OP Malhotra Class 12 Maths Solutions Chapter 12 Maxima and Minima Revision Exercise

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S Chand Class 12 ICSE Maths Solutions Chapter 12 Maxima and Minima Revision Exercise

Question 1.
The sum of three positive numbers is 26. The second number is 3 times as large as the first. If the sum of the squares of these numbers is least, find the numbers.
Solution:
Let the required three numbers are x, y and z. according to given condition, we have
x + y + z = 26 …(i)
y =3x …(ii)
Let S = x² + y² + z²
= x² + (3x²) + (26 – x – 3x)² [using (i) & (ii)]
= 10x² + (26 – 4x)²
= 10x² + 4(13 – 2x)²
∴ \(\frac { dS }{ dx }\) = 20x + 8 (13 – 2x) (-2)
& \(\frac{d^2 S}{d x^2}\) = 20 – 16 (-2)
= 20 + 32 = 52
For maxima/minima, \(\frac { dS }{ dx }\) = 0
⇒ 20x – 16(13 – 2x) = 0
⇒ 4(5x – 52 + 8x) = 0
⇒ 13x = 52
⇒ x = 4
at
x = 4; \(\frac{d^2 S}{d x^2}\) = 52 > 0
Thus S is minimise when x = 4
∴ from (ii); y = 3 × 4 = 12
& from (i); z = 26 – 4 – 12 = 10
Hence the required numbers are 4, 12 & 10.

Question 2.
ABC is a right angled triangle of given area 5. Find the sides of the triangle for which the area of the circumscribed circle is least.
Solution:
Let ABC be the right angled traingle that inscribed in circle with diameter 2a.

i.e. O be the centre of circle with radius a.
∴ x² + y² = 4a²
where ∠ABC = 90°
Then S = area of ∠ABC = \(\frac{1}{2}\)xy …(ii)
Let Δ = area of circumscribed circle

Thus Δ is minimise when x = \(\sqrt{2 S}\)
∴ from (ii); y = \(\sqrt{4 a^2-x^2}\)
or y = \(\frac{2 S}{x}\) = \(\frac{2 \mathrm{~S}}{\sqrt{2 \mathrm{~S}}}\) = \(\sqrt{2 S}\)
and hypotenuse = 2a = \(\sqrt{x^2+y^2}\) = \(\sqrt{2 \mathrm{~S}+2 \mathrm{~S}}\) = \(2 \sqrt{S}\)
Hence the required sides of the triangle are \(\sqrt{2 \mathrm{~S}}\), \(\sqrt{2 \mathrm{~S}}\) and \(2 \sqrt{S}\) respectively.

Question 3.
An open box with a square base is to be made out of a given quantity of cardboard whose area is c² units. Show that the maximum volume of the box is units.
Solution:
Let x be the side of square box x and h be the height of open box
Then V = volume of box = x² h and area of box = c² = x² + 4xh

Question 4.
Three numbers are given whose sum is 180 and the ratio of the first two of them is 1 : 2. If the i of foe numbers is greatest, fend toe monbers.
Solution:
Let the first two numbers be x and 2x, since the ratio of first two of them be 1 : 2.
Also the sum of first three numbers be 180 .
∴ required numbers are x, 2x, 180 – 3x.
Let P = x × 2x × (180 – 3x)
= 2x²(180 – 3x)
= 6 (60x² – x²)
Diff. both sides w.r.t. x, we have
\(\frac{d \mathrm{P}}{d x}\) = 6 [120x – 3x²]; \(\frac{d^2 \mathrm{P}}{d x^2}\) = 6 (120 – 6x)
For maxima/minima, \(\frac { dp }{ dx }\) = 0
⇒ 120x – 3x² = 0
⇒ -3x [x – 40] = 0 ⇒ x = 0, 40
But x > 0 ∴ x = 40
∴ \(\left(\frac{d^2 \mathrm{P}}{d x^2}\right)_{x=40}\) = 60 (120 – 240) = -720 < 0
∴ x = 40 be a point of maxima.
Thus product P is maximum for x = 40.
Hence the required numbers are, 40, 2 × 40 and 180 – 120
i.e. 40,80 and 60.

Question 5.
A closed circular cylinder has a volume of 2156 cm³. What will be the radius of its base so that its total surface area is minimum ? Find the height of the cylinder when its total surface area is minimum.
Solution:
Let r be the radius of base and h be the height of closed circular cylinder respectively.
Then volume of cylinder = 2156
⇒ πr²h = 2156
Let S = total surface area of cylinder = 2πr² + 2πrh
⇒ S = 2πr² + 2πr × \(\frac{2156}{\pi r^2}\) [using eqn. (i)]

Thus S is minimise when r = 7 cm
& from (i); h = \(\frac{2156}{\pi r^2}=\frac{2156}{\frac{22}{7} \times 7^2}\) = 14 cm

Question 6.
Prove that the area of a right angled triangle of a given hypotenuse is maximum when the triangle and is isosceles.
Solution:
Let AC be the hypotenuse of right angled triangle △ABC s.t. AC = a (given)
& ∠ACB = θ
∴ \(\frac{AB}{AC}\) = sin θ ⇒ AB = α sin θ

& \(\frac{BC}{AC}\) = cos θ ⇒ BC = α cos θ
Let Δ = area of △ABC = \(\frac{1}{2}\) × BC × AB
⇒ Δ = \(\frac{1}{2}\) × (α cos θ) (α sin θ)
= \(\frac{a^2}{4}\)sin2θ
∴ \(\frac{d \Delta}{d \theta}\) = \(\frac{a^2}{4}(2 \cos 2 \theta)\) = \(\frac{a^2}{2} \cos 2 \theta\)
& \(\frac{d^2 \Delta}{d \theta^2}\) = -a² sin 2θ
For maxima/minima, \(\frac{d \Delta}{d \theta}\) = 0
⇒ \(\frac{a^2}{2}\)cos 2θ = 0 ⇒ cos 2θ = 0
⇒ 2θ = \(\frac{\pi}{2}\) ⇒ θ = \(\frac{\pi}{4}\)
i.e., ∠ACB = θ = \(\frac{\pi}{4}\) or 45°
∴ ∠BAC = 180° – 90° – θ
= 180° – 90° – 45° = 45°
Thus ∠ACB = ∠BAC
Hence the △ABC is an isosceles triangle.

Question 7.
Find the volume of the largest cone that can be inscribed in a sphere of radius r.
Solution:
Let ABC be the cone of largest volume that can be inscribed in the sphere, it is understandable that for maximum value, the axis of the cone must be along the diameter of sphere. Let DE = x,
∴ radius of cone = \(\sqrt{r^2-x^2}\) = BE
and height of cane = r + x
Then volume of cone
⇒ V = \(\frac { 1 }{ 3 }\)π(r² – x²)(r + x)
⇒ \(\frac{d V}{d x}\) = \(\frac{\pi}{3}\left[r^2-2 r x-3 x^2\right]\)
⇒ \(\frac{d^2 \mathrm{~V}}{d x^2}\) = \(\frac{\pi}{3}[-6 x-2 r]\)
For Max./minima, \(\frac{d V}{d x}\) = 0
⇒ 3x² + 2rx – r² = 0 ⇒ (x + r)(3x – r) = 0
⇒ x = \(\frac { r }{ 3 }\) [∵ r ≠ x]
at x = \(\frac { r }{ 3 }\), \(\frac{d^2 \mathrm{~V}}{d x^2}\) = \(\frac{\pi}{3}(-4 r)<0\)
∴ V is maximise for x = \(\frac { r }{ 3 }\)
∴ Max. volume of cone
= \(\frac{\pi}{3}\left(r^2-\frac{r^2}{9}\right)\left(r+\frac{r}{3}\right)\) = \(\frac{32 \pi r^3}{81}\)

Thus Max. volume of cone = \(\frac{8}{27}\left(\frac{4}{3} \pi r^3\right)\)
= \(\frac { 8 }{ 27 }\) (volume of the sphere)

Question 8.
Show that the semi-vertical angle of a cone of maximum volume and of given slant height is \(\tan ^{-1} \sqrt{2}\).
Solution:
Let r be the radius and h be the height of cone respectively and let α be the semi-vertical angle of cone.

Given l be the start height of cone.
∴ r² + h² = l²
Let V = volume of cone = \(\frac{\pi}{3} r^2 h\)
= \(\frac{\pi}{3} h\left(l^2-h^2\right)\)
∴ \(\frac{d V}{d h}\) = \(\frac{\pi}{3}\left[l^2-3 h^2\right]\)
& \(\frac{d^2 \mathrm{~V}}{d h^2}\) = \(\frac{\pi}{3}(-6 h)\) = -2πh
For maxima/minima, we put \(\frac{d V}{d h}\) = 0
⇒ \(\frac{\pi}{3}\left(l^2-3 h^2\right)\) = 0
⇒ h = \(\frac{l}{\sqrt{3}}\) (∵ h > 0)
∴ At h = \(\frac{l}{\sqrt{3}}\)\(\frac{l}{\sqrt{3}}\);
\(\frac{d^2 \mathrm{~V}}{d h^2}\) = \(-2 \pi\left(\frac{l}{\sqrt{3}}\right)<0\)
Thus V is maximum when h = \(\frac{l}{\sqrt{3}}\)
∴ from (i) ; r = \(\sqrt{l^2-h^2}\)
= \(\sqrt{l^2-\frac{l^2}{3}}\) = \(\sqrt{\frac{2}{3}} l\)
In right angled △OAB, \(\frac { r }{ h }\) = tan α
⇒ \(\frac{\sqrt{\frac{2}{3}} l}{\frac{l}{\sqrt{3}}}\) = tan α
⇒ tan α = \(\sqrt{2}\)
⇒ α = tan^-1\(\sqrt{2}\)

Question 9.
A wire of length 20 m is available to fence off a flower bed in the form of a sector of a circle. What must be the radius of the circle, if we wish to have a flower bed with the greatest possible area ?
Solution:
Let OAB be the sector of circle with radius r and AB = l

∴ l + 2r = 20 …(i)
Let Δ = area of sector of circle
= \(\frac { 1 }{ 2 }\)r²θ = \(\frac{r^2}{2}\) × \(\frac { 1 }{ r }\) = \(\frac { 1 }{ 2 }\)r
= \(\frac { r }{ 2 }\)(20 – 2r) [using eqn. (i)]
∴ \(\frac{d \Delta}{d r}\) = \(\frac { 1 }{ 2 }\)(20 – 4r)
& \(\frac{d^2 \Delta}{d r^2}\) = –\(\frac { 4 }{ 2 }\) = -2
For maxima/minimia, \(\frac{d \Delta}{d r}\) = 0
⇒ 20 – 4r = 0 ⇒ r = 5
∴ \(\left(\frac{d^2 \Delta}{d r^2}\right)_{r=5}\) = -2 < 0
Thus Δ is maximise when r = 5 cm
Hence in this case area of sector of circle be maximise when radius of circle be 5 metre.

Question 10.
A closed right circular cylinder has volume \(\frac { 539 }{ 2 }\) cubic units. Find the radius and the height of the cylinder so that the total surface area is minimum.
Solution:
Let r be the radius and h be the height of closed right circular cylinder respectively given volume of cylinder = \(\frac { 539 }{ 2 }\) cubic units
∴ πr²h = \(\frac { 539 }{ 2 }\) …(i)
Let S = Total surface area of cylinder = 2πrh + 2πr²
⇒ S = 2πr² + 2πr \(\left(\frac{539}{2 \pi r^2}\right)\)
⇒ S = 2πr² + \(\frac { 539 }{ r }\)
∴ \(\frac { dS }{ dr }\) = 4πr – \(\frac{539}{r^2}\)
& \(\frac{d^2 \mathrm{~S}}{d r^2}\) = 4π + \(\frac{1078}{r^3}\)
for maxima/minima, \(\frac { dS }{ dr }\) = 0

Thus S is minimise when r = \(\frac { 7 }{ 2 }\)
∴ total surface area of cylinder is minimise when r = \(\frac { 7 }{ 2 }\)
∴ from (i); h = \(\frac{539}{2 \times \pi r^2}\)
∴ h = \(\frac{539}{2 \times \frac{22}{7} \times\left(\frac{7}{2}\right)^2}\) = \(\frac{539}{11 \times 7}\) = \(\frac{49}{7}\) = 7
∴ h = 7 units.

Question 11.
If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is \(\frac{\pi}{3}\).
Solution:
Let △ACB be the right angled triangle
∴ x² + y² = h² …(i)
also it is given that h + y = k (say)
⇒ h = k – y …(ii)
using eqn. (ii) in eqn. (i), we have

Question 12.
A printed page is to have a total area of 80 sq cm with a margin of 1 cm at the top and on each side and a margin of 1.5 cm at the bottom. What should be the dimensions of the page so that the printed area will be maximum?
Solution:
Let x and y are the dimensions of printed page respectively.
Then total area of printed page = 80 cm²
⇒ xy = 80 …(i)

Let A = area of printed page = (x – 2) (y – 2.5)
= xy – \(\frac { 5 }{ 2 }\)x – 2y + 5
⇒ A = 80 – \(\frac { 5 }{ 2 }\)x – 2 × \(\frac { 80 }{ x }\) + 5
= 85 – \(\frac { 5x }{ 2 }\) – \(\frac { 160 }{ x }\)
∴ \(\frac { dA }{ dx }\) = –\(\frac { 5 }{ 2 }\) + \(\frac{160}{x^2}\)
& \(\frac{d^2 \mathrm{~A}}{d x^2}\) = –\(\frac{320}{x^3}\)
for maxima/minima, we put, \(\frac { dA }{ dx }\) = 0

Thus area A is maxima when x= 8 cm
∴ from (i); y = \(\frac { 80 }{ 8 }\) = 10 cm

Question 13.
Find the maximum volume of the cyclinder which can be inscribed in a sphere of radius \(3 \sqrt{3}\) cm. (Leave the answer in terms of it)
Solution:
Let h be the height and R be the radius of the cylinder that is inscribed in a sphere of radius r.

Question 14.
A wire of length 50 m is cut into two pieces. One piece of the wire is bent in the shape of a square and the other in the shape of a circle. What should be the length of each piece so that the combined area of the two is minimum?
Solution:
Since a wire of length 50 m is cut into two pieces. One piece of the wire is bent in the shape of a square and let xm be the side of square and other in the shape of circle and let r metre be the radius of circle.
∴ Circumference or perimeter of square = 4x
& Circumference of circle = 2πr
∴ 4x + 2πr = 50 …(i)
A₁ = area of square =x²
& A₂ = area of circle = πr²
Let A = combined area
= A₁ + A₂ = x² + πr²
∴ A = x² + \(\pi\left(\frac{50-4 x}{2 \pi}\right)^2\)
= x² + \(\frac{1}{4 \pi}(50-4 x)^2\)
A = x² + \(\frac{1}{4 \pi} 4(25-2 x)^2\)
∴ \(\frac { dA }{ dx }\) = 2x + \(\frac{1}{\pi}\)2(25 – 2x) (-2)
& \(\frac{d^2 \mathrm{~A}}{d x^2}\) = 2 + \(\frac{-4}{\pi}(-2)\) = 2 + \(\frac{8}{\pi}\)
For maxima/minima, we put \(\frac { dA }{ dx }\) = 0
⇒ 2x – \(\frac{4}{\pi}(25-2 x)\)
⇒ 2πr – 100 + 8x = 0
⇒ x = \(\frac{100}{2 \pi+8}\) = \(\frac{50}{\pi+4}\)
At x = \(\frac{50}{\pi+4}\);
\(\frac{d^2 \mathrm{~A}}{d x^2}\) = 2 + \(\frac{8}{\pi}\) > 0
Thus A is minimise when x = \(\frac{50}{\pi+4} m\)
∴ Length of square wire = 4x = \(\frac{200}{\pi+4} \mathrm{~m}\)
∴ from (i) ; r = \(\frac{50-4 x}{2 \pi}\)
= \(\frac{1}{2 \pi}\left[50-\frac{200}{\pi+4}\right]\) = \(\frac{25}{\pi+4}\)
∴ Length of circular wire = 2πr
= \(\frac{50 \pi}{\pi+4}\) metre

Question 15.
Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius 10 cm is a square of side \(10 \sqrt{2}\) cm.
Solution:
Let ABCD be the rectangle inscribed in circle of radius 10 cm with centre O. Let ∠CAB = θ
From right angled △ABC; we have
\(\frac{\mathrm{AB}}{20}\) = cos θ ⇒ AB = 20 cos θ
& \(\frac{\mathrm{BC}}{20}\) = sin θ ⇒ BC = 20 sin θ

Let P = perimeter of rectangle
= 2(AB + BC)
= 2(20 cos θ + 20 sin θ)
= 40(cos θ + sin θ)
∴ \(\frac{d \mathrm{P}}{d \theta}\) = 40(-sin θ + cos θ)
& \(\frac{d^2 \mathrm{P}}{d \theta^2}\) = 40(-cos θ – sin θ)
For maxima/minima, we put \(\frac{d \mathrm{P}}{d \theta}\) = 0
⇒ 40(-sin θ + cos θ) = 0
⇒ tan θ = 1 ⇒ θ = \(\frac{\pi}{4}\)
when θ = \(\frac{\pi}{4}\);
\(\frac{d^2 \mathbf{P}}{d \theta^2}\) = 40\(40\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)\) = –\(40 \sqrt{2}\) < 0
Thus perimeter of rectangle i.e. P is maximise. When θ = \(\frac{\pi}{4}\)
∴ AB = 20 cos\(\frac{\pi}{4}\) = \(\frac{20}{\sqrt{2}}=10 \sqrt{2} \mathrm{~cm}\)
& BC = 20 sin \(\frac{\pi}{4}\) = \(\frac{20}{\sqrt{2}}=10 \sqrt{2} \mathrm{~cm}\)
∴ AB = BC
Hence the rectangle ABCD is a square.

Question 16.
A rectangle is inscribed in a semicircle of radius r with one of its sides on the diameter of the semicircle. Find the dimensjbns of the rectangle to get maximum area. Also, find the maximum area.
Solution:
Let 2x be the length and y be the breadth of the rectangle which is inscribed in a circle of radius r. Let C (x, y) be any point on semi-circle s.t. x = r cos α, y = r sin α

∴ area of rectangle = 2x . y = 2r² sin α cos α = r² sin 2α
Clearly A is maximum when sin 2 α is max. i.e. when sin 2 α = 1
i.e. 2α = \(\frac{\pi}{2}\) ⇒ α = \(\frac{\pi}{4}\)
∴ x = \(\frac{r}{\sqrt{2}}\) and y = \(\frac{r}{\sqrt{2}}\)
∴ the dimensions of the rectangle are
2x = 2 × \(\frac{r}{\sqrt{2}}\) = \(\sqrt{2} r\)
and y = \(\frac{r}{\sqrt{2}}\)
∴ Max. area = 2xy
= 2 × \(\frac{r}{\sqrt{2}}\) × \(\frac{r}{\sqrt{2}}\) = r² sq. units.

Question 17.
The points of local minimum (or relative minimum) are ………
Solution:
A, C, E

Question 18.
The points of local maximum (or relative maximum) are …….
Solutio:
B, D, F

Question 19.
A given function f(x) will have a maximum value for a given value x = c, if f'(x) …… and f ” (c) ………
Solution:
f ‘ (x) = 0 and f ” (c) < 0 Question 20. For a minimum value of f(x) at x = c, the necessary and sufficient conditions are that f ‘ (x) ……. and f ” (c) …. Solution: f ‘ (x) = 0 and f ” (c) > 0

Question 21.
If x is real then the minimum value of x² – 8x + 17 is …….
Solution:
Let y = x² – 8x + 17
⇒ y = x² – 8x + 16 + 1 = (x – 4)² + 1
⇒ y ≥ 1 [∵ (x – 4)² ≥ 0 ∀ x ∈ R]
Thus, minimum value of y be 1 .

Question 22.
The minimum value of 2x + 3y, when xy = 6 is …….
Solution:
Let S = 2x + 3y and xy = 6 ….(1)
⇒ S = 2x + \(\frac{18}{x}\) [using (1)]
∴ \(\frac{d S}{d x}\) = 2 – \(\frac{18}{x^2}\)
and \(\frac{d^2 S}{d x^2}\) = \(\frac{36}{x^3}\)
For maxima/minima, \(\frac{d S}{d x}\) = 0
⇒ 2 – \(\frac{18}{x^2}\) = 0 ⇒ x² = 9 ⇒ x = ± 3
at x = 3 ; \(\frac{d^2 S}{d x^2}\) = \(\frac{36}{9}\) = 4 > 0
Thus S is minimise at, x = 3
∴ from (1) ; y = 2
∴ minimum value of S = 2 × 3 + 3 × 2 = 12

Question 23.
The minimum value of f(x) = | x + 4 | – 3 is ……..
Solution:
Given f(x) = | x + 4 | – 3
since | x + 4 | ≥ 0 ∀ x ∈ R
⇒ | x + 4 | – 3 ≥ – 3 ⇒ f(x) ≥ – 3
∴ Minimum value of f(x) = -3

Question 24.
Consider the function -(x – 3)² + 15. Its maximum value is ……
Solution:
Given f(x) = -(x – 3)² + 15
since (x – 3)² ≥ 0
⇒ – (x – 3)² ≤ 0 ∀ x ∈ R
⇒ f(x) ≤ 15
∴ Maximum value of f(x) = 15

Question 25.
Consider the function -(x – 3)² + 15. Its maximum value is ………
Solution:
Given f(x) = -(x – 3)² + 15
since (x – 3)² ≥ 0
⇒ – (x – 3)² ≤ 0 ∀ x ∈ R
⇒ – (x – 3)² + 15 ≤ 15
⇒ f(x) ≤ 15
∴ Maximum value of f(x) = 15

Question 26.
If f(x) = \(\frac { x }{ 2 }\) – \(\frac { 2 }{ 3 }\), then its maximum value is ……..
Solution:
Given f(x) = \(\frac { x }{ 2 }\) + \(\frac { 2 }{ x }\)
∴ f ‘(x) = \(\frac { 1 }{ 2 }\) – \(\frac{2}{x^2}\)
∴ f ”(x) = \(\frac{4}{x^3}\)
For maxima/minima, f ‘ (x) = 0
⇒ \(\frac { 1 }{ 2 }\) = \(\frac{2}{x^2}\) ⇒ x² = 4
∴ x = ± 2
at x = – 2 ; f ”(-2) = \(\frac{4}{-8}\) = \(\frac{-1}{2}\) < 0
Thus x = – 2 is a point of maxima
∴ Max. value = f (- 2) = \(\frac{-2}{2}\) – \(\frac{2}{2}\) = 2

Question 27.
The profit function of a company is given by p(x) = 41 – 72x – 18x². The maximum profit that it can make is …….
Solution:
p(x) = 41 – 72x – 18x²
∴ p ‘ (x) = – 72 – 36x
For maxima/minima, p ‘ (x) = 0
⇒ – 72 – 36x = 0 ⇒ x = – 2
∴ f ” (x) = – 36 ⇒ f ”(- 2) = -36 < 0
Thus f(x) is maximise when x = – 2
∴ Maximum profit = p (- 2)
= 41 – 72(- 2) – 18(- 2)²
= 41 + 144 – 72 = 113

Question 28.
The sum of two numbers is 10. Their product will be maximum when they are
(a) 3, 7
(b) 4, 6
(c) 5, 5
(d) 8, 2
Solution:
Let the required two numbers are x and y
s.t x + y = 10
and P = xy = x(10 – x) [using (1)]
∴ \(\frac{dP}{dx}\) = 10 – 2x
and \(\frac{d^2 \mathrm{P}}{d x^2}\) = -2
For maxima/minima, \(\frac{dP}{dx}\) = 0
⇒ 10 – 2x = 0 ⇒ x = 5
∴ \(\left(\frac{d^2 \mathrm{P}}{d x^2}\right)_{x=5}\) = – 2 < 0
Thus P is maximum when x = 5
∴ from (1); y = 5
∴ Ans. (c)

Question 29.
The maximum slope of the curve y = – x³ + 3x² – 2x+ 27 is
(a) 5
(b) -5
(c) \(\frac{1}{5}\)
(d) None of these
Solution:
Given eqn. of curve be,
y = – x³ + 3x² – 2x + 27
∴ \(\frac{dy}{dx}\) = – 3x² + 6x – 2
Let m = slope of given curve = -3x² + 6x – 2
∴ \(\frac{dm}{dx}\) = -6x + 6; \(\frac{d^2 m}{d x^2}\) = – 6 < 0
For maxima/minima, \(\frac{dm}{dx}\) = 0
⇒ – 6x + 6 = 0 ⇒ x = 1
∴ \(\left(\frac{d^2 m}{d x^2}\right)_{x=1}\) = – 6 < 0
Thus m is maximise when x = 1
∴ maximum slope = – 3 + 6 – 2 = 1
∴ Ans. (d)

Question 30.
The maximum value of f(x) = \(\frac{\log _e x}{x}\) (x ≠ 0, x ≠ 1) is
(a) e
(b) \(\frac{1}{e}\)
(c) e²
(d) \(\frac{1}{e^2}\)
Solution:

Question 31.
The function f(x) = x + \(\frac{4}{x}\) has
(a) a local maxima at x = 2 and local minima at x = – 2
(b) a local minima at x = 2 and local maxima at x = – 2
(c) absolute maxima at x = 2 and absolute minima at x = – 2
(d) absolute minima at x = 2 and absolute maxima at x = – 2
Answer:
Given f(x) = x + \(\frac{4}{x}\)
∴ f ‘ (x) = 1 – \(\frac{4}{x^2}\)
and f ” (x) = \(\frac{8}{x^3}\)
For maxima/minima, f ‘ (x) = 0
⇒ 1 – \(\frac{4}{x^2}\) = 0 ⇒ x = ± 2
When x = 2; f ” (x) = \(\frac { 8 }{ 8 }\) = 1 > 0
∴ a = 2 is a point of local minima.
when x = -2; f ” (x) = \(\frac { 8 }{ 8 }\) = – 1 < 0
∴ x = -2 is a point of local maxima.
∴ Ans. (b)

Question 32.
The smallest value of polynomial x³ – 18x² + 96x in [0,9] is
(a) 126
(b) 0
(c) 135
(d) 160
Solution:
Given f(x) = x³ – 18x² + 96x
∴ f ‘ (x) = 3x² – 36x + 96
For maxima/minima,
f ‘ (x) = 0 ⇒ 3 (x² – 12x + 32) < 0
⇒ (x – 4)(x – 8) = 0 ⇒ x = 4, 8
When x = 4;
f(4) = 64 – 288 + 384 = 160
When x= 8;
f(8) = 512 – 1152 + 768 = 128
When x = 0 ; f(0) = 0
When x = 9;
f(9) = 729 – 1458 + 864 = 135
∴ smallest value of f(x) be 0 .
∴ Ans. (b)

Question 33.
The maximum value of the function x + cos x is
(a) 1
(b) \(\sqrt{2}\)
(c) 2
(d) None of these
Solution:
Given f(x) = sin x + cos x
∴ f ‘ (x) = cos x – sin x
For maxima/minima, f ‘ (x) = 0
⇒ cos x – sin x = 0
⇒ tan x = 1
⇒ x = \(\frac{\pi}{4}\)
and f ” (x) = sin x – cos x
∴ f ”\(\left(\frac{\pi}{4}\right)\) = – \(\frac{1}{\sqrt{2}}\) – \(\frac{1}{\sqrt{2}}\) = – \(\sqrt{2}\) < 0
Thus x = \(\frac{\pi}{4}\) be a point of maxima.
∴ Max. value of f(x) = f\(\left(\frac{\pi}{4}\right)\)
= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) = \(\sqrt{2}\)
∴ Ans. (b)

Question 34.
The maximum value (in cu.m) of the right circular cone having slant height 3 m is
(a) \(2 \sqrt{3}\) m
(b) \(3 \sqrt{3} \pi\)
(c) 6 π
(d) \(\frac{4}{3} \pi\)
Solution:
Let r be the radius of base of cone and h be the height of cone.
Then V = volume of right circular cone = \(\frac{\pi}{3} r^2 h\)

[l = 3 ; l² = h² + r² ⇒ 9 – h² = r²]
= \(\frac{\pi}{3}\left(9-h^2\right) h\)
∴ \(\frac{d V}{d h}\) = \(\frac{\pi}{3}\left[9-3 h^2\right]\)
and \(\frac{d^2 \mathrm{~V}}{d h^2}\) = \(\frac{\pi}{3}[0-6 h]\) = -2πh
For maxima/minima, \(\frac{d V}{d h}\) = 0
⇒ 9 – 3h² = 0
⇒ h² = 3
⇒ h = ±√3 (∵ h > 0)
⇒ h = √3
Thus, \(\left(\frac{d^2 \mathrm{~V}}{d h^2}\right)_{h=\sqrt{3}}\) = \(-2 \pi \sqrt{3}<0\)
Thus V is maximise when h = √3 and r² = 9 – 3 = 6
∴Maximum volume = \(\frac{\pi}{3}\) × 6 × \(\sqrt{3}\)
= \(2 \pi \sqrt{3}\) cu.m
∴ Ans. (a)

Question 35.
The maximum value of \(\left(\frac{1}{x}\right)^x\), x > 0 is
(a) e
(b) e^e
(c) \(e^{\frac{1}{e}}\)
(d) \(\left(\frac{1}{e}\right)^{\frac{1}{e}}\)
Solution:
Given y = \(\left(\frac{1}{x}\right)^x\)
∴ log y = x log\(\frac{1}{x}\) = – x log x
∴ \(\frac{d y}{d x}\) = – \(y\left[x \times \frac{1}{x}+\log x \times 1\right]\)
= – (1 + log x) y
⇒ \(\frac{d y}{d x}\) = \(-\left(\frac{1}{x}\right)^x[1+\log x]\)
For maxima/minima, \(\frac{d y}{d x}\) = 0
⇒ 1 + log x = 0 ⇒ log x = -log e
⇒ x = \(\frac{1}{e}\)
When x slightly < \(\frac{1}{e}\)
⇒ log x < log e^-1
⇒ log x < – 1
⇒ 1 + log x < 0 ∴ \(\frac{d y}{d x}\) = + ve When x slightly > \(\frac{1}{e}\)
⇒ log x log e^-1 = – 1
⇒ 1 + log x > 0
∴ \(\frac{d y}{d x}\) = – ve
Thus \(\frac{d y}{d x}\) changes its sign from + ve to ve
∴ x = \(\frac{1}{e}\) be a point of maxima.
∴ Maximum value of y = [e]^1/e
∴ Ans. (c)

Question 36.
A point of inflection of the curve given by y = x³ – 6x² + 12x + 50 occurs when
(a) x = \(\frac{2}{3}\)
(b) x = \(\frac{3}{2}\)
(c) x = 2
(d) x = 3
Solution:
Given y = x³ – 6x² + 12x + 50
∴ \(\frac{dy}{dx}\) = 3x² – 12x + 12
= 3 (x² – 4x + 4)
∴ \(\frac{d^2 y}{d x^2}\) = 3(2x – 4)
For maxima/minima,
∴\(\frac{dy}{dx}\) = 0
⇒ 3(x – 2)² = 0 ⇒ x = 2
at x = 2 ; \(\frac{d^2 y}{d x^2}\) = 0
∴ x = 2 is a point of inflexion.
∴ Ans. (c)

Question 37.
The function f(x) = x^x has a stationary point at
(a) x = e
(b) x = \(\frac{1}{e}\)
(c) x = 1
(d) x = √e
Solution:
Given y = x^x ⇒ log y = x log x
∴ \(\frac { 1 }{ y }\) \(\frac { dy }{ dx }\) = x × \(\frac { 1 }{ x }\) + log x × 1
⇒ \(\frac { dy }{ dx }\) = x^x (1 + log x)
For maxima/minima ; \(\frac { dy }{ dx }\) = 0
⇒ x^x (1 + log x) = 0
⇒ log x = -1 = – log e = log e^-1
⇒ x = \(\frac { 1 }{ e }\) [∵ x^x > 0]
When x slightly < \(\frac{1}{e}\)
⇒ log x < log e^-1
⇒ 1 + log x < 0
Thus, \(\frac{dy}{dx}\) < 0
When x slightly > \(\frac { 1 }{ e }\)
⇒ log x > log e^-1 = – log e
⇒ 1 + log x < 0
∴ \(\frac { dy }{ dx }\) > 0
Thus \(\frac { dy }{ dx }\) changes its sign from -ve to +ve.
Thus x = \(\frac { 1 }{ e }\) be a point of minima.
∴ Ans. (b)

Question 38.
The maximum area of a rectangle inscribed in the circle (x + 1)² + (y – 3)² = 64 is
(a) 64 sq. units
(b) 72 sq. units
(c) 128 sq. units
(d) 8 sq. units
Solution:
putting X = x + 1 ; Y = y – 3
The eqn. of circle becomes ;
X² + Y² = 64 …(1)
∴ A = area of rectangle PQRS
= (2X) (2Y)
= 4XY

⇒ A² = 16X²Y² – eX² (64 – X²)
∴ \(\frac { d }{ dx }\) X² – 16 (128X – 4X³)
For maxima/minima, \(\frac { d }{ dx }\) A² = 0
⇒ 128X – 4X³ = 0
⇒ X² = 32
⇒ X = 4√2
and \(\frac{d^2}{d \mathrm{X}^2}\left(\mathrm{~A}^2\right)\) = 128 – 12X²
at X = 4√2;
\(\frac{d^2}{d \mathrm{X}^2}\left(\mathrm{~A}^2\right)\) = 128 – 12 × 32 < 0
Thus X = 4√2 is a point of maxima and Max. value of area = 4XY
= 4 × 4√2 × \(\sqrt{64-32}\)
= 4 × 4√2 × 4√2
=128 sq. units
∴ Ans. (c)

Question 39.
The maximum value of the function 2x³ – 15x² + 36x +4 is attained at
(a) 0
(b) 3
(c) 4
(d) 2
Solution:
Given y = 2x³ – 15x² + 36x + 4
∴ \(\frac { dy }{ dx }\) = 6x² – 30x + 36
= 6 (x² – 5x + 6)
and \(\frac{d^2 y}{d x^2}\) = 6 (2x – 5)
For maxima/minima, \(\frac { dy }{ dx }\) = 0
⇒ x² – 5x + 6 = 0
⇒ (x – 2)(x – 3) = 0
x = 2, 3
at x = 2 ; \(\frac{d^2 y}{d x^2}\) = 6(4 – 5) = -6 < 0
Thus y is maximise when x = 2
∴ y_max = 16 – 60 + 72 + 4 = 32
∴ Ans. (d)

Question 40.
The minimum value of the function f(x) = x log x is
(a) – \(\frac { 1 }{ e }\)
(b) – e
(c) \(\frac { 1 }{ e }\)
(d) e
Solution:
Given f(x) = x log x
∴ f ‘ (x) = x × \(\frac { 1 }{ x }\) + log x × 1 = 1 + log x
and f ” (x) = \(\frac { 1 }{ x }\)
For maxima/minima, \(\frac { dy }{ dx }\) = 0
⇒ 1 + log x = 0 ⇒ x = \(\frac{1}{e}\)
∴ f ” \(\left(\frac{1}{x}\right)\) = \(\frac{1}{1 / e}\) = e > 0
∴ x = \(\frac{1}{e}\) is a point of minima.
∴ Minimum value of f(x) = f\(\left(\frac{1}{e}\right)\)
= \(\frac{1}{e}\) log \(\frac{1}{e}\)
= \(\frac{1}{e}\) log e^-1 = \(\frac{1}{e}\)
∴ Ans. (a)

Question 41.
20 meters of a wire is available for fencing off a flower bed in the form of a circular sector. The maximum area (in sq. m) of the flower bed is
(a) 10
(b) 25
(c) 30
(d) 12.5
Solution:
Given length of wire is available for fencing off a flower bed = 20 m
∴ l + 2r = 20
⇒ rθ + 2r = 20 …(1)

∴ area of flower bed = A = \(\frac{1}{2}\)r²θ
= \(\frac{1}{2} r^2\left(\frac{20-2 r}{r}\right)\) [using (1)]
= \(\frac{1}{2}\left[20 r-2 r^2\right]\)
∴ \(\frac{dA}{dr}\) = \(\frac{1}{2}\) (20 – 4r)
For maxima/minima, \(\frac{dA}{dr}\) = 0
⇒ \(\frac{1}{2}\)(20 – 4r) = 0 ⇒ r = 5
and \(\frac{d^2 \mathrm{~A}}{d r^2}\) = – 2 < 0 for r = 5
Thus A is maximise when r = 5
∴ Max. value of A = \(\frac{1}{2}\) × 5² × \(\left(\frac{20-2 \times 5}{5}\right)\)
= 25 sq m.
∴ Ans. (b)

Question 42.
Which of the following is true about the function f(x) = x⁴ – 4x²?
(a) It has two local minima and one local maxima
(b) It has two local minima and zero local maxima
(c) It has one local minima and one local maxima
(d) It has two local minima and two local maxima
Solution:
Given f(x) = x⁴ – 4x²
∴ f ‘ (x) = 4x³ – 8x
f ” (x) = 12x² – 8
For maxima/minima, f ‘ (x) = 0
⇒ 4x (x² – 2) = 0
⇒ x = 0, ± √2
Here f ” (0) = – 8 < 0
∴ x = 0 is a point of local maxima.
f ”(± √2) = 12(± √2)² – 8 = 16 > 0
Thus x = ± √2 are points of minima.
∴ Ans. (a)

Question 43.
Let f(x) = 2x³ – 9ax² + 12a² x + 1, where a > 0. The minimum value of f(x) is attained at a point q and the maximum value is attained at a point p. If p³ = q, then a is equal to
(a) 1
(b) 3
(c) 2
(d) √2
Solution:
Given f(x) = 2x³ – 9ax² + 12a²x + 1
∴ f ‘ (x) = 6x² – 18ax + 12a²
Now f(x) has minimum value at x = q and maximum value at x = p
∴ f ‘ (p) = 0 =f ‘ (q)
∴ f ‘ (p) = 6p² – 18qp + 12a² = 0
⇒ p² – 3ap + 2a² = 0
⇒(p – a)(p – 2a) = 0
⇒ p = a, 2a
and f ‘ (q) = 0
⇒ 6 q² – 18aq + 12a² = 0
⇒ q² – 3aq + 12a² = 0
⇒ q = a, 2a
For maxima/minima, f ‘ (x) = 0
⇒ 6 (x² – 3ax + 2a²) = 0
⇒ x = a, 2a
and f ” (x) = 6(12x – 18a)
∴ f ” (a) = 6(12a – 18a) = -36a < 0 (∵ a > 0)
∴ x = a is a point of maxima and f(x) has maximum value at x = p
∴ p = a
and f ” (2a) = 6(24a – 18a) = 36a > 0
(∵ a > 0 )
∴ x = 2 a is a point of minima and f(x) has minimum value at x = q
∴ q = 2a
Also, p³ = q
⇒ a³ = 2a
⇒ (a² – 2) = 0
⇒ a = ±√2, 0 but a > 0
∴ a = √2
∴ Ans. (d)

Question 44.
A missile is fired from the ground level. It rises x meters vertically upwards in t seconds, where x = 100t – \(\frac { 25 }{ 2 }\)t². The maximum height reached is
(a) 200 m
(b) 125 m
(c) 160 m
(d) 190 m
Solution:
Given x = 100t – \(\frac { 25 }{ 2 }\)t²
∴ \(\frac { dx }{ dt }\) = 100 – 25t
and \(\frac{d^2 x}{d t^2}\) = -25
For maxima/minima, \(\frac { dx }{ dt }\) = 0
⇒ 100 – 25t = 0 ⇒ t = 4
∴ at t = 4 ; \(\frac{d^2 x}{d t^2}\) = – 25 < 0
Thus height x is maximise when t = 4 seconds
∴ Maximum height = 100 × 4 – \(\frac { 25 }{ 2 }\) × 16
= 200 m.
∴ Ans. (a)

Question 45.
The point on the curve x² = 2y which is nearest to the point (0, 5) is
(a) (2√2, 4)
(b) (2√2, 0)
(c) (0, 0)
(d) (2, 2)
Solution:
Let P(x, y) be any point on given curve x² = 2y …(i)
and given point is A(0, 5)
∴ AP² =(x – 0)² + (y – 5)²
= 2y + (y – 5)² [using (1)]
To minimis/maximise AP it is convenient to minimise or maximise AP² and Let Z = AP²
∴ \(\frac { dz }{ dy }\) = 2 + 2(y – 5)
and \(\frac{d^2 z}{d y^2}\) = 2
For maxima/minima, \(\frac { dz }{ dy }\) = 0
⇒ 2 + 2y – 10 = 0
⇒ y = 4
at y = 4 ; \(\frac{d^2 z}{d y^2}\) = 2 > 0
∴ z is minimise when y = 4
∴ from (1); x² = 8
⇒ x = ±2√2
Thus required coordinates of any point on given curve are (±2√2, 4) .
∴ Ans. (a)

Question 46.
If f(x) = \(\frac{-3}{4}\)x⁴ – 8x³ – \(\frac{45}{2}\) x² – 105, then which of the following holds ?
(i) f(x) has local maxima at x = 2
(ii) f(x) has local maxima at x = 5
(iii) f(x) has local minima at x = -3
(a) only (i) is true
(b) only (iii) is true
(c) Both (i) and (ii) are true
(d) All (i), (ii) and (iii) are true
Solution:
Given f(x) = – \(\frac{3}{4}\)x⁴ – 8x³ – \(\frac{45}{2}\)x² – 105
∴ f ‘ (x) = – 3x³ – 24x² – 45x
= – 3 (x³ + 8x² + 15x)
and f ” (x) = – 3 (3x² + 16x + 15)
For maxima/minima, f ‘ (x) = 0
⇒ – 3 (x³ + 8x² + 15x) = 0
⇒ x (x² + 8x + 15) = 0
⇒ x(x + 5)(x + 3) = 0
⇒ x = 0,- 5, – 3
Here f ” (-5) = -3(75 – 80 + 15) = – 30 < 0
∴ x = – 5 is a point of maxima.
and f ” (-3) = – 3(27 – 48 + 15) = – 30 < 0
∴ x = – 5 is a point of maxima. and f ” (-3) = – 3(27 – 48 + 1 5) = 18 > 0
∴ x = – 3 is a point of local minima.
∴ Ans. (b)

Question 47.
If y = a log x + bx² + x has its extremum at x = -1 and x = 2, then
(a) a = 2, b = \(\frac{1}{2}\)
(b) a = 2, b = \(\frac{-1}{2}\)
(c) a = \(\frac{-1}{2}\), b = 2
(d) a = \(\frac{1}{2}\), b = 2
Solution:
Given y = a log x + bx² + x
∴ \(\frac{dy}{dx}\) = \(\frac{a}{x}\) + 2bx + 1
It is given that y has its extremum at x = -1 and x = 2
∴ \(\left(\frac{d y}{d x}\right)_{x=-1}\) = 0
⇒ – a – 2b + 1 = 0 …(1)
and \(\left(\frac{d y}{d x}\right)_{x=2}\) = 0
⇒ \(\frac{a}{2}\) + 4b + 1 = 0
⇒ a + 8b + 2 = 0 …(2)
On solving eqn. (1) and eqn. (2) ; we have
6b +3 = 0 ⇒ b = \(\frac{-1}{2}\)
∴ from (1); a = +2
∴ Ans. (b)