Monday, 24 October 2016

Elitmus Previous Year Question on Functions

Elitmus Previous Year Question on 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) = 2x- 4x + 3(b)  f(x)=4x4- 3x+5(c)  f(x) = x- 2x – 6(d) None of these
03.Find the maximum value of the functions 1/(x- 3x + 2) ?
(a) 11/4(b) 1/4(c) 0(d) None of these
04.Find the minimum value off function f(x)= log(x- 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 + x, 6 – 3x), when x > 0 is  
(a) 1(b) 2(c) 3(d) 4
Answers : Find Detailed Solutions at the end of the the page.
1. D2. D
3. D4. B
5. D6. D
7. D8. B
9. D10. B    
11. C
1. Take different values of n to check each option. you can take n=3, 4 etc. option (a), (b), (c) will be rules out.
(d)
2. Differentiate the function with respect to x and equate it to 0 for getting the maximum and minimum value of the functions.
Step 1 : Differentiate with respect to x.
Step 2: Equate to 0
Step 3: Find the value of x
In option (a), (b), (c) none of the  three options will get you a value of x = -3  as its solution.
(d)
3. The denominator x-3x + 2 has real roots. Hence the maximum value of the function f(x) will be infinity.
(d)
4. The minimum value of the function would occur at the minimum value of (x- 2x + 5) as this quadratic equation function has imaginary roots.
y= x- 2x + 5
Step 1  : Differentiate with respect to x
Step 2 : Equate to 0
Step 3 : Find the value of x
dy/dx=2x-2 =0 implies x=1
Hence f(1)= 1- 2 + 5= 4
Thus minimum value of the argument of the log is 4.
So minimum value of the function is log 4 (base 2) =2
(b)
5. Given function = f(1) + f(2) + f(3) + f(4)+ ……. =nf(n)
Given f(1)= 3600
For n=2,
f(1) + f(2) =22f(2)
i.e. 22f(2)-f(2)=f(1)
f(2)= f(1)/(22-1)  —(1)
For n=3
f(1)+f(2)+f(3)=32f(3)
put the value of f(2) from (1)
» f(1)+ f(1)/(22-1)=32f(3)-f(3)
» f(1)+ f(1)/(22-1)=(32-1)f(3)
now take f(1) in left side
i.e.f(1)=[ 1+ 1/(2^2-1)]=f(3)(3^2-1)
i.e.f(3)=f(1) x 2^2/(2^2-1) x 1/(3^2-1)
f(3)=600
Similarly
f(9)=f(1)x (2^2 x3^2 x4^2………..8^2)/((2^2-1)(3^2-1)(4^2-1)…………………….(9^2-1))
f(9)=80
(d)
6. As f(x)= max( 2x + 1, 3 – 4x )
A function would b minimum at the point of intersection of these curves.
i.e. 2x + 1 = 3 – 4x
x=1/3
Hence, minimum value of f(x) is 5/3
(d)
7. g(x+1) + g(x-1)=g(x)
g(x+2)+g(x)=g(x+1)
Adding these two equation, we get
g(x+2)+g(x-1)=0
» g(x + 3) + g(x)=0 —- (1)
» g(x + 4) + g(x + 1)=0
» g(x + 5) + g(x + 2)=0
» g(x + 6) + g(x + 3)=0
» g(x+6)-g(x)=0  (From (1))
(d)
8. f(x)=x3- 4x + p
f(0)=p
f(1)=p-3
Given f(0) and f(1) are of opposite signs
therefore p(p – 3)<0
If p < 0 then p – 3 is also less than 0.
Hence, p(p – 3)>0 i.e. p cannot be negative
option (a), (b), (d) are eliminated.
0 < p < 3
(b)
9. g(x)= max ( 5 – x, x + 2)
We have to draw graph and the find the point of intersection.
y= 5 – x
y= x + 2
Hence at the point of intersection of two straight line.
Smallest of g(x)=3.5
(d)
10.
f(x)= |x – 2| + |2.5 – x| + |3.6 – x| can attain minimum value when either of the terms =0.
Case 1 :
When |x – 2|=0 i.e. x=2
f(x)= 0.5+1.6=2.1
Case 2 :
When |2.5 – x|=0 i.e.  x=2.5
f(x)= 0.5+0+1.1=1.6
Case 3 :
When |3.6 – x|=0  i.e. x=3.6
f(x)= 1.6 + 1.1 + 0 = 2.7
Hence, the minimum value of f(x) is 1.6 at x=2.5.
(b)
11. Equating  2 + x= 6 – 3x
» x+ 3x – 4 = 0
» x+ 4x – x – 4 = 0
»(x + 4)(x – 1) = 0
» x= -4 or x=1
But x > 0 so x = 1 so LHS = RHS  i.e. 2 + 1 = 3
Hence, largest value of function min(2 + x2, 6 – 3x) is 3.
(c)


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