); \(\frac{d^2 y}{d x^2}\) = 24 × \(\frac { 1 }{ 3 }\) + 38 = 46 > 0
∴ x = \(\frac { 1 }{ 3 }\) be a point of minima.
∴ min value of y = 4\(\left(\frac{1}{3}\right)^3\) + 19\(\left(\frac{1}{3}\right)^2\) – \(\frac { 14 }{ 3 }\) + 3
= \(\frac { 4 }{ 27 }\) + \(\frac { 19 }{ 9 }\) – \(\frac { 14 }{ 3 }\) + 3
= \(\frac{4+57-126+81}{27}\) = \(\frac{16}{27}\)
when x = –\(\frac{7}{2}\); \(\frac{d^2 y}{d x^2}\) = 24\(\left(-\frac{7}{2}\right)\) + 38
= – 84 + 38 = -46 < 0
∴ x = –\(\frac { 7 }{ 2 }\) be a point of maxima
& maximum value of y
= 4\(\left(-\frac{7}{2}\right)^3\) + 19\(\left(-\frac{7}{2}\right)^2\) – 14\(\left(-\frac{7}{2}\right)\) + 3
= \(\frac{-343}{2}\) + \(\frac{931}{4}\) + 49 + 3
= \(\frac{-686+931+208}{4}\) = \(\frac{453}{4}\) = 113\(\frac{1}{4}\)
(ii) Let y = 2x3 + 3x2 – 12x + 7
∴ \(\frac{d y}{d x}\) = 6x2 + 6x – 12
& \(\frac{d^2 y}{d x^2}\) = 12x + 6
for manima/minima, \(\frac{d y}{d x}\) = 0
⇒ 6(x2 + x – 2) = 0
⇒ x = 1, – 2
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=1}\) = 12 + 6 = 18 > 0
Thus x = 1 be a point of minima
& min value = 2 + 3 – 12 + 7 = 0
& \(\left(\frac{d^2 y}{d x^2}\right)_{x=-2}\) = -24 + 6 = – 18 < 0
∴ x = -2 be a point of maxima
& Maximum value
= 2(-2)3 + 3(-2)2 – 12(-2) + 7
= -16 + 12 + 24 + 7
= 27
(iii) Let y = 3x4 + 8x3 + 6x2
∴ \(\frac{d y}{d x}\) = 12x3 + 24x2 + 12x
& \(\frac{d^2 y}{d x^2}\) = 36x2 + 48x + 12
for maxima/minma, \(\frac{d y}{d x}\) = 0
⇒ 12x(x2 + 2x + 1) = 0
⇒ x (x + 1)2 = 0
⇒ x = 0, -1
Now \(\left(\frac{d^2 y}{d x^2}\right)_{x=0}\) =12 > 0
∴ x = 0 be a point of minima.
∴ min value = 0 + 0 + 0 = 0
& \(\left(\frac{d^2 y}{d x^2}\right)_{x=-1}\) = 36 – 48 + 12 = 0
Now, \(\frac{d^3 y}{d x^3}\) = 72x + 48
∴ \(\left(\frac{d^3 y}{d x^3}\right)_{x=-1}\) = -72 + 48 = -24 ≠ 0
∴ x = -1 be a point of inflexion
(iv) Let y = x3 – 2x2 – 4x – 1
∴ \(\frac{d y}{d x}\) = 3x2 – 4x – 4
& \(\frac{d^2 y}{d x^2}\) = 6x – 4
For maxima/minima, \(\frac{d y}{d x}\) = 0
⇒ 3x2 – 4x – 4 = 0
⇒ (x – 2)(3x + 2) = 0
⇒ x = 2, –\(\frac{2}{3}\)
Now, \(\left(\frac{d^2 y}{d x^2}\right)_{x=2}\) = 6 × 2 – 4 = 8 > 0
∴ x = 2 be a point of minima & min value of f(x) = f(2)
= 23 – 2 × 22 – 4 × 2 – 1
= 8 – 8 – 8 – 1 = – 9
∴ at \(\left(\frac{d^2 y}{d x^2}\right)\)x = –\(\frac{2}{3}\) = 6\(\left(-\frac{2}{3}\right)\) – 4 = -8 < 0
∴ x = –\(\frac{2}{3}\) be a point of maxima
& maximum value = – \(\frac{8}{27}\) – \(\frac{8}{9}\) + \(\frac{8}{3}\) – 1
= \(\frac{-8-24+72-27}{27}\) = \(\frac{13}{27}\)
Question 2.
If V = 2x2 (6 – x), where x is (positive, determine the greatest value of V.
Solution:
Given V = 2x2(6x – x)
∴ \(\frac{dV}{dx}\) = 2(12x – 3x2)
& \(\frac{d^2 \mathrm{~V}}{d x^2}\) = 2(12 – 6x)
for maxima/minima, \(\frac{d V}{d x}\) = 0
⇒ 2(12x – 3x2) = 0
⇒ 2x(12 – 3x) = 0
⇒ x = 0, 4
∴ \(\left(\frac{d^2 \mathrm{~V}}{d x^2}\right)_{x=0}\) = 2(12 – 0) = 24 > 0
∴ x = 0 be a point of minima & min value = 0
& \(\left(\frac{d^2 \mathrm{~V}}{d x^2}\right)_{x=4}\) = 2(12 – 24) = -24 < 0
∴ x = 4 be a point of maxima
& greatest value of V = 2 × 42(6 – 4) = 64
Question 3.
Find the co-ordinates of the turning points on the curve y = x3 – 3x2 – 9x + 7, distinguishing between maximum and minimum points.
Solution:
Given y = x3 – 3x2 – 9x + 7
∴ \(\frac{dy}{dx}\) = 3x2 – 6x – 9x
& \(\frac{d^2 \mathrm{~V}}{d x^2}\) = 6x – 6
for maxima/minima, \(\frac{dV}{dx}\) = 0
⇒ 3(x2 – 2x – 3) = 0
⇒ (x + 1)(x – 3) = 0
⇒ x = -1, 3
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=-1}\) = – 6 – 6 = – 12 < 0 ∴ x = -1 be a point of mixima & maximum value of y = – 1 – 3 + 9 + 7 = 12 & \(\left(\frac{d^2 y}{d x^2}\right)_{x=3}\) = 18 – 6 = 12 > 0
∴ x = 3 is a point of minima
& minimum value = 27 – 27 – 27 + 7 = -20
thus the turning point are (-1, 12) & (3, – 20).
Question 4.
Find the minimum value of \(\left(x+\frac{4}{x^2}\right)\).
Solution:
Let y = x + \(\frac{4}{x^2}\)
∴ \(\frac{dy}{dx}\) = 1 – \(\frac{8}{x^3}\) & \(\frac{d^2 y}{d x}\) = \(\frac{24}{x^4}\)
for maxima/minima, put \(\frac{dy}{dx}\) = 0
⇒ 1 – \(\frac{8}{x^3}\) = 0 ⇒ x3 = 8
⇒ (x – 2) (x2 + 2x + 4) = 0
⇒ x = 2
while the other two values of x are non-real.
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=2}\) = \(\frac{24}{2^4}\) = \(\frac{3}{2}\) > 0
∴ x = 2 be a point of minima & minimum value of y = 2 + \(\frac{4}{2^2}\) = 3
Question 5.
If y = \(\frac { 1 }{ 4 }\)x4 – \(\frac { 2 }{ 3 }\)x3 + \(\frac { 1 }{ 2 }\)x2 + \(\frac { 11 }{ 2 }\), show that the ordinate at the point x = 1 is neither a maximum nor a minimum, though \(\frac{dy}{dx}\) = 0, when x = 1.
Solution:
Given y = \(\frac{x^4}{4}\) – \(\frac{2}{3}\)x3 + \(\frac{1}{2}\)x2 + \(\frac{11}{2}\)
\(\frac{dy}{dx}\) = x3 – 2x2 + x
& \(\frac{d^2 y}{d x^2}\) = 3x2 – 4x + 1
for maxima/minima, we put \(\frac{dy}{dx}\) = 0
⇒ x (x2 – 2x – 1) = 0
⇒ x (x – 1)2 = 0
⇒ x = 0, 1
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=0}\) = 1 > 0 ∴ x = 0 be a point of minima
∴ \(\left(\frac{d^2 y}{d x^2}\right)_{x=1}\) = 3 – 4 + 1 = 0
& \(\frac{d^3 y}{d x^3}\) = 6x – 4
∴ at x = 1, \(\frac{d^3 y}{d x^3}\) = 6 – 4 = 2 ≠ 0
Thus, x = 1, be a point of neither maxima nor minima.
Question 6.
Find the points at which the function f given by f(x) = (x – 2)4 (x + 1)3 has
(i) local maxima
(ii) local minima
(iii) point of inflexion
Solution:
Given, f(x) = (x – 2)4(x + 1)3 Diff. both sides w.r.t. x; we have
f ‘ (x) = (x – 2)4 3 (x + 1)2 + (x + 1)3 4(x – 2)3
= (x + 1)2 (x – 2)3 [3(x – 2)+ 4(x + 1)]
= (x + 1)2 (x – 2)3 (7x – 2)
For critical points, f ‘ (x) = 0
⇒ (x + 1)2 (x – 2)3 (7x – 2) = 0
⇒ x = -1, 2, \(\frac{2}{7}\)
Case – I : at x = -1
When x slightly < – 1 ⇒ x + 1 < 0
also x < -1 < 2 ⇒ x – 2 < 0 ∴ f ‘ (x) = (+ve) (- ve)(- ve) = + ve When x slightly > – 1
⇒ x + 1 > 0, x – 2 < 0
∴ f ‘ (x) = (+ ve) (- ve) (- ve) = + ve
So f ‘ (x) does not changes its sign as we move from slightly < – 1 to slightly > – 1 .
∴ x = – 1 be a point of neither maxima nor minima.
Hence x = -1 be a point of inflexion.
Case-II : at x = 2
When x slightly < 2
⇒ x – 2 < 0 but 7x – 2 > 0
∴ f ‘ (x) = (+ Ve) (- Ve) (+ Ve) = – Ve
When x slightly > 2
⇒ x – 2 > 0 and 7x – 2 > 0
∴ f ‘ (x) = (+ Ve) (+ Ve) (+ Ve) = + Ve
Thus, f ‘ (x) changes its sign from -ve to +ve as we move from slightly <2 to slightly >2
∴ x = 2 is a point of minima.
Case-III : at x = \(\frac{2}{7}\)
When x slightly < \(\frac{2}{7}\)
⇒ 7x – 2 < 0 and x – 2 < 0 ∴ f ‘ (x) = (+ ve) (- ve) (- ve) = + ve ⇒ 7 x – 2 > 0 and x – 2 < 0
∴ f ‘ (x) = (+ve) (-ve) (+ve) = -ve
Thus, f ‘ (x) changes its sign from + ve to – ve as we move from slightly < \(\frac{2}{7}\) to slightly > \(\frac{2}{7}\).
∴ x = \(\frac{2}{7}\) be a point of local maxima.
Question 7.
Discuss the maxima and minima of the expression \(\frac{6 x^3-45 x^2+108 x+2}{2 x^3-15 x^2+36 x+1}\).
Solution:
Question 8.
Find the turning values of the function – x3 + 12x2 – 5, distinguishing whether the value is a maximum, minimum or inflexional.
Solution:
Let y = -x3 + 12 x2 – 5
∴ \(\frac{dy}{dx}\) = -3x2 + 24x
& \(\frac{d^2 y}{d x^2}\) = -6x + 24
for maxima minima, we put \(\frac{dy}{dx}\) = 0
⇒ -3x2+ 24x = 0 ⇒ -3x(x – 8) = 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 of y = 0 + 0 – 5 = – 5
& \(\left(\frac{d^2 y}{d x^2}\right)_{x=8}\) = -6 × 8 + 24 = -24 < 0
∴ x = 8 be a point of maxima.
Thus maximum value of y
= -512 + 768 – 5 = 251