Maxima and minima is an important topic in the application of derivatives. Maxima and minima are the maximum or the minimum value of a function within the given set of ranges. As far as the JEE Advanced exam is concerned, students can definitely expect questions from this topic. Here, we have given JEE Advanced previous year questions on maxima and minima along with the solutions. Students are recommended to learn the answers to these questions so that they can easily crack the JEE Advanced exam.
Question 1: A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume. The lengths of the sides of the rectangular sheet are
(a) 24
(b) 32
(c) 45
(d) 60
Solution:
Let a be the side of the square, 15x be the length of the rectangle and 8x be the breadth of the rectangle.
4a2 = 100
=>a = 5
Volume = (15x-2a)×(8x-2a)×a
= 4a3 + 120x2a – 16a2x- 30a2x
= 4a3 – 46a2x + 120x2a
dV/da = 12a2 – 92ax + 120x2 …(i)
d2V/da2 = 24a – 92x
Substitute a = 5 in (i)
dV/da = 12×25 – 92×5×x + 120x2
= 300 – 460x + 120x2
= 120x2 – 460x + 300
= 6x2 – 23x + 15 (divide by 20)
For maximum volume dV/da = 0
=> 6x2 – 23x + 15 = 0
Solving the above equation using the quadratic formula, we get
x = 3, or x = 5/6
When x = 3,
d2V/da2 <0, hence volume is maximum.
When x = 5/6,
d2V/da2 >0, so x = â…š is rejected.
Length of rectangle = 15x = 15×3 = 45
Breadth of rectangle = 8x = 8×3 = 24
Hence option a and option c are the correct answers.
Question 2: A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed inner volume of V mm3, has a 2 mm thick solid wall, and is open at the top. The bottom of the container is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the container. If the volume of the material used to make the container is minimum when the inner radius of the container is 10 mm, then the value of V/250 π is
Solution:
Let h be the height of the cylinder and r be the inner radius.
Then volume, V = πr2h
Let V1 be volume of the material.
Given that the bottom of the container is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the container
V1 = Ï€(r+2)2h – Ï€r2h + Ï€(r+2)22
= Ï€h[(r+2)2 – r2] + Ï€(r+2)22
= Ï€h[r2+4r + 4 – r2] + Ï€(r+2)22
= πh(4r + 4) + π(r+2)22
= 4Ï€rh + 4Ï€h + 2Ï€(r+2)2
= (4V/r) + (4V/r2) + 2Ï€(r+2)2
For minimum material required, dV/dr = 0
=> (-4V/r2) – (8V/r3) + 4Ï€(r+2) = 0
Put r = 10
=> (-4V/100) – (8V/1000) + 48Ï€ = 0
=> (-40V – 8V)/1000 = -48Ï€
=> -48V/1000 = -48Ï€
=> V/Ï€ = 1000
Divide both sides by 250
=> V/250 π = 4
Question 3: Let p(x) be a real polynomial of least degree which has a local maximum at x = 1 and a local minimum at x = 3. If p(1) = 6 and p(3) = 2, then p’(0) is
Solution:
Let p’(x) = k(x-1)(x-3)
= k(x2 – 4x + 3)
p(x) = k(x3/3 – 2x2 + 3x) + c (Integrating)
Given p(1) = 6.
=> k(â…“ – 2 + 3) + c = 6
=> 4k/3 + c = 6 …(i)
Also given p(3) = 2
=> k(9 – 18 + 9) + c = 2
=> c = 2
Put c in (i), we get
4k/3 + 2 = 6
=> 4k/3 = 4
=> k = 3
So p’(x) = 3(x-1)(x-3)
p’(0) = 3(0-1)(0-3)
= 9
Hence the value of p’(0) is 9.
Question 4: Let f: R → R be defined as f(x) = |x| + |x2 – 1|. The total number of points at which f attains either a local maximum or a local minimum is
Solution:
Given f(x) = |x| + |x2 – 1|
When x < -1, f(x) = x2 – x – 1
When x ≥ 1, f(x) = x2 + x – 1
When -1 ≤ x < 0, f(x) = -x2 – x + 1
When 0 ≤ x < 1, f(x) = -x2 + x + 1
From the graph, it is clear that the total number of maxima and minima is 5.
Question 5: A box is made with a square base and an open top. The area of the material used is 192 sq. cms. If the volume of the box is maximum, the dimensions of the box are
(a) 4, 4, 8
(b) 2, 2, 4
(c) 8, 8, 4
(d) 2, 2, 2
Solution:
Area of the material used = 192 cm2
Let a be the side of square base and h be height of the box.
The box is open on top.
Surface area = a2 + 4ah = 192
=> h = (192 – a2)/4a ..(i)
Volume of box = a2h
= a2 (192 – a2)/4a
= a(192 – a2)/4
For maximum volume, dV/da = 0
=> (192 – 3a2)/4 = 0
=> 3a2 = 192
=> a2 = 64
=> a = 8
d2V/da2 = -6a/4 < 0
Put a in (i)
=> h = (192 – 64)/32
= 4
So the dimensions of the box are 8, 8, 4.
Hence option c is the answer.
Question 6: The least value of α∈R for which 4αx2 + 1/x ≥ 1, for all x>0, is
(a) 1/64
(b) 1/32
(c) 1/27
(d) 1/25
Solution:
Let f(x) = 4αx2 + 1/x
Differentiate w.r.t.x
f’(x) = 8αx – 1/x2
f’(x) = 0 => 8αx – 1/x2 = 0
=> 8αx = 1/x2
=> x3= 1/8α
=> x = 1/2α1/3
f(x) > 1
=> 4αx2 + 1/x > 1
Put x = 1/2α1/3 in above equation
α/α2/3 + 2α1/3 > 1
=> α1/3 + 2α1/3 > 1
=> 3α1/3 > 1
=> α1/3 > 1/3
=> α > 1/27
So the least value of α is 1/27.
Hence option c is the answer.
Question 7: The area (in sq.units) of the largest rectangle ABCD whose vertices are A and B lie on the x axis and vertices C and D lie on the parabola, y = x2-1 below the x-axis is
(a) 4/3
(b) 2/3√3
(c) 4/3√3
(d) 1/3√3
Solution:
Area of ABCD = AB×BC
= 2a(a2-1) (using distance formula, we can find AB and BC)
= 2a3-2a
dA/da = 6a2-2
d2A/da2 = 12a
dA/da = 0
=> 6a2-2 = 0
a2 = â…“
So a = ±1/√3
d2A/da2 = 12a
At a = -1/√3, d2A/da2 < 0
So area is maximum at a = -1/√3
Area = 2×(-1/√3)×(-⅔)
= 4/3√3 sq. units
Hence option c is the answer.
Question 8: Suppose f(x) is a polynomial of degree four, having critical points at -1, 0, 1. If T = {x∈R| f(x) = f(0)}, then the sum of squares of all the elements of T is:
(a) 6
(b) 2
(c) 8
(d) 4
Solution:
Let f’(x) = k(x+1)x(x-1)
f’(x) = k[ x3-x]
Integrating both sides
f(x) = k[(x4/4) – (x2/2)] + c
f(0) = c
f(x) = f(0)
=> k[(x4/4 )- (x2/2)] + c = c
=> (x4/4 )- (x2/2) = 0
=> x2(x2 – 2) = 0
=> x = 0, √2, -√2
Sum of squares of all elements = 02+(√2)2+(-√2)2
= 4
Hence option d is the answer.
Question 9: Let a1, a2, a3 … be an A.P with a6 = 2. Then the common difference of this A.P, which maximises the product a1a4a5 is
(a) 3/2
(b) 8/5
(c) 6/5
(d) 2/3
Solution:
Let the first term of the AP be a and d be the common difference.
Given a6 = 2
=> a + 5d = 2
=> a = 2-5d
Product, a1a4a5 = a(a+3d)(a+4d)
f(d) = (2-5d)(2-2d)(2-d)
= (4 – 10d – 4d + 10d2)(2-d)
= (4 – 14d + 10d2)(2-d)
= 8 – 28d + 20d2 – 4d + 14d2 – 10d3
= 8 – 32d + 34d2– 10d3
f’(d) = -32 + 68d – 30d2
f’(d) = 0
=> d = 8/5, 2/3
f’’(d) = 68-60d
For maximise the product, f’’(d) <0.
At d = 8/5, f’’(d) = -28 <0
Hence option b is the answer.
Question 10: Let P = x3-1/x3, Q= x-1/x and a is the minimum value of P/Q2. Then the value of [a] is where [x] = greatest integer less than or equal to x.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
Given P = x3-1/x3
Q= x-1/x
P/Q2 = (x3– 1/x3)/(x – 1/x)2
= (x – 1/x)(x2 + 1 + 1/x2)/(x – 1/x)2
= (x2 + 1 + 1/x2)/(x – 1/x)
= [(x – 1/x)2 + 3]/(x – 1/x)
= (x – 1/x) + 3/(x – 1/x)
Use AM ≥ GM
[(x – 1/x) + 3/(x – 1/x)]/2 ≥ 31/2 [(x – 1/x) + 3/(x – 1/x)] ≥ 2√3P/Q2 ≥ 2√3
≥ 3.464 = a
[a] = [3.464] = 3Hence option c is the answer.
Question 11: If f(x) = x5 – 5x4 + 5x3 – 10 has local maxima and minima at x = k and x = m respectively, then 2k+m is equal to
Solution:
Given f(x) = x5 – 5x4 + 5x3 – 10
Differentiate with respect to x.
f’(x) = 5x4 – 20x3 + 15x2
To find extreme points, equate f’(x) to zero.
=> 5x4 – 20x3 + 15x2= 0
5x2(x2 – 4x + 3) = 0
=> 5x2(x-3)(x-1) = 0
=>x = 0, 3, 1
We use second derivative test to find maxima and minima.
f’’(x) = 20x3 – 60x2 + 30x
f’’(0) = 0
f’’(1) = 20 – 60 + 30 = -10 < 0
f’’(3) = 20(27) – 60(9) + 30(3)
= 540 – 540 + 90
= 90 >0
So x = 1 is a maxima and x = 3 is a minima.
Thus k = 1 and m = 3
2k+m = 2 + 3
= 5
Hence the value of 2k+m = 5.
Question 12: If f(x) = x3 + x2 f’(1) + xf’’(2) + f’’’(3) for all x belongs to R. Then the value of f’(1) + f’’(2) + f’’’(3) is
Solution:
f(x) = x3 + x2 f’(1) + xf’’(2) + f’’’(3)
f’(x) = 3x2 + 2x f’(1) + f’’(2)
f’’(x) = 6x + 2f’(1)
f’’’(x) = 6
f’(1) = 3 + 2 f’(1) + f’’(2)
=> f’(1) = -3 – f’’(2)
f’’(2) = 12 + 2f’(1)
= 12 + 2(-3 – f’’(2))
= 12 – 6 – 2f’’(2)
3f’’(2) = 6
=> f’’(2) = 2
So f’(1) = -3 – 2 = -5
f’’’(3) = 6
f’(1) + f’’(2) + f’’’(3) = -5 + 2 + 6
= -5 + 8
= 3
Hence the value of f’(1) + f’’(2) + f’’’(3) is 3.
Question 13: From a given solid cone of height H, another inverted cone is carved such that its volume is maximum. If the height of the inscribed cone is H/k, find k.
Solution:
Let h be the height of the inverted cone and r be its radius. Let R be the radius of the given cone of height H.
r/R = (H-h)/H
=> r = R(H-h)/H
Volume of inverted cone = ⅓ πr2h
= ⅓ π(R(H-h)/H)2h
= ⅓ π(R2(H-h)2h/H2)
= ⅓ ×(πR2/H2)×(H-h)2h
= P(H-h)2h, where P = ⅓ ×(πR2/H2)
dV/dh =P[(H-h)2 – 2(H-h)h]
dV/dh = 0 => [(H-h)2 – 2(H-h)h] = 0
=> (H-h)[ (H-h) – 2h] = 0 (H is not equal to h)
=> (H-h) – 2h = 0
=> H – 3h = 0
=> H = 3h
=> h = H/3
=> k = 3
Hence the value of k is 3.
Question 14: The maximum volume in (cu.m) of the right circular cone having slant height 3 m is
(a) 6Ï€
(b) 4Ï€/3
(c) 2√3 π
(d) 3√3 π
Solution:
Let r be the radius, h be the height of the right circular cone.
Given slant height = 3 m
So h2 + r2 = 32 …(i)
=> r2 = 9-h2
Volume of cone, V = ⅓ πr2h
= ⅓ π (9-h2)h
dV/dh = â…“ Ï€ [9 – 3h2]
d2V/dh2 = -2Ï€h
dV/dh = 0 => â…“ Ï€ [9 – 3h2] = 0
=> 9 – 3h2 = 0
3h2 = 9
h2 = 3
=> h = ±√3
d2V/dh2 < 0, when h = √3
When h = √3, r2 = (9 – 3) = 6 (from (i))
Volume = ⅓ πr2h
= ⅓ π×6×√3
= 2π√3
Hence option c is the answer.
Question 15: If ab = 2a + 3b, a > 0, b > 0 then the minimum value of ab is
(a) 12
(b) 24
(c) 1/4
(d) 2/3
Solution:
Given ab = 2a + 3b
=> ab – 3b = 2a
=> b(a-3) = 2a
=> b = 2a/(a-3)
Let k = ab = 2a2/(a-3)
dk/da = [(a – 3) 4a – 2a2] / (a – 3)2
= (4a2 – 12a – 2a2)/(a-3)2
= (2a2 – 12a)/(a-3)2
= 2 [a2 – 6a] / (a – 3)2
dk/da = 0 => a2 – 6a = 0
=> a = 6
d2k/da2 = 2a – 6 = 12-6 = 6 > 0
So b = 2a/(a-3) = 12/3 = 4
The minimum value of ab = 6×4 = 24
Hence option b is the answer.
Also Read
JEE Main Maths Applications of Derivatives Previous Year Questions With Solutions
Recommended Videos
Application of Derivatives JEE Solved Questions Part – 1
Application of Derivatives JEE Solved Questions Part – 2
Comments