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Let f and g be the function from the set of integers to itself, defined by f(x) = 2x + 1 and g(x) = 3x + 4. Then the composition of f and g is ____________
The composition of f and g is given by f(g(x)) which is equal to 2(3x + 4) + 1.
A certain function always obeys the rule: If f (x.y) = f(x). f(y) where x and y are positive realnumbers. A certain Mr. Mogambo found that the value of f (128) = 4, then find the value of thevariable M = f (0.5). f (1). f (2). f (4). f (8). f (16). f (32). f (64). f (128). f (256)
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.
The graph of y = (x + 3)^{3} + 1 is the graph of y = x^{3} shifted
(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.
(x^{2} + 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)).
f(x) is any function and f^{–1}(x) is known as inverse of f(x), then f–1(x) of f(x) = e^{x} is
y = e^{x}
fi log_{e} y = x.
fi f ^{–1}(x) = log_{e} x.
Which of the following functions will have a minimum value at x = –3?
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.
Define the following functions:
f(x, y, z) = xy + yz + zx
g(x, y, z) = x^{2}y + y^{2}z + z^{2}x and
h(x, y, z) = 3 xyz
Q.
Find the value of the following expressions:37. h[f(2, 3, 1), g(3, 4, 2), h(1/3, 1/3, 3)]
The given function would become h[ 11, 80, 1] = 2640.
Define the following functions:
f(x, y, z) = xy + yz + zx
g(x, y, z) = x^{2}y + y^{2}z + z^{2}x and
h(x, y, z) = 3 xyz
Find the value of the following expressions:
Q.
f[ f (1, 1, 1), g(1, 1, 1), h(1, 1, 1)]
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.
If R(a/b) = Remainder when a is divided by b;
Q(a/b) = Quotient obtained when a is divided by b;
SQ(a) = Smallest integer just bigger than square root of a.
Q.
If a = 12, b = 5, then find the value of SQ[R {(a + b)/b}].
SQ [R[(a + b)/b]] = SQ [R[17/5]] fi SQ [2] = 2.
If R(a/b) = Remainder when a is divided by b;
Q(a/b) = Quotient obtained when a is divided by b;
SQ(a) = Smallest integer just bigger than square root of a.
Q.
If a =18, b = 2 and c = 7, then find the value of Q [{SQ(ab) + R(a/c)}/b].
Q [[SA (36) + R (18/7)]/2] = Q [(7 + 4)/2] = Q [11/2] = 5.
Read the following passage and try to answer questions based on
them.
[x] = Greatest integer less than or equal to x
{x} = Smallest integer greater than or equal to x.
Q.
If x is not an integer, then ({x} + [x]) is
[x] + {x} will always be odd as the values are consecutive integers.
If f(t) = t^{2} + 2 and g(t) = (1/t) + 2, then for t = 2, f [g(t)] – g[f(t)] = ?
f(g(t)) – g(f(t)) = f(2.5) – g(6) = 8.25 – 2.166 = 6.0833.
Let F(x) be a function such that F(x) F(x + 1) = – F(x – 1)F(x–2)F(x–3)F(x–4) for all x ≥ 0.Given the values of If F (83) = 81 and F(77) = 9, then the value of F(81) equals to
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.
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