JEE Advanced Level Test: Relations and Functions- 4 - JEE MCQ

# JEE Advanced Level Test: Relations and Functions- 4 - JEE MCQ

Test Description

## 30 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - JEE Advanced Level Test: Relations and Functions- 4

JEE Advanced Level Test: Relations and Functions- 4 for JEE 2024 is part of Chapter-wise Tests for JEE Main & Advanced preparation. The JEE Advanced Level Test: Relations and Functions- 4 questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Level Test: Relations and Functions- 4 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Level Test: Relations and Functions- 4 below.
JEE Advanced Level Test: Relations and Functions- 4 - Question 1

### The critical point of f (x) = |x- 1| is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 1

f (x) = |x- 1| is not diff at x = 1 critical point is at x = 1

JEE Advanced Level Test: Relations and Functions- 4 - Question 2

### IF f (x) = a log x + bx2 + x has extreme values at x = -1, x = 2 then a = ...., b = .....

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 2

 1 Crore+ students have signed up on EduRev. Have you?
JEE Advanced Level Test: Relations and Functions- 4 - Question 3

### If  x is real then the minimum value of

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 3

Min. value = f(1)
Max. Value = f(-1)

JEE Advanced Level Test: Relations and Functions- 4 - Question 4

The value of “a” for which the sum of the squares of the roots of the equation x2 - (a - 2) x - a - 1 = 0 assume the least value is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 4

JEE Advanced Level Test: Relations and Functions- 4 - Question 5

Greatest value of (1/x)x is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 5

Let y = (1/x)x
Taking ln both sides, we get ln y =x ln 1/x = x ln (x)-1
= -x ln x {Since,ln xa = a ln x  }
Now, differentiating both sides with respect to x we get,
1/y(dy/dx) = -1-ln x
⇒  (dy/dx) =(1/x)x [-1-ln x} = 0 {Condition for y to be maximum}

⇒ln x = -1  {Since, (1/x)x = 0}

⇒ x = e-1  {Converting logarithmic equation to exponential equation}

⇒ So,  y(max) = (e)1/e

JEE Advanced Level Test: Relations and Functions- 4 - Question 6

A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x unit and circle of radius = r units. If the sum of the area of the square and the circle so found is minimum then

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 6

Given 4x + 2πr = 2

For minimum value of area A

x = 2r

JEE Advanced Level Test: Relations and Functions- 4 - Question 7

The function f (x) = 4x5 - 25x4 + 40x3 - 10 has

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 7

f '(x) = 20x4 -100x3 + 120x2 f '(x) = 0
x = 2, 3 one maximum and one minimum

JEE Advanced Level Test: Relations and Functions- 4 - Question 8

The function x4 - 62x2 + ax + 9 attains its maximum value at x = 1, on the interval [0, 2] then value of ‘a’ is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 8

f (x) = x4 - 62x2 + ax + 9
f '(x) = 4x3 -124x + a
f '(1) = 0 at x =1 4 -124 + a = 0
x = 120

JEE Advanced Level Test: Relations and Functions- 4 - Question 9

The longest distance of the point (a, 0) from the curve 2x2 + y2- 2x is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 9

f (x) = (PA)2 = (x - a)2 +2x - 2x2
f '(x) = 0x = (1-a)

JEE Advanced Level Test: Relations and Functions- 4 - Question 10

The point on the curve y = x2 which is nearest to (3, 0) is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 10

Let Z=(x-3)2+y= x2-6x+9+x4
Z is minimum dz/dz = 2x-6+4x3 = 0
∴ x = 1 ∴ y = 1, (1,1)

JEE Advanced Level Test: Relations and Functions- 4 - Question 11

The difference between the greatest and least value of the function f (x) = sin 2x - x on

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 11

JEE Advanced Level Test: Relations and Functions- 4 - Question 12

The sides of a rectangle of the greatest area which can be inscribed into an ellipse

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 12

JEE Advanced Level Test: Relations and Functions- 4 - Question 13

A window is in the shape of a rectangle surmounted by a semi circle. If the perimeter of the window is of fixed length „l‟ then the maximum area of the window is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 13

JEE Advanced Level Test: Relations and Functions- 4 - Question 14

The sum of the hypotenuse and a side of a right triangle is constant. If the area of the triangle is maximum then the angle between the hypotenuse and the given side is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 14

z+x=k  z+zsinθ = k   z = k/1+sinθ

JEE Advanced Level Test: Relations and Functions- 4 - Question 15

The number of linear functions which map from [-1,1] onto [0, 2] is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 15

Let f (x) = ax+ b be the required liner function.
If a > 0 then f is increa sin g, f(-1) = 0,f(1) = 2 ⇒ -a + b = 0,a +b = 2 ⇒ a = -1,b = 1
If a < 0 then f decrea sin g, f(-1) = 2,f (1) ⇒ = 0 -a + b = 2,a+b = 0 = ⇒ a = -1,b = 1
∴ The required linear function are f(x) = x+1,f(x) = -x+1.

JEE Advanced Level Test: Relations and Functions- 4 - Question 16

for x ∈ R Then f (2002) =

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 16

(cos2x + sin4x) - (sin2x + cos4x) = (cos2x - sin2x) - (cos4x - sin4x)
= cos2x - cos2x = 0

JEE Advanced Level Test: Relations and Functions- 4 - Question 17

If f (0) = 0, f (1) = 1, f (2) = 2 and f (x) = f (x - 2) + f (x - 3) for x = 3, 4, 5,&.., then f(9) =

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 17

Givenf (x) = f(x-2)+f(x-3)
∴ f(3) = f(1)+f(0) = 1 + 0 = 1,
f(4) = f(2)+f(1) = 2 + 1 = 3
f(5) = f(3)+f(2) = 1+ 2 = 3,f(6). f (4) + f(3) = 3+1 = 4,
f(7) = f(5)+f(4) = 3+3 = 6,f(8) = f(6)+f(5) = 4+3 =7,
f(9) = f(7)+f(6) = 6+4 = 10.

JEE Advanced Level Test: Relations and Functions- 4 - Question 18

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 18

JEE Advanced Level Test: Relations and Functions- 4 - Question 19

If f (x) = cos [π2] x + cos [- π2] x where [x] is the step function, then

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 19

cos 9x + cos (-10) x = cos 9x + cos10x

f (0) = cos0 + cos0 = 2,f (π/4) = cos9 π/4 + cos5π/2 = cos π/ 4 =
f (π/2) = cos9π/2 + cos5π = 0 -1 = -l,f (π) = cos9π + cos10π = - l +1 = 0.

JEE Advanced Level Test: Relations and Functions- 4 - Question 20

If f (x) is a polynomial fuction such that f (x) f (1/x) = f (x)+ f (1/x) and f (2) = 33 then f (x) =

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 20

f (x) f (1/x) = f (x) + f (1/x) ⇒ f(x) = xn +1.
f (2) = 33 ⇒ 2n +1= 33 ⇒ 2n = 32 = 25 ⇒ n = 5 ⇒ f (x) = x5 +1.

JEE Advanced Level Test: Relations and Functions- 4 - Question 21

If f (x) is a function such that f (x + y) = f (x) f (y) and f (3) = 125 then f (x) =

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 21

f(x+y) + f(x) + f(y) ⇒ f(x) = ax.f(3)
= 125 ⇒ a3 = 125 ⇒ a = 5.
∴ f (x) = 5x

JEE Advanced Level Test: Relations and Functions- 4 - Question 22

If f : [2, ∞) → B defined by f (x) = x2 - 4x + 5is a bijection, then B =

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 22

f (x) = x2 -4x +5 = (x -2)2 +1
x ∈ [2,∞) ⇒ x > 2 ⇒ x - 2 > 0 ⇒ (x - 2)2 +1 > 1 ⇒ B = [1,∞).

JEE Advanced Level Test: Relations and Functions- 4 - Question 23

The function f : C → C defined by f (x)  for x ∈ C where bd ≠ 0 reduces to a constant function if

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 23

JEE Advanced Level Test: Relations and Functions- 4 - Question 24

A function f : N → Z defined by f(n) = (n - 1)/2 when n is odd and f(n) = -n/2 when n is even, is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 24

Explanation : f(n-1)/2 ; n is odd        -n/2 ; n is even

f(1) = 0                f(3) = 1

f(2) = -1               f(4) = -2

f(5) = 2                f(6) = -3

f(n) + f(n+1) = -1    n is odd

if = 0     n is even

Range of f(n) = Z

f(n) is one-one, therefore f(n1) = f(n2)

(n1-1)/2 = -n2/2

=> 1 < (n1 + n2)/2 is not equal to 1/2

Therefore f(n) is onto.

JEE Advanced Level Test: Relations and Functions- 4 - Question 25

A = {x : |x| < 1|} and f : A → A such that f (x) = x |x|, then

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 25

Graph of y = f (x) is given in the figure.
∴ f is bijective.

JEE Advanced Level Test: Relations and Functions- 4 - Question 26

The domain of the function f = {(1, 3), (3, 5), (2, 6)} is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 26

Domain = {1, 3, 2}

JEE Advanced Level Test: Relations and Functions- 4 - Question 27

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 27

∴ f is an even function.

JEE Advanced Level Test: Relations and Functions- 4 - Question 28

The range of f(x)

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 28

If x > 0 then f x = 0. If x < 0 then f (x) = tanh x. If f (x) = -1then
ex - e|x| = - ex - e|x| ⇒ex = 0 ⇒ x ∉ R.
Range does not contain -1.
∴ Range = (-1, 0].

JEE Advanced Level Test: Relations and Functions- 4 - Question 29

The domain of log (x - 3) (5- x) is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 29

log (x - 3) (5 - x) exists ⇒ (x - 3) (5 - x) > 0
⇒ (x -3)(x -5) < 0 ⇒ 3< x < 5
∴ Domain = (3, 5).

JEE Advanced Level Test: Relations and Functions- 4 - Question 30

The domain of log10 (x3 - x) is

Detailed Solution for JEE Advanced Level Test: Relations and Functions- 4 - Question 30

log10 (x3 - x)is defmed ⇒ x3 - x > 0 ⇒ x(x2 - 1) > 0 ⇒ (x +1)x(x - 1) > 0
x < -1 ⇒ (x+1)x(x-1) < 0; -1 < x < 0 ⇒ (x+1)x(x-1) > 0
0 < x < 1 ⇒ (x+1)x(x-1) < 0; x > 1 ⇒ (x+1)x(x-1) > 0.
∴ Domain = (-1,0)∪(1, ∞)

## Chapter-wise Tests for JEE Main & Advanced

447 docs|930 tests