Test: Functions- 4 - CAT MCQ

The composition of f and g is given by f(g(x)) which is equal to 2(3x + 4) + 1.

Since f (128) = 4, we can see that the product of f (256). f (0.5) = f (256 × 0.5) = f (128) = 4.
Similarly, the products f (1). f (128) = f (2). f (64)
= f (4). f (32) = f (8). f (16) = 4.
Thus, M = 4 × 4 × 4 × 4 × 4 = 1024.
Option (d) is the correct answer.

f(0) = 1, f(1) = 2 and f(2) = 4
Hence, they are in G.P.

(x + 3)³ would be shifted 3 units to the left and hence (x + 3)³ + 1 would shift 3 units to the left
and 1 unit up. Option (c) is correct.

(x² + loge x) would be neither odd nor even since it obeys neither of the rules for even function
(f(x) = f(–x)) nor for odd functions (f(x) = –f(–x)).

y = e^x
fi log_e y = x.
fi f ^–1(x) = log_e x.

If you differentiate each function with respect to x, and equate it to 0 you would see that for none
of the three options will get you a value of x = –3 as its solution. Thus, option (d) viz. None of
these is correct.

The given function would become h[ 11, 80, 1] = 2640.

The given function would become f[3, 3, 3] = 27.

The number of g’s and f’s should be equal on the LHS and RHS since both these functions are
essentially inverse of each other.
Option (c) is correct.

SQ [R[(a + b)/b]] = SQ [R[17/5]] fi SQ [2] = 2.

Q [[SA (36) + R (18/7)]/2] = Q [(7 + 4)/2] = Q [11/2] = 5.

[x] + {x} will always be odd as the values are consecutive integers.

f(g(t)) – g(f(t)) = f(2.5) – g(6) = 8.25 – 2.166 = 6.0833.

When the value of x = 81 and 82 is substituted in the given expression, we get,
F (81) F (82) = – F (80) F (79) F(78) F(77)
F (82) F (83) = – F (81) F (80) F(79) F(78)
On dividing (i) by (a), we get

Option (a) is the correct answer.