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 |
Answers :
1. | D | 2. | D |
3. | D | 4. | B |
5. | D | 6. | D |
7. | D | 8. | B |
9. | D | 10. | B |
11. | C |
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Contents fulfil the need of Elitmus’s exam……….Thanks for providing it…….great stufff
Can anyone explain second question please …..
Hello Sneha,
Please check the below article. It might help you in understanding the maxima and minima of the quadratic equations.
https://www.mathsisfun.com/calculus/maxima-minima.html
Best Regards,
ElitmusZone
please give some more question on this topic for exam preparation ….
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
Q6 anyone??
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
u r absolutely right. the answer should be D