Test Level 2: Functions - 2 - CAT MCQ

Putting x = 1/2,
We get f(2) + 2f(-1/2) = 1
Or, f(2) = 1 - 2f(-1/2)........(i)
Putting x = -2,
We get f(-1/2) - (1/2)f(2) = -4
Or, f(-1/2) = -4 + (1/2)f(2).........(ii)
Putting (ii) in (i),
We get f(2) = 1 - 2 (-4 + (1/2)f(2))
Or, f(2) = 1 + 8 - f(2) 
Or, 2f(2) = 9
Or, f(2) = 9/2 = 4.5

If the answer is a constant, this implies that the answer does not depend on the variable i.e. x.
Thus, if we take only constant terms that we would get after writing all the functions, we will get the required answer.
Thus, 7(8³ - 7³ - 6³ + 5³ - 4³ + 3³ + 2³ - 1³) + 18(1 - 1 - 1 + 1 - 1 + 1 + 1 - 1) + 23(8 - 7 - 6 + 5 - 4 + 3 + 2 - 1) = 336.
Hence, option d is correct.

Let A be the common value of f(1), f(2), f(3), f(4), and f(5).
Then f(x) - A has the roots 1, 2, 3, 4, and 5.

Since f(x) - A = x⁵ + ax⁴ + bx³ + cx² + dx + (e - A), f(x) - A is a polynomial of degree 5 and, hence, the five numbers 1, 2, 3, 4, and 5 must be the five roots of f(x).

It follows that the sum of the roots of f(x) is 1 + 2 + 3 + 4 + 5 = 15.
Since in general the sum of the n roots of a polynomial a_nxⁿ + a_n - 1^{xn - 1} + …….. + a₁x + a₀ of degree n is , we get that the sum of the roots of f(x) - A is - a.
Thus, - a = 15 so that a = - 15.
(Comment: When we say that in general the sum of the n roots above is, we mean that the polynomial may have "multiple" roots and we are to count the roots with multiplicity. For example, (x - 1)² (x - 2)³ = x⁵ - 8x⁴ + …. Where 8 represents the sum of the roots counting 1 twice and 2 thrice.)

Since h(x) = 
h(x) = 
h(1) == 5.
(Observe that (x + 1) (x² - x + 1) = x³ + 1, so h(x) = x⁴ + x³ + x2 + x + 1. Computing h(x) is not necessary, but it is certainly a reasonable thing to do.)

C(n, 0) = n + 1 ……….. (1)
C(0, i) = C(1, i - 1)  ……….. (2)
C(n, i) = C(C (n - 1, i), (i - 1)) ……….. (3)
From (3), put n = 1 and i = 2
C(1, 2) = C[C(0, 2), 1]  ……….. (4)
From (2), 
C(0, 2) = C(1, 1) ……….. (5)
From (3) and (5),           
C(1, 1) = C[C(0, 1), 0] ……….. (6)
Now, C(0, 1) = C(1, 0) [From (2)]
Also, C(1, 0) = 1 + 1 = 2 [From (1)]
Put this in (6)
C(1, 1) = C(2, 0) = 2 + 1 = 3 [From (1)]
Put this value in (5)
C(0, 2) = 3
Put this value in (4)
C(1, 2) = C(3, 1) ……….. (7)
Now, from (3),
C(3, 1) = C[C(2, 1), 0] ……….. (8)
C(2, 1) = C[C(1, 1), 0]
= C(3, 0) = 4
Put in (8)
C(3, 1) = C(4, 0) = 5 ………. (9)
From (7) to (9),
C(1, 2) = 5.

f(g(x)) = x + 2
Also f(g(x)) = 
xg(x) + x + 2g(x) + 2 = 4 - g(x)
g(x) =
g(t) =
g(t) =

f(x) = 1 - 
f(4) = 1 - 
f(f(4)) = f = 1 -= 4.
f(f(4)) = 4
f(f(f(4))) = f(4) = 4/3
So, if there even no. of f's in the composition, output is 4, otherwise 4/3.
As there are 1997 f's in the composition, answer is 4/3.

f(n) = 1 + 4x + 7x² + 10x³ + …..(1)
xf(n) = x + 4x² + 7x³ + 10x⁴ + ……∝ …(2)
Subtracting (1) and (2), we get
(1 - x) f(n) = 1 + 3x + 3x² + 3x³ +......
⇒ (1 - x) f(n) = 1 + 3x/1-x
⇒ (1 - x) × 35 x 16 = 1 + 2x/1 - x
⇒ 35(1 - x)² = 16 + 32x
⇒ 35x² - 102x + 19 = 0
⇒ (7x - 19)(5x - 1) = 0
x ≠ 19/7 (∴ For infinity series, common ratio < 1)
∴ x = 1/5

Given: f(x + y) = f(x) + f(y)
f(1 + 1) = f(2) = f(1) + f(1) = 2f(1)
f(2 + 1) = f(3) = f(2) + f(1) = 3f(1)
Similarly, f(4) = 4f(1), f(5) = 5f(1) … f(10) = 10f(1)
Given expression: 
f(1) + f(2) + f(3) … f(10) = 1
f(1) + 2f(1) + 3f(1) + ... + 10f(1) = 1
f(1)(1 + 2 + 3 + ... + 10) = 1
f(1) = 1/55

f(x + 2) = 2f(x) - f(x + 1) ... (i)
Put x = 0 in (i),
f(2) = 2f(0) - f(1) = 2 × 2 - 3 = 1
Put x = 1 in (i),
f(3) = 2f(1) - f(2) = 2 × 3 - 1 = 5
Put x = 2 in (i),
f(4) = 2f(2) - f(3) = 2 × 1 - 5 = -3
Put x = 3 in (i),
f(5) = 2f(3) - f(4) = 2 × 5 - (-3) = 13

y = f(x) is transformed to the graph of 2y - 6 = -4f(x - 3). 
(A, B) = ?
2y - 6 = -4f(x - 3)
2y = -4f(x - 3) + 6
y = -2f(x - 3) + 3
The graph of y = -2f(x - 3) + 3 is that of y = f(x) shifted 3 units to the right, stretched vertically by a factor of 2, reflected on the x-axis and shifted up by 3 units. A point of y = f(x) will undergo the same transformation. Hence, point (a, b) on the graph of y = f(x) becomes:
(a + 3, b) on the graph of y = f(x - 3) (shifted 3 units to the right)
(a + 3, 2b) on the graph of y = 2f(x - 3) (stretched vertically by 2)
(a + 3, -2b) on the graph of y = -2f(x - 3) (reflected on the x-axis)
(a + 3, -2b + 3) on the graph of y = -2f(x - 3) + 3 (shifted up 3 units)

The following table shows the values of the graph of the functions f(x) and f(-x) at different values of x:

From the above table, we get
f(x) = f(-x)
f(x) - f(-x) = 0

First note that (fog)(-1) = f(g(- 1)).
This means that you first need to find g(-1).
So, look at the graph of g(x) and find x = -1.
Tracing up to the graph of g(x), you get x = -1, y = 3, i.e. the point (-1, 3) is on the graph of g(x).
So, g(-1) = 3. Now, plug this value into f(x).
To do this, look at the graph of f(x) and find x = 3.
Tracing up to the graph of f(x), you will get x = 3, y = 3, i.e. the point (3, 3) is on the graph of f(x).
So, f(3) = 3. Then, (fog)(-1) = f(g(- 1)) = f(3) = 3

= 10,560 cm²/s

 Longest diagonal of a cube = D = √3  a, where a is the length of each side.

Profit = P = Total SP - Total CP
= 500 X - (100 + 5X²) = - 5X² + 500X - 100
Here, X = Production level
 dP/dX = 0
 500 - 10X = 0  X = 50 units
The most profitable production level is 50 units.

Let X kg be the most profitable level. 
⇒ Profit = P = Total SP - Total CP = 155X - (X² + 5) x ⇒ dP/dX = 0 ⇒ 155 - 3X² - 5 = 0
⇒ x = √50 kg

F(x) = (2000 - x)(280 + x)
F`(x) = (2000 - x) + (280 + x)(-1)
= 2000 - x - 280 - x = 1720 - 2x
F"(x) = -2 (which is negative)
So,
1720 - 2x = 0
1720 = 2x
x = 860
 x = Rs. 860

Area will be maximum if it is a square of edge = 12/4 = 3 or area = 3² = 9 sq. units.

Put m = 0 and n = 0, the least value of f(x) = n - 28 + m - 16  = - 28 + (- 16) = - 44 and the greatest value of f(x) = 6 - 1 = 5.