Test: Functions- 5 - CAT MCQ

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

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.

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.

fog = f (log_ex) = e^logex = x.

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)

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.

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.

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.

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.

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.

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.

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 (e) is the correct answer.

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 (e) is the correct answer.

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.