FUNCTIONS SOLVED QUESTIONS

FUNCTIONS QUESTIONS

  1. If f(x) = 1/x, g(x) = 1/(1 – x) and h(x) = x² , then find f(g(h(2))).

      A. -1

      B. 1

      C. 1/2

      D. none

Solution : we first resolve the function h(2). Since

 h(2) = 4 we will get

 f(g(h(2))) = f(g(4)) = f(–1/3) = –3. Hence, the option (d)

 

   2. Find the domain of the definition of the function y = (x – 2)^1/2+ (8 – x)^1/2

       All the real values except 2 <= x <= 8

A.         A.  2 <= x

B.         B. 2 <= x <= 8

C.         C.  x <= 8

Solution :  For (x – 2)^1/2

to be defined x ≥ 2.

For (8 – x)^1/2

to be defined x <= 8.

Thus, 2 <= x <= 8

   3. Find the minimum value of the function f(x) = log₂(x² – 2x + 5)

      A. -4

      B. 2

      C. 4

      D. -2

   Solution : The minimum value of the function would occur at the minimum value of (x² –   2x + 5) as this

quadratic function has imaginary roots.

For y = x² – 2x + 5

dy/dx = 2x – 2 = 0

fi x = 1

fi x² – 2x + 5 = 4.

Thus, minimum value of the argument of the log is 4.

So minimum value of the function is log₂ 4 = 2

 

  4. IF f(x,y)= x+y+x³y+y²x ; then find the value of f(1,2)

Solution : x,y are independent

=> f(1,2) = 1+2+2+4 = 9

 Directions for questions 5 to 10: Read the instructions and solve the questions accordingly

f(x, y) = x² – 2y

g(x, y) = x + y³

h(x, y) = x² + y – 2xy

k(x, y) = x + y

  5. Find f(2,3) – h(1, 5)

        A. 3

        B. 4

        C. 15

        D. 34

 Solution : Option B

 f(2,3) = 2² – 2(3) = 4 – 6 = -2

 h(1,5) = 1² + 5 – 2(1)(5) = -4

  Now, f(2,3) – h(1, 5) = -2 – (-4) = -2 + 6 = 4

6.  Find h(3, 4) – g(5,2) + k(13, 4)

     A. -7

     B. -11

     C. -13

     D. -17

Solution : Option A

h(3,4) = 3² + 4 – 2(3)(4) = 9 + 4 – 24 = -11

g(5,2) = 5 + 2³ = 5 + 8 = 13

k(13, 4) = 13 + 4 = 17

Now, h(3, 4) – g(5,2) + k(13, 4) = -11 – 13 + 17 = -7

7. Find k[g(2,3) , k(11, 9)])

       A. 56

       B. 35

       C. 67

       D. 49

Solution: Option D

g(2,3) = 2 + 3³ = 2 + 27 = 29

k(11, 9) = 11 + 9 = 20

Now, k[g(2,3) , k(11, 9)] = k[29 + 20] = 49

8. Find h(3,4) – g(2,3)

       A. 58

       B. 54

       C. -40

       D. -23

Solution : Option C

H(3,4) = 3² + 4 – 2(3)(4) = 9 + 4 -24 = -11

G(2,3) = 2 + 3³ = 2 + 27 = 29

Now, h(3,4) – g(2,3) = -11 – 29 = -40

9. Find the value of h(1,3) – h (3,1)

       A. -6

       B. -12

       C. -17

       D. -3

Solution : Option A

 H(1,3) = 1² + 3 – 2(1)(3) = 1 + 3 – 6 = -2

H(3,1) = 3² + 1 – 2(3)(1) = 10 – 6 = 4

Therefore, -2 -4 = -6

 10. Given f(x) = x3 + 1,

    g(x) = 2x – 5

    h(x) = [f(x)]² – g(x)

    Find h(-2)

      A. 34

      B. -23

      C. -12

      D. 58

Solution: Option D

h(-2) = [f(-2)]² – g(-2)

    = [(-2)³ + 1]² – [2(-2) -5] [-8 + 1]² – [-4-5]

    = [-7]² – [-9]

    = 49 + 9 = 58

11. The range of the function f(x)= x² - 4x + 9 is:

     A.  [5 , 9]

     B.  [-5 , +infinity)

     C.  [5 , + infinity)

     D.  None of these

Solution: correct answer is C

h(x) = x²-4x+9 = x²-4x+4-4+9 = (x - 2)² + 5

(x - 2)² >= 0

(x - 2)²+5 >= 5

Hence minimum value h(x) can have is 5 and maximum can go upto + infinity. So the range is [5 , + infinity)

 12.  If 2*f(x) - f(1/x) = x² , then f(x) is?

     A. f(x) = 3{2*x² + (1/x)² }
     B. 2 f(x) = {2*x²+(1/x)²}
     C. f(x) = {2*x² +(1/x)² }
     D. 3 f(x) = {2*x+2 +(1/x)² }
     E. None of these
Solution : 2*f(x) - f(1/x) = x²  ..............(1)
Take x = (1/x),
2*(1/x) - f(x) = (1/x)²  ...............(2)
Add: 2 (equation 1) + equation 2
3*f(x) = 2*x²+ (1/x)²
f(x) = {2*x² + (1/x)²}/3

  13.  A function f(x) is said to reflect on itself if, y = f(x) and x = f(y). Which of the       following function reflects on itself?

    A. f(x) = x
    B. f(x) = (2x + 1) / (x - 2)
    C. f(x) = (3x + 2) / (2x - 3)
    D. f(x) = (4x + 3) / (3x - 4)
    E. All of these
Solution :

(a) f(x) = x = f(y)
y = x = f(y) Hence (a) is true.
(b) f(x) = 2x+1/x−2= y
xy - 2y = 2x + 1
xy - 2x = 1 + 2y
x(y - 2) = 1 + 2y
x = 1+2y/y−2 Hence (b) is also true.
Similarly if we check for options (c) & (d), they will also hold true, hence t he answer is option (e).

  14.  If f(x) = 2x + 3, & g(x) = (x - 3)/2, then what is the value of,     fo(fo(go(go(fo(fo(go(go................(fo(fo(go(gof(x) )))).....)))))) ?

     A. (x - 3) / 2
     B. x
     C. 2x - 3
     D. 2x + 3
     E. None of these

 Solution : f(x) = 2x + 3
  g(x) = (x - 3)/2
  gof(x) = 2x+3−3/2=x
  gogof(x) = x−32
  fogogof(x) = 2(x−32)+3 = 2x−6+6/2 = x
  fofogogof(x) = 2x + 3
 This means when we apply two times g(x) and two-time f(x) on f(x) we get f(x).  This pattern continues, Hence the answer is option (d).

  15.  f1(x) = 2x - 1 and fn(x)=f1(fn−1(x))  for n ≥≥ 2. Find f5(2)

     A. 30  

     B. 33 

     C. 31   

     D. None of these

 Solution:
  f1(2) = 2*2 - 1 = 3
  f2(2)=f1(f1(2))=f1(3)= 2*3 - 1 = 5
  f3(2)=f1(f2(2))=f1(5)= 2*5 - 1 = 9
  f4(2)=f1(f3(2))=f1(9)= 2*9 - 1 = 17
  f5(2)=f1(f4(2))=f1(17)= 2*17 - 1 = 33
 In fact, The general term of the above function is 2n2n+1

  16. Given that f(1) = 1 and f(2) = 1.  If f(n) = f(n+1) - f(n - 1), then find the value of {f(8)−f(7)+f(5)} / {f(7)−f(6)−f(4)}

 Solution:

  f(n) = f(n+1) - f(n - 1)
  ⇒⇒ f(n - 1) = f(n + 1) = f(1)
   f(8) - f(7) = f(6) and f(7) - f(6) = f(5)
  ⇒⇒f(n + 1) = f(n) + f(n - 1)
  ⇒⇒f(3) = f(2) + f(1) = 1 + 1 = 2
  ⇒⇒f(4) = f(3) + f(2) = 2 + 1 = 3
  ⇒⇒f(5) = f(4) + f(3) = 3 + 2 = 5
  ⇒⇒ f(6) = f(5) + f(4) = 5 + 3 = 8
  {f(8)−f(7)+f(5)} / {f(7)−f(6)−f(4)} = {f(6)+f(5)} / {f(5)−f(4)} = (8+5)/(5−3)= 13/2

  17.  Given g(x) is a function such that g(x+1) + g(x-1) = g(x).  For what minimum     value of P does the relation g(x+p) = -g(x) necessarily hold true?

        A.  2

        B. 3  

        C.  5 

        D. 6

 Solution : given g(x+1) + g(x-1) = g(x) ......(1)
  g(x + 2) + g(x) = g(x+1) .........(2)

  adding

  g(x+2) + g(x-1) = 0

  g(x+3) + g(x) = 0

  So g(x+3) = - g(x)

  P =3

Directions for questions 17 to 22: Read the instructions and solve the questions accordingly

1. x%y means x+ y

2. x&y means x – y

3. x$y means x² + y²

4. x@y means x²– y²

  18. Find 3%(4&5)

         A. 5

         B. 18

         C. 3

         D. 2

Solution : Option D
Explanation: : 3% (4-5) = 3% (-1) = 3+ (-1) = 2

     19. Find 2$(4@2)

               a) 132

               b) 148

               c) 149

               d) 134

Solution : Option B
Explanation:2$ (4² – 2²) = 2$ (16-4) = 2$ 12 = 2² + 12² = 4 + 144 = 148

20. Find the value of (9@3) % (3&9)

a) 98

b) 97

c) 72

d) 73

Solution : Option C
Explanation: (9² – 3²) % (3-9) = (81-3) % (-6) 78%(-6) = 78 + (-6) = 72

21. Find the value of (2&5) % (3@7)

a) -45

b) -35

c) -43

d) -38

Solution : Option C
(2-5) % (3²– 7²) = (-3) % (-40) = -3-40 = -43

22. Find the value of (4&3) + (3@2)

a) 6

b) 11

c) 13

d) 15

Solution : Option A
Explanation:(4-3) +(3² – 2²) = 1 + 5 = 6

Directions for questions 23 to 26: Read the instructions and solve the questions accordingly

x # y = x + y
x @ y = x – y
x & y = x²– y²
x $ y = (x + y)²

23. Find 3 # (4 $ 6)

a) 23

b) 256

c) 103

d) 109

Solution : Option C
Explanation:3 # (4+6)² = 3 # 100 = 3 + 100 = 103

24. Find [3 @ (7 # ( 5 & 3) ]

a) -20

b) -23

c) -34

d) -45

Solution : Option A
Explanation: [ 3 @ (7 # (5² – 3²)] = [ 3 @ (7 # 16)] [3 @ (7 + 16)] = [ 3 @ 23] = [ 3 – 23] = -20

25. Find -3 $ (6 & (4 # 3))

a) 361

b) 121

c) 144

d) 987

Solution : Option A
Explanation: -3 $ (6 & (4 + 3)) = -3 $ (6 & 7) -3 $ (6² – 7²) = -3 $ (36 – 49) = -3 $ -13 = (-3-16)² = (-19)² = 361

26. Find the 3 $ (4 & (5 & 6))

a) -102

b) -56

c) -34

d) -23

Solution : Option A
Explanation: 3 $ (4 & (5² – 6²)) = 3 $ (4 & (25 – 36)) 3 $ (4 & -11) = 3 $ (16 -121) = 3 $ -105 = 3 -105 = -102

27: How many onto functions can be defined from the set A = {1, 2, 3, 4} to {a, b, c}?

A. 81

B. 79

C. 36

D. 45

Correct Answer: Option C

the number of possible functions = 3⁴ = 81

Total number of onto functions = (Total no. of functions) – (No. of functions where one element from the co-domin remains without a pre-image) - (No. of functions where 2 elements from the co-domin remain without a pre-image.)

=> Total number of onto functions = 81 – 42 – 3 = 81 – 45 = 36

28: Find the maximum value of f(x); if f(x) is defined as the Min {-(x – 1)² + 2, (x – 2)² + 1}

A. 1

B. 2

C. 0

D. 3

Correct Answer: Option B

First let us find the range where Min (-(x – 1)² + 2, (x – 2)² + 1) is – (x – 1)² + 2.
In other words, in which range is – (x – 1)=2 + 2 < (x – 2)² + 1.
–(x²– 2x +1) + 2 < x² – 4x + 4 + 1
0 < 2x² – 6x + 4
x² – 3x + 2 > 0
(x – 1) (x – 2) > 0
=> x > 2 or x < 1
So, for x ∈ (1, 2) , f(x) = (x – 2)² + 1
And f(x) = –(x – 1)² + 2 elsewhere.
Let us also compute f(1) and f(2)
f(1) = 2, f(2) = 1
For x ∈ (-∞, 1), f(x) = –(x – 1)² + 2
f(1) = 2
For x ∈ (1, 2), f(x) = (x – 2)² + 1
f(2) = 1
For x ∈ (2, ∞), f(x) = –(x – 1)² + 2
For x < 1 and x > 2, f(x) is -(square) + 2 and so is less than 2.
When x lies between 1 and 2, the maximum value it can take is 2. f(1) = 2 is the highest value f(x) can take.

29: Consider functions f(x) = x² + 2x, g(x) = √(x + 1) and h(x) = g(f(x)). What are the domain and range of h(x)?

h(x) = g(f(x)) = g(x² + 2x) = √(x² + 2x + 1) = √(x + 1)² = |x + 1|
This bit is very important, and often overlooked.
√(9) = 3, not ±3
If x² = 9, then x can be ±3, but √(9) is only +3.
So, √(x²) = |x|, not +x, not ±x
Domain of |x + 1| = ( -∞, +∞), As far as the range is concerned, |x + 1| cannot be negative. So, range = [0, ∞)

30: [x] = greatest integer less than or equal to x. If x lies between 3 and 5, 5 inclusive, what is the probability that [x²] = [x]²?

A. Roughly 0.64

B. Roughly 0.5

C. Roughly 0.14

D. Roughly 0.36

Correct Answer: Option C

[3²]= [3]²
[3.5²] = 12 ; [3.5]² = 9
[4²] = 16 ; [4]² = 16
For x ∈ (3, 5). [x]² can only take value 9, 16 and 25.

Let us see when [x²] will be 9, 16, or 25.
If [x²] = 9,
x² ∈ [9, 10)
=> x ∈ [3, √10)
[x²] = 16
x² ∈ [16, 17)
=> x ∈ [4, √17)
In the given range [x²] = 25 only when x = 5
So [x²] = [x]² when x ∈ [3, √10] or [4, √10) or 5.
Probability = (10−3+√17−4) / 2
(3.16−3+4.12−4) / 2 = 0.28 / 2 = 0.14

31: Give the domain and range of the following functions:

A. f(x) = x2 + 1

B. g(x) = log(x + 1)

C. h(x) = 2^x

D. f(x) = 1 / x+1

E. p(x) = |x + 1|

F. q(x) = [2x], where [x] gives the greatest integer less than or equal to x

Correct Answer:

A. f(x) = x² + 1
Domain = All real numbers (x can take any value)
Range [1, ∞). Minimum value of x² is 0.

B. g(x) = log (x + 1)
Domain = Log of a negative number is not defined so (x + 1) > 0 or x > -1
Domain ( -1, ∞)
Range = (-∞, +∞)
Note: Log is one of those beautiful functions that is defined from a restricted domain to all real numbers. Log 0 is also not defined. Log is defined only for positive numbers

C. h(x) = 2^x Domain - All real numbers.
Range = (0, ∞)
The exponent function is the mirror image of the log function.

D. f(x) = 1/x+1
Domain = All real numbers except -1
Range = All real numbers except 0.

E. p(x) = |x + 1|
Domain = All real numbers
Range = [0, ∞) Modulus cannot be negative

F. q(x) = [2x], where [x] gives the greatest integer less than or equal to x
Domain = All real numbers
Range = All integers

32: How many elements are present in the domain of ^9–xC_x+1?

A. 5

B. 6

C. 4

D. 7

Correct Answer : Option B

For ⁿC_r to be defined, we should have r greater than or equal to zero and, n greater than or equal to r.
Therefore 9 – x ≥ x + 1 or 4 ≥ x and
also x + 1 > 0 or x > –1
Therefore x takes on the values {–1,0,1,2,3,4}
Therefore domain = {–1,0,1,2,3,4} and range = {¹⁰C₀,⁹C₁,⁸C₂,⁷C₃,⁶C₄,⁵C₅}
There are 6 elements in the domain.

33: f(x + y) = f(x)f(y) for all x, y, f(4) = + 3 what is f(–8)?

A. 1 / 3

B. 1 / 9

C. 9

D. 6

Correct Answer: Option B

f(x + 0) = f(x) f(0)
f(0) = 1
f(4 + – 4) = f(0)
f(4 + – 4) = f(4) f(–4)
1 = +3 * f(–4)
f(-4) = 1/3
f(– 8) = f(– 4 + (– 4)) = f(– 4) f(– 4)
f(– 8) = 1/3 * 1/3 = 1/9

34: If f(x – 3) = 2x³ + p – qx and f(x² – 4) = x² – 8q + 6p, then what is the value of p – q?

A. 5

B. 10

C. 6

D. Cannot determine

Correct Answer: Option B

f(x – 3) = 2x³ + p – qx
Let x = 3, then f(0) = 54 + p – 3q ---- (1)
f(x² – 4) = x² – 8q + 6p
Let x² = 4, then f(0) = 4 – 8q + 6p --- (2)

From (1) and (2)
54 + p – 3q = 4 – 8q + 6p
50 = 5p – 5q
p - q = 10

35: Given that x is real and f(x) = f(x + 1) + f(x – 1). Determine the value of ‘a’ that will satisfy f(x) + f(x + a) = 0

A. -1

B. -2

C. 1

D. 3

Correct Answer: Option D

f(x) = f(x + 1) + f(x – 1)

Let f(x) = p and f(x – 1) = q
f(x + 1) = f(x + 1 + 1) + f(x + 1 – 1) (Put x = x + 1 here!)
= f(x + 2) + f(x)
p – q = f(x + 2) + p
Or f(x + 2) = -q

f(x + 2) = f(x + 2 + 1) + f(x + 2 – 1) (Put x = x + 2 here!)
= f(x + 3) + f(x + 1)
– q = f(x + 3) + p – q

Or f(x + 3) = -p
At this point we notice that f(x) = p and f(x+3) = -p
f(x) + f(x + 3) = 0 (This is the condition to be satisfied to determine a)
Hence ‘a’ = 3

36: x is a real number such that f(x) = 1/x when x > 0 and f(x) = 1/x+1 otherwise. Also fn(x) = f(fn - 1 (x)). What is f(3) + f2(-3) + f3(3) + f4(-3)?

A. −2/3

B. 14/3

C. 0

D. 3

Correct Answer: Option B

Substitute x = 3 or x = -3 as appropriate for the function of a function and get the values.

f(x) = 1/x when x > 0; f(x) = 1/x+1 otherwise

f(3) = 1/3
f(-3) = −1/2
f2(3) = f(f(3) = f(1/3) = 3
f2(-3) = f(f(-3) = f(−1/2) = 1/(−1/2+1) = 2

f(3) + f2(-3) + f3(3) + f4(-3)
= 1/3 + f(f1(-3)) + f(f2(3)) + f(f3(-3))
= 1/3 + f(−1/2) + f(3) + f(f(f2(-3)))
= 1/3 + 2 + 1/3 + f(f(2))
= 8/3 + f[1/2]
= 8/3 + 2
= 14/3

 37: Which of the following functions are identical?
f(x) = x³/x²
g(x) = (√x)²
h(x) = x

A. f(x) and g(x)

B. f(x) and h(x)

C. All 3 are identical

D. None of these are identical

Correct Answer: Option D

For functions to be identical, their domains should be equal
f(x) – x can’t be zero
g(x) – x can’t be negative
h(x) – x can take all possible values

38: The value of fogoh(9) could be, if
f(x) = 1/x
g(x) = 1/(x−2)
h(x) = √x

A. 3

B. 1/3

C. -5

D. None of these

Correct Answer: Option D

fogoh(9) means f(g(h(9)))
Start by solving h(9) = √9 = 3 and not (3, -3) as square root of a number is always positive.
Taking the value to be 3, g(3) = 1/(3−2) = 1 => f(1) = 1

39: Find the domain of: 1/(1−log(9−x)) + √(x+1)?

A. (-∞,9)

B. [-1,9)

C. [-1,9) excluding 0

D. (-1,9)

Correct Answer: Option D

In the expression,
9-x > 0
=> x < 9

Also, 1-log ( 9-x) ≠ 0
=> log (9 – x) ≠ 1
=> 9-x ≠ 10
=> x ≠ -1
And, x + 1 > 0
=> x > -1

40: If [X] – Greatest integer less than or equal to x. Find the value of
[√1] + [√2] + [√3] +……………………………………………………+ [√100]

A. 615

B. 625

C. 5050

D. 505

Correct Answer: Option B

Thus, the series becomes:
3 * 1 + 5 * 2 + ……………………………………+ 19 * 9 + 10
=> ∑(2n+1)n + 10 (where, n = 1 to 9)
=> ∑ 2n² + ∑n + 10
=> 2 * n (n+1) * (2n+1)/6 + n * (n+1)/2 + 10
=> 2 * 9 * 10 * 19/6 + 9 * 10/2 + 10
=> 570 + 45 + 10 => 625

41: Find the value of x for which x[x] = 39?

A. 6.244

B. 6.2

C. 6.3

D. 6.5

Correct Answer: Option D

When x = 7,
x[x] = 7 * 7 = 49
When x = 6,
x[x] = 6 * 6 = 36
Therefore, x must lie in between 6 and 7 => [x] = 6
=> x = 45/[x] = 39/6 = 6.5

 42: Find the value of x for which x[x] = 15?

A. 3.5

B. 5

C. 6.1

D. None of these

Correct Answer: Option D

Using similar approach as previous question, [x] = 3
=> x = 15/3 => x = 5 which is not possible since 3 < x < 4