1. The following values define the relationship between x and y:
Find the value of: |
Easy | ||||||||||
| A. 101, ±8 B. 101, ±6 C. 121, ±8 D. None of these |
View Answer
Answer: Option A
Explanation:
From the table, x = 3, y = 10 -> y = (3)2 +1
x = 6, y = 37 -> y = (6)2 + 1
x = -4, y = 17 -> y = (-4)2 +1
x = -7, y = 50 -> y = (-7)2 +1
So, the relationship between x and y is y = x2 +1
i. When x = -10, y = (-10)2 +1 = 101
ii. When y = 65, 65 = x2 + 1 -> x = √64 = ±8
| 2. If g(x) = x-(1/(x+1)), find the value of: i. g(-2) ii. g(b-1) |
Easy |
| A. 1, (b2-b-1)/b B. -1, (b2-b-1)/b C. -1, (b2-b+1)/b D. None of these |
View Answer
Answer: Option B
Explanation:
i. Given, g(x) = x-(1/(x+1)) = (x2+x-1)/(x+1)
-> g(-2) = (-22-2-1)/(-2+1)
= -1
ii. g(b-1) = ((b-1)2+(b-1)-1)/((b-1)+1)
= (b2-b-1)/b
| 3. If g(x) = 3x + 8, find the: i. maximum integer value of x where g(x) is negative. ii. minimum integer value of x when g(x) is positive. |
Easy |
| A. -4, -4 B. -4, -3 C. -3, -2 D. None of these |
View Answer
Answer: Option C
Explanation:
i. g(x) = 3x + 8 ˂ 0
-> 3x ˂ -8
-> x ˂ -(8/3)
-> The maximum integer value of x is -3.
ii. g(x) = 3x + 8 > 0
-> 3x > -8
-> x > -(8/3)
-> The minimum integer value of x is -2.
| 4. If g(x) = (x+1)/x for all non-zero x and g(k+1) =3, then what is the value of k? | Easy |
| A. -1/2 B. -1/4 C. -1/3 D. None of these |
View Answer
Answer: Option A
Explanation:
g(x) = (x+1)/x
-> g(k+1) = (k+1+1)/k+1 = (k+2)/(k+1)
Given, g(k+1) = 3
-> (k+2)/(k+1) = 3
-> k = -1/2
| 5. Which of the following functions is odd? i. f(x) = x6 – x4 + 5 ii. f(x) = 7x3 – 3x iii. f(x) = x8 – 6x3 |
Medium |
| A. i. B. ii. C. iii. D. None of these |
View Answer
Answer: Option B
Explanation:
i. f(x) = x6 – x4 + 5
-> f(-x) = (-x)6 –(-x)4 + 5 = x6 – x4 + 5 = f(x)
Thus, by definition, it is even as f(-x) = f(x)
ii. f(x) = 7x3 – 3x
-> f(-x) = 7(-x)3 -3(-x) = -7x3 + 3x = -(7x3-3x) = -f(x)
Thus, by definition, it is odd as f(-x) = -f(x)
iii. f(x) = x8 – 6x3
-> f(-x) = (-x)8 -6(-x)3 = x8 + 6x3`, which is neither equal to f(x) nor equal to -f(x). Thus, by definition, it is neither even nor odd.
| 6. Which of the following equations could represent a quadratic function which does not cross the x-axis and has a maximum at the point (0,–5)? | Difficult |
| A. f(x) = -(3/5)x2 – 5 B. f(x) = 5x2 – 5 C. f(x) = -(2/5)x2 + 5 D. f(x) = x2+5 |
View Answer
Answer: Option A
Explanation:
The given quadratic function does not cross the x-axis but has a maximum at (0,-5), which means that the quadratic must open down, which implies that the coefficient of x2 must be negative (i.e., a<0 in the form ax2+bx+c) -> Only options A and C are possible. When x=0, f(x)=-5, between options A and C, only A satisfies this condition.
| 7. If f(x) = x2 – 5x, g(x) = x – 2, h(x) = x + 3, then find the value of: i. f(g(h(x))) ii. h(g(f(x))) |
Medium |
| A. x2 -3x -4, x2 -5x +1 B. x2 -3x +4, x2 -5x -1 C. x2 -3x -9, x2 -3x +9 D. None of these |
View Answer
Answer: Option A
Explanation:
Given, f(x) = x2 – 5x, g(x) = x – 2, h(x) = x + 3
i. f(g(h(x)))
= f(g(x + 3))
= f(x+3-2)
= f(x +1)
= (x + 1)2 – 5(x + 1)
= x2 +2x + 1 – 5x – 5
= x2 -3x-4
ii. h(g(f(x))) = h(g(x2 – 5x))
= h(x2 – 5x -2)
= x2 – 5x -2 +3
= x2 – 5x +1
| 8. If k is a natural number constant, and f(x) = 2x, which of the following functions yields the greatest value for f(g(x)), for all integers x>1? | Medium |
| A. g(x) = k/x B. g(x) = kx C. g(x) = x-k D. g(x) = x/k |
View Answer
Answer: Option B
Explanation:
g(x) = k/x -> f(g(x)) = f(k/x) = 2k/x
g(x) = kx -> f(g(x)) = f(kx) = 2kx
g(x) = x-k -> f(g(x)) = f(x-k) = 2(x-k)
g(x) = x/k -> f(g(x)) = f(x/k) = 2x/k
The greatest value of g(x) is 2kx because k and x are integers greater than 1.
| 9. If f(x) = x3-4x+p and f(0) and f(1) are of opposite signs, then which of the following is necessarily true? | Medium |
| A. -1<p<2 B. 0<p<3 C. -2<p<1 D. -3<p<0 |
View Answer
Answer: Option B
Explanation:
f(0) = 03-4(0)+p = p and f(1) = 13-4(1)+p = p-3
Now take values of p which correspond to the given options and verify for the range of p with the condition that f(0) and f(1) are of opposite signs i.e., p & p-3 are of opposite signs.
If p = 2 then f(0) = 2, f(1) = 2-3 = -1 (opposite signs)
If p = 1 then f(0) = 1, f(1) = 1-3 = -2 (opposite signs)
If p = 1/2 then f(0) = 1/2, f(1) = (1/2)-3 = -2.5 (opposite signs)
If p = -1/2 then f(0) = -1/2, f(1) = -1/2 – 3 = -7/2 (not of opposite signs) -> Options A, C and D are eliminated.
| 10. A function f(x) is defined for real values of x as: f(x) = 1/log(5-|x|)√(x3-7x2+14x-8). What is the domain of f(x)? | Difficult |
| A. x € (0, ∞) B. x € (-5, 4) ∪ (-4, 4) ∪ (4, 5) C. x € (1, 2) ∪ (4, 5) D. x € (1, 2) ∪ (4, ∞) |
View Answer
Answer: Option C
Explanation:
Given, f(x) = 1/log(5-|x|)√(x3-7x2+14x-8)
Factorize x3-7x2+14x-8; by trial & error, we find that x = 1 satisfies x3-7x2+14x-8 =0, or (x-1) is a factor of x3-7x2+14x-8. To find other roots of this expression, divide (x3-7x2+14x-8) by (x-1) to get (x2-6x+8) as the quotient. Now (x2-6x+8) = (x-2)(x-4)
-> x3-7x2+14x-8 = (x-1)(x-2)(x-4) —– ❶
The function f(x) is defined for all values of x for which the logarithmic function is defined (because denominator can’t be zero or infinite), which can be analysed from two dimensions:
A. The argument of the logarithm: x3-7x2+14x-8
Since logarithm is not defined for 0 and negative values, x3-7x2+14x-8 > 0
->(x-1)(x-2)(x-4) > 0 —– (from ❶)
Case I: x < 1
Here (x-1)(x-2)(x-4) > 0, so f(x) is not defined in this interval
Case II: 1<x<2
Here (x-1)(x-2)(x-4) < 0, so f(x) is defined in this interval
Case III: 2<x<4
Here (x-1)(x-2)(x-4) < 0, so f(x) is not defined in this interval
Case IV: x>4
Here (x-1)(x-2)(x-4) > 0, so f(x) is defined in this set of values.
B. The base of the logarithm: 5-|x|
Since base of a logarithm can’t be zero or negative
5-|x| > 0 -> |x| < 5 -> -5 < x < 5
From B and case II & IV of A, we conclude that
1<x<2 —– ❷
4<x<∞ —– ❸
and -5<x<5 —– ❹
So, the domain for x will be x € (1, 2) ∪ (4, 5), as it satisfies all three inequalities ❷, ❸ and ❹