JEE Advanced Level Test: Maxima and Minima - JEE MCQ

# JEE Advanced Level Test: Maxima and Minima - JEE MCQ

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## 30 Questions MCQ Test Mathematics (Maths) Class 12 - JEE Advanced Level Test: Maxima and Minima

JEE Advanced Level Test: Maxima and Minima for JEE 2024 is part of Mathematics (Maths) Class 12 preparation. The JEE Advanced Level Test: Maxima and Minima questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Level Test: Maxima and Minima MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Level Test: Maxima and Minima below.
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JEE Advanced Level Test: Maxima and Minima - Question 1

### f(x) = 1 + 2x2 + 4x4 + 6x6 + ................+ 100x100 is polynomial in a real variable x, then f(x) has

Detailed Solution for JEE Advanced Level Test: Maxima and Minima - Question 1

f(x) = 1 + 2x2 + 4x4 +...............+ 100x100
= f’(x) = 4x + (4)2 x3 + (6)2 x5 +......................+ (100)2 x99
f’(x) = x[(2)2 + (4)2 x2 + (6)2 x4 +....................+ (100)2 x98
f’(x) = 0(only one minimum)

JEE Advanced Level Test: Maxima and Minima - Question 2

### On the interval [0, 1] the function x25(1 – x)75 takes its maximum value at

Detailed Solution for JEE Advanced Level Test: Maxima and Minima - Question 2

Let f(x)=x25(1−x)75,xϵ[0,1]
⇒f′(x)=25x24(1−x)75−75x25(1−x)74
=25x24(1−x)74{(1−x)−3x}
=25x24(1−x)74(1−4x)
We can see that f′(x) is positive for x<1/4 and f'(x) is negative for x>1/4
Hence, f(x) attains maximum at x= 1/4

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JEE Advanced Level Test: Maxima and Minima - Question 3

### The product of minimum value of xx and maximum value of  is

Detailed Solution for JEE Advanced Level Test: Maxima and Minima - Question 3

f(x) = xx     f(x) = x(-x)
f’(x) = xx(1+ln x)          f’(x) = -x(-x)(1+ln x)
1+ln x = 0                   x = 1/e
x = 1/e
Min value = (1/e)(1/e)        f(1/e) = e(1/e)
Product = (e-1/e)(e)1/e = 1

JEE Advanced Level Test: Maxima and Minima - Question 4

The minimum value of the function defined by f(x) = max (x, x + 1, 2 – x) is

Detailed Solution for JEE Advanced Level Test: Maxima and Minima - Question 4

f(x) = max(x, x+1, 2-x)
y = 2-x
y = x+1
Comparing both the equations, we get
2 - x = x + 1
x = ½
y = 2 - x
y = 2 - (½)
y = 1/3

JEE Advanced Level Test: Maxima and Minima - Question 5

Let f(x) =  then

Detailed Solution for JEE Advanced Level Test: Maxima and Minima - Question 5

f(x) has local maxima at x = 1

JEE Advanced Level Test: Maxima and Minima - Question 6

The greatest and the least values of the function, f(x) = 2 – , x ∈ [–2, 1] are

JEE Advanced Level Test: Maxima and Minima - Question 7

Let f(x) = {x}, For f(x), x = 5 is (where {*} denotes the fractional part)

JEE Advanced Level Test: Maxima and Minima - Question 8

The critical points of f(x) =  lies at

JEE Advanced Level Test: Maxima and Minima - Question 9

The difference between the greatest and least values of the function f(x) = sin 2x – x on [–p/2, p/2] is

JEE Advanced Level Test: Maxima and Minima - Question 10

The radius of a right circular cylinder of greatest curved surface which can be inscribed in a given right circular cone is

Detailed Solution for JEE Advanced Level Test: Maxima and Minima - Question 10

OE = x, OC = r, AO = h
∠AOC ≅ ∠QEC
AO / OC = QE/EC
(QE) = AO/OC(EC)
=h / r (r − x)
δx = 2πx (h / r(r − x))
=2 πhr (rx − x2)
δ'(x) = 0
x = r/2
derivating again
`delta''(r/2) = (2pih) / r(-2) < 0
r/2 is at maxima

JEE Advanced Level Test: Maxima and Minima - Question 11

The dimensions of the rectangle of maximum area that can be inscribed in the ellipse (x/4)2 + (y/3)2 = 1 are

JEE Advanced Level Test: Maxima and Minima - Question 12

The largest area of a rectangle which has one side on the x–axis and the two vertices on the curve y =  is

JEE Advanced Level Test: Maxima and Minima - Question 13

The co–ordinates of the point on the curve x2 = 4y, which is at least distance from the line

y = x – 4 is

JEE Advanced Level Test: Maxima and Minima - Question 14

f(x) =  then

JEE Advanced Level Test: Maxima and Minima - Question 15

Let f(x) =  the set of values of b for which f(x) has greatest value at x = 1 is given by

JEE Advanced Level Test: Maxima and Minima - Question 16

The set of values of p for which the extrema of the function, f(x) = x3 – 3px2 + 3(p2 – 1) x + 1 lie in the interval (–2, 4) is

Detailed Solution for JEE Advanced Level Test: Maxima and Minima - Question 16

f(x)=x3−3px2+3(p2−1)x+1
f′(x)=3{x2−2px+(p−1)(p+1)}
f′(x)=3{x−(p−1)}{x−(p+1)}
Using given condition,
−2<p−1<4 and −2<p+1<4
⇒−1<p<5 and −3<p<3
∴p∈(−1,3)

JEE Advanced Level Test: Maxima and Minima - Question 17

Four points A, B, C, D lie in that order on the parabola y = ax2 + bx + c. The co–ordinates of A, B & D are known as A(–2, 3); B(–1, 1) and D(2, 7). The co–ordinates of C for which the area of the quadrilateral ABCD is greatest is

JEE Advanced Level Test: Maxima and Minima - Question 18

In a regular triangular prism the distance from the centre of one base to one of the vertices of the other base is l. The altitude of the prism for which the volume is greatest is

JEE Advanced Level Test: Maxima and Minima - Question 19

Two vertices of a rectangle are on the positive x–axis. The other two vertices lie on the lines y = 4x and y = –5x + 6. Then the maximum area of the rectangle is

Detailed Solution for JEE Advanced Level Test: Maxima and Minima - Question 19

If the base of the rectangle is the segment (a,0) to (b,0) then the upper vertices are (a,4a) and (b,6-5b).
Since it is a rectangle, the height is the same, so 4a = 6-5b.
That is, b = (6-4a)/5
Now, the area is (b-a)(4a) = ((6-4a)/5-a)(4a) = 12a/5 (2-3a)
This is a parabola with vertex at (1/3, 4/5).
check: So, b=14/15 and the maximum area is
(14/15 - 1/3)(4/3) = (9/15)(4/3) = 4/5

JEE Advanced Level Test: Maxima and Minima - Question 20

A variable point P is chosen on the straight line x + y = 4 and tangents PA and PB are drawn from it to circle x2 + y2 = 1. Then the position of P for the smallest length of chord of contact AB is

JEE Advanced Level Test: Maxima and Minima - Question 21

The maximum area of the rectangle whose sides pass through the angular points of a given rectangle of sides a and b is

Detailed Solution for JEE Advanced Level Test: Maxima and Minima - Question 21

A = Area EFGH = (bsinθ + acosθ) (acosθ + bcosθ)
= ab(a2 + b2)sinθcosθ
d2A/dθ2 = 1-(2)(a2 + b2)sin2θ,
so d2A/dθ2|<0    {θ = pi/4}
Hence A(max) = ½(a+b)2

JEE Advanced Level Test: Maxima and Minima - Question 22

If p and q are positive real numbers such that p2 + q2 = 1, then the maximum value of (p + q) is

JEE Advanced Level Test: Maxima and Minima - Question 23

The function f(x) =  has a local minimum at

Detailed Solution for JEE Advanced Level Test: Maxima and Minima - Question 23

f(x)=x/2 + 2/x
f`(x)=1/2-2/x²
For extremum f ′(x)=0
>0 for x=2
⇒ 1/2-2/x²=0
⇒ x=±2
f′′(x)=4/x³>0 for x=2
∴ x=2 is the point of minima
∴ Given function has local minima at x=2.

JEE Advanced Level Test: Maxima and Minima - Question 24

If x is real, the maximum value of  is

JEE Advanced Level Test: Maxima and Minima - Question 25

If the function f(x) = 2x3 – 9ax2 + 12a2x + 1, where a > 0, attains its maximum and minimum at p and q respectively such that p2 = q, then a equals

JEE Advanced Level Test: Maxima and Minima - Question 26

The maximum value x3 – 3x in the interval [0, 2] is

JEE Advanced Level Test: Maxima and Minima - Question 27

Minimum value of    is

JEE Advanced Level Test: Maxima and Minima - Question 28

The minimum value of (x – p)2 + (x – q)2 + (x – r)2 will be at x equals to

JEE Advanced Level Test: Maxima and Minima - Question 29

The number of values of x where f(x) = cos x + cos  x attains its maximum value is

Detailed Solution for JEE Advanced Level Test: Maxima and Minima - Question 29

f′(x) =−sinx− (2)1/2sin((2)1/2x)
Hence x = 0
Hence maximum value will be attained at x=0
f(0)=2
The maximum value of f(x) = cos x + cos (√2x) is 2 which occurs at x = 0. Also, there is no other value of x for which this value will be attained again.

JEE Advanced Level Test: Maxima and Minima - Question 30

The maximum value of cos a1 . cos a2 . cos a3....cos an under the restriction 0 £ a1, a2,.....an £and cot a1 cot a.... cot an=1 is

Detailed Solution for JEE Advanced Level Test: Maxima and Minima - Question 30

Given condition cotα1.cotα2.............cotαn=1
⇒cosα1.cosα2.......cosαn
​=sinα1.sinα2.........sinαn
​This is possible only when  α1=α2=α3=....=αn
​=45o
⇒cosα1.cosα2.......cosαn = 1/(2)1/2 . 1/(2)1/2 ......... 1/(2)1/2 =1/ (2)n/2

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