Functions Test

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Functions Test - 5

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Weekly Quiz Competition

Question 1
3 / -1

The minimum value of ax² + bx + c is 7/8 at x = 5/4. Find the value of the expression at x = 5, ifthe value of the expression at x = 1 is 1.

A
75

B
29

C
121

D
129

Solution

Correct Answer :- B
Explanation : ax² + bx + c
a[x² + b/a x + c/a]
a[x² + 2b/2a x + c/a + (b/2a)² - (b/2a)²]
a[(x+b/2a)² + c/a - (b/2a)²]
Minimum value occurs when
x = - b/2a
=> -b/2a = 5/4
=> b/a = -5/2
And, minimum value is a[(c/a - (b/2a)²]
a[(c/a - (b/2a)²] = 7/8
=> c - 25s/16 = 7/8
=> 16c - 25a = 14.....(1)
Again value of the expression at x = 1 is 1
a * 1² + b * 1 + c = 1
a - 5a/2 + c = 1
=> 2a - 5a + 2c = 2
=> - 3a + 2c = 2.......(2)
Multiplying eq (2) by 8 and subtracting it from eq (1)
-a = -2
or a = 2
Thus, b = -5/2 × 2 = -5
Putting the value of a in eq (2)
-6 + 2c = 2
2c = 8
c = 4
Thus, the expressions is 2x² - 5x + 4
Value of the expression at x = 5 is
2 * 5²- 5 * 5 + 4
= 50 - 25 + 4
= 29
Question 2
3 / -1

A function a(x) is defined for x as 3a(x) + 2a (2 – x) = (x + 3)². What is the value of [G (–5)]
where [x] represents the greatest integer less than or equal to x?

A
37

B
–38

C
–37

D
Cannot be determined

Solution

The equation given in the question is: 3a(x) + 2a (2–x) = (x + 3)²
Replacing x by (2–x) in the above equation, we get
3a(2–x) + 2a(x) = (5–x)²
Solving the above pairs of equation, we get
a(x) = (x2 + 38 – 23)/5
Thus, G(–5) = –188/5 = –37.6. The value of [–37.6] = –37. Hence, Option (c) is the correct
answer.
Question 3
3 / -1

Find the domain of the definition of the functiony = [(x – 3)/(x + 3)]^1/2 + [(1 – x)/(1 + x)]1/2.

A
x > 3

B
x < –3

C
-3 £ x £ 3

D
Nowhere

Solution

Both the brackets should be non-negative and neither (x + 3) nor (1+ x) should be 0.
For (x – 3)/(x + 3) to be non negative we have x>3 or x< – 3.
Also for (1– x)/(1+ x) to be non-negative –1 < x < 1. Since there is no interference in the two
ranges, Option (d) would be correct.
Question 4
3 / -1

If f(x) = e^x and g(x) = log_ex then value of fog will be

A
x

B
0

C
1

D
e

Solution

fog = f (log_ex) = e^logex = x.
Question 5
3 / -1

The function y = 1/x shifted 1 unit down and 1 unit right is given by

A
y – 1 = 1/ (x + 1)

B
y – 1 = 1/(x – 1)

C
y + 1 = 1/(x – 1)

D
y + 1 = 1/(x + 1)

Solution

Looking at the options, one unit right means x is replaced by (x – 1). Also, 1 unit down means –1
on the RHS.
Thus, (y + 1) = 1/(x – 1)
Question 6
3 / -1

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

Solution

Take different values of n to check each option. Each of Options (a), (b) and (c) can be ruled out.
Hence, Option (d) is correct.
Question 7
3 / -1

Read the instructions below and solve:
f(x) = f(x – 2) – f(x – 1), x is a natural number
f(1) = 0, f(2) = 1
Q.
The value of f[f(6)] is

A
5

B
– 1

C
– 3

D
– 2

Solution

f(1) = 0, f(2) = 1,
f(3) = f(1) – f(2) = –1
f(4) = f(2) – f(3) = 2
f(5) = f(3) – f(4) = –3
f(6) = f(4) – f(5) = 5
f(7) = f(5) – f(6) = –8
f(8) = f(6) – f(7) = 13
Ans:-
f(f(6) = f(5) = –3.
Question 8
3 / -1

If f(x) is a function satisfying f(x). f(1/x) = f(x) + f(1/x) and f(4) = 65, what will be the value off(6)?

A
37

B
217

C
64

D
None of these

Solution

We have f(x) ◊ f(1/x) = f(x) + f(1/x)
fi f(1/x) [f(x) – 1] = f(x)
For x = 4, we have f(1/4) [f(4) – 1] = f (4)
fi f(1/4) [64] = 65
fi f(1/4) = 65/64 = 1/64 + 1
This means f(x) = x3 + 1
For f(6) we have f(6) = 216 + 1 = 217.
Question 9
3 / -1

Define the functions:
A(x, y, z) = Max (max (x, y), min (y, z) min (x, z))
B(x, y, z) = Max (max (x, y), min (y, z) max (x, z))
C(x, y, z) = Max (min (x, y), min (y, z) min (x, z))
D(x, y, z) = Min (max (x, y), max (y, z) max (x, z))
Max (x, y, z) = Maximum of x, y and z.
Min (x, y, z) = Minimum of x, y and z.
Assume that x, y and z are distinct integers.
Q.
For what condition will A(x, y, z) be equal to Max (x, y, z)?

A
When x is maximum

B
When y is maximum

C
When z is maximum

D
Either (a) or (b)

Solution

A(x, y, z) = Max (max (x, y), min (y, z) min (x, z))
B(x, y, z) = Max (max (x, y), min (y, z) max (x, z))
C(x, y, z) = Max (min (x, y), min (y, z) min (x, z))
D(x, y, z) = Min (max (x, y), max (y, z) max (x, z))
Max (x, y, z) = Maximum of x, y and z.
Min (x, y, z) = Minimum of x, y and z.
Assume that x, y and z are distinct integers.
Ans:-
When either x or y is maximum.
Question 10
3 / -1

Define the functions:
A(x, y, z) = Max (max (x, y), min (y, z) min (x, z))
B(x, y, z) = Max (max (x, y), min (y, z) max (x, z))
C(x, y, z) = Max (min (x, y), min (y, z) min (x, z))
D(x, y, z) = Min (max (x, y), max (y, z) max (x, z))
Max (x, y, z) = Maximum of x, y and z.
Min (x, y, z) = Minimum of x, y and z.
Assume that x, y and z are distinct integers.
Q.
For what condition will A(x, y, z) not be equal to B (x, y, z)?

A
x > y > z

B
y > z > x

C
z > y > x

D
None of these

Solution

A(x, y, z) = Max (max (x, y), min (y, z) min (x, z))
B(x, y, z) = Max (max (x, y), min (y, z) max (x, z))
C(x, y, z) = Max (min (x, y), min (y, z) min (x, z))
D(x, y, z) = Min (max (x, y), max (y, z) max (x, z))
Max (x, y, z) = Maximum of x, y and z.
Min (x, y, z) = Minimum of x, y and z.
Assume that x, y and z are distinct integers.
Ans:-
When z is maximum, A and B would give different values. Thus, option (c) is correct.
Question 11
3 / -1

Define the functions:
A(x, y, z) = Max (max (x, y), min (y, z) min (x, z))
B(x, y, z) = Max (max (x, y), min (y, z) max (x, z))
C(x, y, z) = Max (min (x, y), min (y, z) min (x, z))
D(x, y, z) = Min (max (x, y), max (y, z) max (x, z))
Max (x, y, z) = Maximum of x, y and z.
Min (x, y, z) = Minimum of x, y and z.
Assume that x, y and z are distinct integers.
Q.
The highest value amongst the following will be

A
Max/Min

B
A/B

C
C/D

D
Cannot be determined

Solution

A(x, y, z) = Max (max (x, y), min (y, z) min (x, z))
B(x, y, z) = Max (max (x, y), min (y, z) max (x, z))
C(x, y, z) = Max (min (x, y), min (y, z) min (x, z))
D(x, y, z) = Min (max (x, y), max (y, z) max (x, z))
Max (x, y, z) = Maximum of x, y and z.
Min (x, y, z) = Minimum of x, y and z.
Assume that x, y and z are distinct integers.
Ans:-
We cannot determine this because it would depend on whether the integers x, y, and z are positive
or negative.
Question 12
3 / -1

A0, A1, A2,...... is a sequence of numbers withA0 = 1, A1 = 3, and At = (t +1) At–1 – t At–2 = 2, 3, 4,....Conclusion I. A8 = 77Conclusion II. A10 = 121Conclusion III. A12 = 145

A
Using the given statement, only Conclusion I can be derived.

B
Using the given statement, only Conclusion II can be derived.

C
Using the given statement, only Conclusion III can be derived.

D
Using the given statement, none of the three Conclusions I, II and III can be derived.

Solution

This question is based on the logic of a chain function. Given the relationship
At = (t + 1) At – 1 – t At – 2
We can clearly see that the value of A2 would depend on the values of A₀ and A₁. Putting t = 2 in
the expression, we get:
A₂ = 3A1 – 2A₀ = 7; A₃ = 19; A₄ = 67 and A₅ = 307. Clearly, A6 onwards will be larger than 307
and hence none of the three conclusions are true. Option (d) is the correct answer.
Question 13
3 / -1

The figure below shows the graph of a function f (x). How many solutions does the equation f ( f(x)) = 15 have?

A
5

B
6

C
Cannot be determined from the given graph

D
8

Solution

f ( f (x) = 15 when f (x) = 4 or f (x) = 12 in the given function. The graph given in the figure
becomes equal to 4 at 4 points and it becomes equal to 12 at 2 points in the figure. This gives us 6
points in the given figure when f (f (x) =15. However, the given function is continuous beyond the
part of it which is shown between –10 and +13 in the figure. Hence, we do not know how many
more solutions to f (f (x) = 15 would be there. Hence, Option (c) is the correct answer.
Question 14
3 / -1

For all real numbers x, except x = 0 and x = 1, the function F is defined by
If 0 < a < 90° then F((cosec a)2) =

A
(sin a)²

B
(cos a)²

C
(tan a)²

D
(cot a)²

Solution
Question 15
3 / -1

where [x] is defined as integral part of x and f is a fraction, then x(1 – f) equals–

A
1

B
Less than 1

C
More than 1

D
Between 1 and 2

Solution

retained the even terms which are integral. Hence, the value of x+y is an integer.
Further, x+y = [x] + f and hence, if x+y is an integer, [x] + f + y would also be an integer. This
automatically means that f+y must be an integer (as [x] is an integer).
Now, the value of y is between 0 to 1 and hence when we add the fractional part of x i.e. ‘f ’ to y,
and we need to make it an integer, the only possible integer that f + y can be equal to is 1.
Thus, if f + y = 1 Æ y = (1 – f ).
In order to find the value of x(1 – f ) we can find the value of x × y.