#### ElitmusZone

ElitmusZone » Functions

# Functions

 01. If  f(x)=| x – 2 |,  then which of the following is always true ? (a) f(x) = (f(x))2 (b) f(x) = f(-x) (c) f(x) = x – 2 (d) None of these
 02. Which of the following functions will have a minimum value at x = -3 ? (a)f(x) = 2x3 – 4x + 3 (b)  f(x)=4x4– 3x+5 (c)  f(x) = x6 – 2x – 6 (d) None of these
 03. Find the maximum value of the functions 1/(x2 – 3x + 2) ? (a) 11/4 (b) 1/4 (c) 0 (d) None of these
 04. Find the minimum value off function f(x)= log(x2 – 2x + 5) (base 2) ? (a) -4 (b) 2 (c) 4 (d) -2
 05. A function f(x) satisfies f(1)=3600  and f(1) + f(2) +……f(n) =n2f(n), for all positive integers n>1. What is the value of f(9) ? (a) 200 (b) 100 (c) 120 (d) 80
 06. Let f(x)= max( 2x + 1, 3 – 4x), where x is any real number. Then, the minimum possible value of f(x) is (a) 4/3 (b) 1/2 (c) 2/3 (d) 5/3
 07. Let g(x) be a function such that g(x + 1) + g(x – 1) = g(x) for every real x. Then, for what value of p is the relation g(x + p)= g(x) necessarily true for every real x ? (a) 5 (b) 3 (c) 2 (d) 6
 08. If f(x)=x3– 4x + p  and f(0) and f(1) are of opposite signs, then which of the following is necessarily true ? (a) -1 < p < 2 (b) 0 < p < 3 (c)  -2 < p < 1 (d) -3 < p < 0
 09. Let g(x) =  max ( 5 – x , x + 2 ). The smallest possible value of g(x) is ? (a) 4.0 (b) 4.5 (c) 1.5 (d) None of these
 10. Let f(x)= |x – 2| + |2.5 – x| + |3.6 – x|, where x is a real number, attains a minimum at ? (a) x = 2.3 (b) x = 2.5 (c) x = 2.7 (d) None of these
 11. Largest value of min ( 2 + x2 , 6 – 3x), when x > 0 is (a) 1 (b) 2 (c) 3 (d) 4

 1 D 2. D 3 D 4. B 5 D 6. D 7 D 8. B 9 D 10. B 11 C

Detailed Solution

1. ankit jain

Contents fulfil the need of Elitmus’s exam……….Thanks for providing it…….great stufff

2. sneha

Can anyone explain second question please …..

3. krishnkantnayak

please give some more question on this topic for exam preparation ….

4. Rajat Tripathi

f(1) = 3600
3600 + f(2)= 4f(2)–>f(2)= 1200
4800 + f(3)= 9f(3)–>f(3)= 600
5400 + f(4)= 16f(4)–>f(4)= 360
5760 + f(5)= 25f(5)–>f(5)= 240
6000 + f(6)= 36f(6)–>f(6)= 171.428
6171.428 + f(7)= 49f(7)–> f(7)=128.571
6300 + f(8)=64f(8)–>f(8)= 100
6400 + f(9)= 81f(9)–>f(9)= 80

5. Mahendra Kumar

Q6 anyone??

6. Jasvinder Singh

answer for Q.3 should be d.
to get max. value denominator should be min. and min. value can be calculated by putting x=-b/2a

1. shehzad

u r absolutely right. the answer should be D