Algebra - Quadratic Equations - Previous Year CAT/MBA Questions
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Let a, b and c be non-zero real numbers such that b2 < 4ac, and f(x) = ax2 + bx + c. If the set S consists of all integers m such that f(m) < 0, then the set S must necessarily be
- A.
the empty set
- B.
the set of all integers
- C.
either the empty set or the set of all integers
- D.
the set of all positive integers
Answer: Option C
Explanation :
Since b2 < 4ac ⇒ D < 0, hence, the roots of f(x) are imaginary.
Case 1: a > 0
Since a > 0 and D < 0 ⇒ f(x) > 0
Hence, there is no value of m for which f(m) < 0
⇒ m is an empty set.
Case 2: a < 0
Since a < 0 and D < 0 ⇒ f(x) < 0
Hence, for all values of m, f(m) < 0
⇒ m can take any integral value.
Hence, option (c).
Workspace:
Let f(x) be a quadratic ploynomial in x such that f(x) ≥ 0 for all real numbers x. If f(2) = 0 and f(4) = 6, then f(-2) is equal to
- A.
24
- B.
6
- C.
36
- D.
12
Answer: Option A
Explanation :
Let f(x) = ax2 + bx + c
Since, f(x) ≥ 0, and f(2) = 0, it means the graph of the quadratic lies above x-axis and touches x-axis at x = 2.
Since x = 2 is the only root of the equation (i.e., graph is symmetric about x = 2),
⇒ f(2 + a) = f(2 – a)
⇒ f(2 + 2) = f(2 - 2)
⇒ f(4) = f(0)
⇒ f(0) = 6
⇒ c = 6
f(2) = 0
⇒ 4a + 2b + c = 0
⇒ 4a + 2b = -6 …(2)
f(4) = 6
⇒ 16a + 4b + c = 6
⇒ 16a + 4b = 0 …(3)
Solving (2) and (4), we get
a = -3/2 and b = -6
⇒ f(-2) = (-2)2 – 6 × - 2 + 6 = 6 + 12 + 6 = 24
Hence, option (a).
Workspace:
Let r and c be real numbers. If r and -r are roots of 5x3 + cx2 - 10x + 9 = 0, then c equals
- A.
4
- B.
-9/2
- C.
-4
- D.
9/2
Answer: Option B
Explanation :
In a polynomial ax3 + bx2 + cx + d = 0, whose roots are α, β and γ.
⇒ α + β + γ = -b/a
⇒ αβ + βγ + γα = c/a
⇒ αβγ = -d/a
If roots of 5x3 + cx2 – 10x + 9 = 0, r, -r and γ
⇒ r – r + γ = -c/5
⇒ γ = -c/5 …(1)
⇒ r × -r + r × γ – r × γ = (-10)/5 = -2
⇒ r2 = 2 …(2)
⇒ r × - r × γ = -(9/5)
⇒ r2 × γ = 9/5
⇒ γ = 9/10 …(3)
From (1) and (3)
⇒ -c/5 = 9/10
⇒ c = -9/2
Hence, option (b).
Workspace:
The number of integral solutions of the equation = 1 is
Answer: 4
Explanation :
Given, (x2 -
For this to be equal to 1
Case 1: x2 - 10 = 1
⇒ x =
Rejected as x is not an integer.
Case 2: x2 – 3x – 10 = 0
⇒ (x – 5)(x + 2) = 0
⇒ x = 5 or -2 i.e., 2 integral values of x.
Case 3: x2 – 10 = -1 and x2 – 3x – 10 = even
If x2 – 10 = -1 ⇒ x = ±3
For both x = +3 and -3 x2 – 3x – 10 is even, hence, 2 integral values of x.
⇒ Total 4 integral values of x are possible.
Hence, 4.
Workspace:
Suppose k is any integer such that the equation 2x2 + kx + 5 = 0 has no real roots and the equation x2 + (k - 5)x + 1 = 0 has two distinct real roots for x. Then, the number of possible values of k is
- A.
8
- B.
7
- C.
9
- D.
13
Answer: Option C
Explanation :
2x2 + kx + 5 = 0 has no real roots ⇒ D < 0
⇒ k2 – 4 × 2 × 5 < 0
⇒ k2 < 40
⇒ -√40 < k < √40
∴ Possible integral values of k are -6, -5, -4, …, 0, …4, 5, 6 …(1)
Also, x2 + (k - 5)x + 1 = 0 has two distinct roots ⇒ D > 0
⇒ (k - 5)2 – 4 × 1 × 1 > 0
⇒ k2 + 25 – 10k – 4 > 0
⇒ k2 – 10k + 21 > 0
⇒ (k - 7)(k - 3) > 0
⇒ k ∈ (-∞, 3) ∪ (7, ∞) …(2)
The integral value of k satisfying both (1) and (2) are
-6, -5, -4, -3, -2, -1, 0, 1, 2 i.e., 9 values.
Hence, option (c).
Workspace:
If (3 + 2√2) is a root of the equation ax2 + bx + c = 0, and (4 + 2√3) is a root of the equation ay2 + my + n = 0, where a, b, c, m and n are integers, then the value of is
- A.
1
- B.
0
- C.
4
- D.
3
Answer: Option C
Explanation :
(3 + 2√2) is a root of the equation ax2 + bx + c = 0 whose coefficients are integers
⇒ Since coefficients are rational the other root will be (3 - 2√2)
∴ Sum of roots = 6 = -b/a
⇒ b = -6a …(1)
∴ product of roots = 1 = c/a
⇒ c = a …(2)
(4 + 2√3) is a root of the equation ay2 + my + n = 0 whose coefficients are integers
⇒ Since coefficients are rational the other root will be (4 - 2√3)
∴ Sum of roots = 8 = -m/a
⇒ m = -8a …(3)
∴ product of roots = 4 = n/a
⇒ n = 4a …(4)
Now,
=
= = 4
Hence, option (c).
Workspace:
If r is a constant such that |x2 – 4x - 13| = r has exactly three distinct real roots, then the value of r is?
- A.
21
- B.
15
- C.
17
- D.
18
Answer: Option C
Explanation :
Given, |x2 – 4x - 13| = r
∴ x2 – 4x - 13 = ± r
We have two quadratic equations here but only three distinct roots it means one of the quadratic equations will have equal roots.
Case 1: x2 – 4x - 13 = r has equal roots, i.e., Discriminant = 0
⇒ x2 – 4x – 13 - r = 0 had D = 0
⇒ D = 16 – 4(-13 - r) = 0
⇒ 16 + 52 + 4r = 0
⇒ r = - 17
Case 2: x2 – 4x - 13 = - r has equal roots, i.e., Discriminant = 0
⇒ x2 – 4x – 13 + r = 0 had D = 0
⇒ D = 16 – 4(-13 + r) = 0
⇒ 16 + 52 - 4r = 0
⇒ r = 17
Hence, option (c).
Workspace:
Suppose one of the roots of the equation ax2 – bx + c = 0 is 2 + √3, where a, b and c are rational numbers and a ≠ 0. If b = c3 then |a| equals
- A.
2
- B.
4
- C.
3
- D.
1
Answer: Option A
Explanation :
Since coefficients of the quadratic are rational numbers, hence the roots will be conjugate of each other.
If 2 + √3 is one of the roots, the other root will be 2 - √3.
∴ Sum of the roots = -(-b)/a = (2 + √3) + (2 - √3)
⇒ b/a = 4
⇒ b = 4a
Also, product of the roots = c/a = (2 + √3) × (2 + √3)
⇒ c/a = 4 – 3
⇒ c = a
Now, b = c3
⇒ 4a = a3
⇒ a2 = 4
⇒ a = ±2
∴ |a| = 2
Hence, option (a).
Workspace:
For all real values of x, the range of the function f(x) = is
- A.
- B.
- C.
- D.
Answer: Option A
Explanation :
Given, f(x) =
⇒ f(x) =
⇒ f(x) =
⇒ f(x) =
f(x) will be minimum when (2x2 + 4x + 9) is minimum.
Now, 2x2 + 4x + 9 will be minimum when x = -(4)/2 × 2 = -1
∴ Minimum value of 2x2 + 4x + 9 = 2(-1)2 + 4(-1) + 9 = 7
∴ Minimum value of f(x) = =
f(x) will be maximum when (2x2 + 4x + 9) is maximum.
Maximum value of (2x2 + 4x + 9) will be ∞.
∴ Maximum value of f(x) = =
∴ Range of f(x) =
Upper value of ½ is in open bracket as value of (2x2 + 4x + 9) will never actually be ∞.
Hence, option (a).
Workspace:
A tea shop offers tea in cups of three different sizes. The product of the prices, in INR, of three different sizes is equal to 800. The prices of the smallest size and the medium size are in the ratio 2 : 5. If the shop owner decides to increase the prices of the smallest and the medium ones by INR 6 keeping the price of the largest size unchanged, the product then changes to 3200. The sum of the original prices of three different sizes, in INR, is
Answer: 34
Explanation :
Let the price of small, medium and large cups be Rs. 2x, 5x and p respectively.
⇒ 2x × 5x × p = 800 …(1)
Also, (2x + 6) × (5x + 6) × p = 3200 …(2)
(2) = (1) × 4
⇒ (2x + 6) × (5x + 6) × p = 2x × 5x × p × 4
⇒ 10x2 + 42x + 36 = 40x2
⇒ 30x2 - 42x - 36 = 0
⇒ 5x2 – 7x – 6 = 0
⇒ 5x2 – 10x + 3x – 6 = 0
⇒ (5x + 3)(x - 2) = 0
⇒ x = 2
∴ Price of small cup = 2x = 4
Price of medium cup = 5x = 10
Price of large cup = 800/(4 × 10) = 20
⇒ Sum of the prices of three cups = 4 + 10 + 20 = 34.
Hence, 34.
Workspace:
A shop owner bought a total of 64 shirts from a wholesale market that came in two sizes, small and large. The price of a small shirt was INR 50 less than that of a large shirt. She paid a total of INR 5000 for the large shirts, and a total of INR 1800 for the small shirts. Then, the price of a large shirt and a small shirt together, in INR, is
- A.
175
- B.
200
- C.
150
- D.
225
Answer: Option B
Explanation :
Let the price of a smaller shirt be Rs. p. Hence, price of a larger shirt will be Rs. (p + 50).
⇒ + = 64
⇒ 6800p + 90000 = 64p2 + 3200p
⇒ 4p2 - 225p - 5625 = 0
⇒ 4p2 - 300p + 75p - 5625 = 0
⇒ (4p + 75)(p - 75) = 0
⇒ p = 75 or -75/4 (rejected)
⇒ Price of a smaller and a larger shirt = 75 + (75 + 50) = 200
Hence, option (b).
Workspace:
How many distinct positive integer-valued solutions exist to the equation = 1?
- A.
2
- B.
8
- C.
4
- D.
6
Answer: Option D
Explanation :
Given, = 1?
This is possible when either (x² - 7x + 11) = 1 or
(x² - 13x + 42) = 0 or
(x² - 7x + 11) = -1 and (x² - 13x + 42) = even number.
Case 1 : x² - 7x + 11 = 1
⇒ x² - 7x + 10 = 0
⇒ (x - 5)(x - 2) = 0
⇒ x = 5 or 2.
Case 2 : (x² - 13x + 42) = 0
⇒ (x - 6)(x - 7) = 0
⇒ x = 6 or 7.
Case 3 : x² - 7x + 11 = -1 and (x² - 13x + 42) = even
⇒ x² - 7x + 12 = 0
⇒ (x - 3)(x - 4) = 0
⇒ x = 3 or 4.
Now, (x² - 13x + 42) is even for all values of x.
∴ We have a total of 6 possible values of x i.e. 5 or 2, 6 or 7 and 3 or 4.
Hence, option (d).
Workspace:
The number of distinct real roots of the equation = 0 equals
Answer: 1
Explanation :
Let x + 1/x = y
⇒ y2 – 3y + 2 = 0
⇒ (y – 1)(y – 2) = 0
⇒ y = 1 or 2
We know, sum of a number and its reciprocal are either ≤ 2 or ≥ 2.
∴ y = 2
⇒ x + 1/x = 2
This is only possible when x = 1 hence, only one real root.
Hence, 1.
Workspace:
In how many ways can a pair of integers (x , a) be chosen such that x2 − 2|x| + |a - 2| = 0?
- A.
4
- B.
5
- C.
6
- D.
7
Answer: Option D
Explanation :
Given, x2 − 2|x| + |a - 2| = 0
This can be written as
|x|2 − 2|x| + 1 – 1 + |a - 2| = 0
⇒ (|x| - 1)2 + |a - 2| - 1 = 0
Now, ((|x| - 1)2 ≥ 0 and |a - 2| - 1 will be an integer since a is an integer.
∴ |a - 2| - 1 can take only take non-positive values i.e., 0 or -1.
Case 1: |a - 2| - 1 = 0 and (|x| - 1) = 0
⇒ |a – 2| = 1 and |x| = 1
⇒ a = 1 or 3 and x = ± 1
∴ 4 possible combinations of (x, a)
Case 2: |a - 2| - 1 = -1 and (|x| - 1) = ±1
|a - 2| = 0 and |x| = 0 or 2
⇒ a = 2 and x = 0 or ± 2
∴ 3 possible combinations of (x, a)
∴ Total 7 possible combination of (x, a) are there.
Hence, option (d).
Workspace:
The number of integers that satisfy the equality (x² - 5x + 7)x+1 = 1 is
- A.
5
- B.
2
- C.
4
- D.
3
Answer: Option D
Explanation :
For (x² - 5x + 7)x+1 = 1, either
Case 1: x + 1 = 0
⇒ x = -1
Case 2: (x² - 5x + 7) = 1
⇒ x² - 5x + 6 = 0
⇒ x = 2 or 3
Case 3: (x² - 5x + 7) = -1 and x + 1 is even
⇒ x² - 5x + 8 = 0
No integral value of x is possible in this case.
∴ Possible values of x are -1, 2 and 3 i.e., 3 values.
Hence, option (d).
Workspace:
Let m and n be positive integers, If x² + mx + 2n = 0 and x² + 2nx + m = 0 have real roots, then the smallest possible value of m + n is
- A.
8
- B.
7
- C.
5
- D.
6
Answer: Option D
Explanation :
First, we have the quadratic equation, x² + mx + 2n = 0
Since its roots are real, that means D ≥ 0.
⇒ m2 – 8n ≥ 0
⇒ m ≥ 2√(2n) …(1)
Also, we have the quadratic equation, x² + 2nx + m = 0
Since its roots are real, that means D ≥ 0.
⇒ (2n)2 – 4m ≥ 0
⇒ 4n2 – 4m ≥ 0
⇒ n2 ≥ m …(2)
From (1) and (2)
⇒ n2 ≥ m ≥ 2√(2n)
⇒ n3/2 ≥ 23/2
⇒ n ≥ 2
Hence, least possible value of n = 2.
⇒ m ≥ 2√(2n)
m will be least when n is least
⇒ m ≥ 2√(4)
⇒ m ≥ 4
Hence, least possible value of m = 4.
So, the smallest possible values of m and n are 4 and 2 respectively.
Hence, smallest value of m + n = 4 + 2 = 6.
Hence, option (d).
Workspace:
The number of solutions to the equation |x|(6 + 1) = 5 is
Answer: 5
Explanation :
Since we have |x|, we will have to consider cases where, x < 0 or x = 0 or x > 0.
For x > 0: 6 + 1 = 5x ⇒ 6 − 5x + 1 = 0. The two roots of this equation are 1/2 and 1/3.
For x = 0, LHS = RHS, ∴ x = 0 is a root of the equation.
For x < 0: 6 + 1 = −5x ⇒ 6 + 5x + 1 = 0. The two roots of this equation are −1/2 and −1/3.
∴ Number of roots = 2 + 1 + 2 = 5.
Answer: 5.
Workspace:
The product of the distinct roots of ∣x2 − x − 6∣ = x + 2 is
- A.
-16
- B.
-4
- C.
-8
- D.
-24
Answer: Option A
Explanation :
∣x2 − x − 6∣ = x + 2
∴ ∣(x − 3)(x + 2)∣ = x + 2
Case 1: x < −2. i.e. (x − 3)(x + 2) > 0
(x - 3)(x + 2) = x + 2
∴ x = 4. (which is rejected since 4 is not less than −2)
Case 2: x = −2.
This is a real root of this equation.
Case 3: −2 < x < 3. i.e. (x − 3)(x + 2) < 0
- (x - 3)(x + 2) = x + 2
∴ x = 2.
Case 4: x = 3.
This does not satisfy the equation so x = 3 is not a root of this equation.
Case 5: x > 3. (x − 3)(x + 2) > 0
(x − 3)(x + 2) = x + 2
∴ x = 4.
Required product = (−2) × 2 × 4 = −16.
Hence, option (a).
Workspace:
The real root of the equation 26x + 23x+2 - 21 = 0 is
- A.
- B.
log29
- C.
- D.
log227
Answer: Option A
Explanation :
Given: 26x + 23x+2 - 21 = 0
Replace 23x with y
So, 26x = y2
Now, 26x + 23x+2 - 21 = 0 can be rewritten as y2 + 22 × 23x - 21 = 0
y2 + 4y - 21 = 0
Solving the above quadratic equation,
(y + 7) (y - 3) = 0
So, y = -7 or +3
y = - 7 is rejected (since, y = 23x which should always be positive)
⇒ 23x = 3
Taking log on both sides,
log23 = 3x
x =
Hence, option (a).
Workspace:
Let A be a real number. Then the roots of the equation x2 - 4x - log2A = 0 are real and distinct if and only if
- A.
A < 1/16
- B.
A < 1/8
- C.
A > 1/8
- D.
A > 1/16
Answer: Option D
Explanation :
For roots of a quadratic equation to be real and distinct, Discriminant > 0
So, for x2 − 4x – log2A = 0,
D = (-4)2 - (4 × 1 × (-log2A)) > 0
⇒ 16 + 4 × log2A > 0
⇒ log2A > -4
⇒ A > 2-4
⇒ A > 1/16
Hence, option (d).
Workspace:
The quadratic equation x2 + bx + c = 0 has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of b2 + c?
- A.
3721
- B.
549
- C.
361
- D.
427
Answer: Option B
Explanation :
Given quadratic equation is x2 + bx + c = 0
Sum of roots = -b
⇒ -b = 4a + 3a
⇒ -b = 7a
Product of the roots = c
⇒ c = 4a × 3a
⇒ c = 12a2
Now, b2 = 49a2 and c =12a2
Hence, b2 + c = 49a2 + 12a2
⇒ b2 + c = 61a2
Final answer must be a multiple of 61 & the multiple should be a perfect square.
Going through the options,
3721 = 61 × 61 (multiple is not a perfect square)
361 (not a multiple of 61)
427 = 61 × 7 (multiple is not a perfect square)
549 = 61 × 9 = 61 × 32
Thus, possible value of b2 + c = 549.
Hence, option (b).
Workspace:
If u2 + (u−2v−1)2 = −4v(u + v), then what is the value of u + 3v?
- A.
1/2
- B.
-1/4
- C.
0
- D.
1/4
Answer: Option B
Explanation :
u2 + (u – 2v – 1)2 = −4v(u + v)
u2 + (u – 2v – 1)2 + 4v(u + v) = 0
(u – 2v – 1)2 + u2 + 4uv + 4v2 = 0
(u – 2v – 1)2 + (u + 2v)2 = 0
(u – 2v – 1)2 & (u + 2v)2 are both non-negative terms. Sum of two negative terms will be zero only when both the terms are equal to zero.
∴ u – 2v – 1 = 0 and u + 2v = 0
Solving the two equations, we get
u = 1/2 and v = -1/4
u + 3v = 1/2 - 3/4 = -1/4
Hence, option (b).
Workspace:
If a and b are integers such that 2x2 − ax + 2 > 0 and x2 − bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a − 6b is
Answer: 36
Explanation :
We have to maximize the value of 2a - 6b. Therefore let us look for the largest possible value of a and the smallest possible value of b.
If 2x2 – ax + 2 > 0 for all value of x, the graph of 2x2 – ax + 2 is above the x-axis or the roots of the quadratic equations 2x2 – ax + 2 = 0 are imaginary or its discriminant is less than 0.
∴ We have a2 – 16 < 0 or a2 < 16 or –4 < a < 4.
⇒ The largest possible value of a is 3.
If x2 – bx + 8 ≥ 0, the discriminant of the equation is less than or equal to zero.
⇒ b2 - 32 ≤ 0
⇒ -4√2 ≤ b ≤ 4√2
⇒ -5.64 ≤ b ≤ 5.64
∴ The smallest possible value of b is –5.
Therefore the maximum possible value of 2a - 6b = 2(3) – 6(–5) = 36.
Hence, 36.
Workspace:
The smallest integer n such that n3 – 11n2 + 32n – 28 > 0 is
Answer: 8
Explanation :
The sum of the coefficients of n3 -11n2 + 32n - 28 is not zero.
Similarly the sum of the odd coefficients of n3 -11n2 + 32n - 28 is not equal to its even coefficients. Therefore it is not divisible by (n - 1) and (n + 1).
If we put n = 2, n3 – 11n2 + 32n – 28 = 8 – 44 + 64 – 28 = 0.
Therefore (n – 2) is a factor of the polynomial n3 - 11n2 + 32n – 28.
We have, (n - 2) is a factor of n3 – 11n2 + 32n – 28
∴ n3 – 11n2 + 32n – 28 = (n - 2)(n2 – 9n + 14) = (n – 2)(n – 2)(n - 7) = (n – 2)2 (n – 7).
Now, we have (n – 2)2 (n – 7) > 0
(n - 2)2 is always greater than 0, when n > 2.
⇒ For (n – 2)2 (n – 7) > 0, (n - 7) should be greater than 0.
⇒ n > 7.
Therefore the smallest integer for which n3 - 11n2 + 32n - 28 > 0 is 8.
Hence, 8.
Workspace:
If f1(x) = x2 + 11x + n and f2(x) = x, then the largest positive integer n for which the equation f1(x) = f2(x) has two distinct real roots, is :
Answer: 24
Explanation :
x2 + 11x + n = x
⇒ x2 + 10x + n = 0
For roots to be real and distinct, Discriminant > 0
⇒ D = (10)2 - 4 × 1 × n > 0
⇒ 4n < 100
⇒ n < 25
Therefore, highest possible integral value of n is 24.
Hence, 24.
Workspace:
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