Algebra - Quadratic Equations - Previous Year CAT/MBA Questions

Answer: 38

Text Explanation :

Answer: Option B

Text Explanation :

Answer: Option B

Text Explanation :

Answer: 3

Video Explanation

Text Explanation :

Case 1: x ≥ 0 ⇒ |x| = x
∴ 2 × x × (x² + 1) = 5x²
⇒ 2x(x² + 1) = 5x² 
⇒ 2x(x² + 1) - 5x² = 0
⇒ x[2(x² + 1) - 5x] = 0
⇒ x(2x² – 5x + 2) = 0 
⇒ x(2x² – 4x - x + 2) = 0 
⇒ x[2x(x – 2) - (x - 2)] = 0 
⇒ x(2x - 1)(x - 2) = 0
⇒ x = 0 or ½ or 2.
We need only integral solutions hence acceptable answers are 0 and 2.

Case 2: x < 0 ⇒ |x| = -x
∴ 2 × -x × (x² + 1) = 5x²
⇒ -2x(x² + 1) = 5x² 
⇒ 2x(x² + 1) + 5x² = 0
⇒ x[2(x² + 1) + 5x] = 0
⇒ x(2x² + 5x + 2) = 0
⇒ x(2x² + 4x + x + 2) = 0
⇒ x[(2x(x + 2) + (x + 2)] = 0
⇒ x(2x + 1)(x + 2) = 0
⇒ x = 0 or -1/2 or -2
We need only integral solutions hence acceptable answers are 0 and -2.

∴ Acceptable integral solutions are -2, 0 and 2, i.e., 3 integral solutions.

Hence, 3.

Concept:

Solving a Quadratic Equation

Answer: 6

Video Explanation

Text Explanation :

α and β be two distinct root of 2x² – 6x + k = 0,
∴ α + β = -(-6)/2 = 3
And, α × β = k/2 = k/2

Now, 3 and k are the roots of the equation x² + px + p = 0.
∴ Sum of the roots = 3 + k/2 = -(p)/1 = -p    …(1)
∴ Product of the roots = 3 × k/2 = (p)/1 = p    …(2)

(1) + (2)
⇒ 3 + k/2 + 3k/2 = p – p = 0
⇒ k = -3/2

⇒ p = -9/4    [from (2)]

Now, we need to find 8(k - p)
= $8\left(\frac{-3}{2}-\frac{-9}{4}\right)$
= $8\left(\frac{3}{4}\right)$
= 6

Hence, 6.

Concept:

Sum & Product of roots of a Quadratic Equation

Answer: 2

Video Explanation

Text Explanation :

Let p and q be the other two real roots of the given cubic equation.

Product of the three roots of the given cubic equation = - $\frac{2}{1}$ = -2 × p × q
⇒ q = 1/p

∴ The three roots are -2, p and 1/p

Sum of the three roots of the given cubic equation = - $\frac{2\mathrm{r}+1}{1}$ = -2 + p + $\frac{1}{\mathrm{p}}$
⇒ p + $\frac{1}{\mathrm{p}}$ = -2r + 1

We know than sum of a number and its reciprocal is either less than or equal to - 2 or greater than or equal to 2.

⇒ 2 ≤ p + $\frac{1}{\mathrm{p}}$ ≤ -2

⇒ 2 ≤ -2r + 1 ≤ -2

⇒ 1 ≤ -2r ≤ -3

⇒ -1/2 ≥ r ≥ 3/2

∴ Least non-negative integral value of r is 2.

Hence, 2.

Concept:

Sum & Product of the roots of a Polynomial Equation

Answer: 6

Video Explanation

Text Explanation :

The given equation can be written as: x² - 2x + 1 + 2kx + 11 = 0
⇒ x² + (2k - 2)x + 12 = 0

The given quadratic has no real roots, hence discriminant is less than 0.
⇒ (2k - 2)² - 4 × 1 × 12 < 0
⇒ 4(k - 1)² - 48 < 0
⇒ (k - 1)² < 12

Largets integral value of k satisying above inequality is 4.

Now, we have k/4y + 9y
= 4/4y + 9y
= 1/y + 9y

We know AM ≥ GM
⇒ (1/y + 9y)/2 ≥ $\sqrt{\frac{1}{y}\times 9y}$
⇒ (1/y + 9y)/2 ≥ 3
⇒ 1/y + 9y ≥ 6

∴ Least possible value of 1/y + 9y = 6.

Hence, 6.

Concept:

No Real Roots of a Quadratic

Answer: Option A

Video Explanation

Text Explanation :

Given, x⁸ + ${\left(\frac{1}{\mathrm{x}}\right)}^8$ = 47

⇒ x⁸ + ${\left(\frac{1}{\mathrm{x}}\right)}^8$ + 2 = 49

⇒ (x⁴)² + ${\left(\frac{1}{{\mathrm{x}}^4}\right)}^2$ + $2\times {\mathrm{x}}^4\times \frac{1}{{\mathrm{x}}^4}$ = 49

⇒ ${\left({\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}\right)}^2$ = 49

⇒ ${\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}$ = 7 [-7 will be rejected as LHS should be positive]

Now, ${\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}$ + 2 = 9

⇒ ${\left({\mathrm{x}}^2\right)}^2+{\left(\frac{1}{{\mathrm{x}}^2}\right)}^2$ + $2\times {\mathrm{x}}^2\times \frac{1}{{\mathrm{x}}^2}$ = 9

⇒ ${\left({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}\right)}^2$ = 9

⇒ ${\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}$ = 3

Similarly, $\mathrm{x}+\frac{1}{\mathrm{x}}$ = $\sqrt{5}$

Cubing both sides, we get

⇒ x³ + $\frac{1}{{\mathrm{x}}^3}$ + $3\times \mathrm{x}\times \frac{1}{\mathrm{x}}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)$ = 5√5

⇒ x³ + $\frac{1}{{\mathrm{x}}^3}$ = 2√5

Again cubing both sides, we get

⇒ x⁹ + $\frac{1}{{\mathrm{x}}^9}$ + $3\times {\mathrm{x}}^3\times \frac{1}{{\mathrm{x}}^3}\left({\mathrm{x}}^3+\frac{1}{{\mathrm{x}}^3}\right)$ = 40√5

⇒ x⁹ + $\frac{1}{{\mathrm{x}}^9}$ = 34√5

Hence, option (a).

Answer: 9

Video Explanation

Text Explanation :

Let the roots of the given quandratic equation be p and q.

⇒ $\frac{1}{\mathrm{p}}-\frac{1}{\mathrm{q}}$ = $\frac{1}{3}$   ...(1)

⇒ $\frac{1}{{\mathrm{p}}^2}+\frac{1}{{\mathrm{q}}^2}$ = $\frac{5}{9}$   ...(2)

Squaring (1) we get

⇒  $\frac{1}{{\mathrm{p}}^2}+\frac{1}{{\mathrm{q}}^2}-\frac{2}{\mathrm{pq}}$ = $\frac{1}{9}$

⇒  $\frac{5}{9}-\frac{2}{\mathrm{pq}}$ = $\frac{1}{9}$

⇒ pq = 9/2 = c

From (2) we get

⇒ 9(p² + q²) = 5p²q² 

⇒ 9(p² + q²) = 5(pq)²  

⇒ 9((p + q)² - 2pq) = 5(9/2)²  

⇒ (p + q)² - 9 = 45/4

⇒ (p + q)² = 81/4

⇒ (p + q) = ± 9/2 = - b

largest possible value of b = 9/2

∴ Largest possible value of a + b = 9/2 + 9/2 = 9.

Hence, 9.

Answer: Option C

Video Explanation

Text 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).

Answer: Option A

Video Explanation

Text 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) = $\frac{3}{2}$(-2)² – 6 × - 2 + 6 = 6 + 12 + 6 = 24

Hence, option (a).

Answer: Option B

Video Explanation

Text 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).

Answer: 4

Video Explanation

Text Explanation :

Given, (x² - $10{)}^{x^2-3x-10}$

For this to be equal to 1

Case 1: x² - 10 = 1

⇒ x = $\sqrt{11}$
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.

Answer: Option C

Video Explanation

Text 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).

Answer: Option C

Video Explanation

Text 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, $\left(\frac{\mathrm{b}}{\mathrm{m}}+\frac{\mathrm{c}-2\mathrm{b}}{\mathrm{n}}\right)$ 

= $\left(\frac{-6\mathrm{a}}{-8\mathrm{a}}+\frac{\mathrm{a}+12\mathrm{a}}{4\mathrm{a}}\right)$

= $\frac{3}{4}+\frac{13}{4}$ = 4

Hence, option (c).

Answer: Option C

Video Explanation

Text 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).

Answer: Option A

Video Explanation

Text 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).

Answer: Option A

Video Explanation

Text Explanation :

Given, f(x) = $\frac{x^2+2x+4}{2x^2+4x+9}$

⇒ f(x) = $\frac{1}{2}\left[\frac{2x^2+4x+8}{2x^2+4x+9}\right]$

⇒ f(x) = $\frac{1}{2}\left[\frac{2x^2+4x+9-1}{2x^2+4x+9}\right]$

⇒ f(x) = $\frac{1}{2}\left[1-\frac{1}{2x^2+4x+9}\right]$

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) = $\frac{1}{2}\left[1-\frac{1}{7}\right]$ = $\frac{3}{7}$

f(x) will be maximum when (2x² + 4x + 9) is maximum.

Maximum value of (2x² + 4x + 9) will be ∞.

∴ Maximum value of f(x) = $\frac{1}{2}\left[1-\frac{1}{\infty }\right]$ = $\frac{1}{2}$

∴ Range of f(x) = $\left[\frac{3}{7},\frac{1}{2}\right)$

Upper value of ½ is in open bracket as value of (2x² + 4x + 9) will never actually be ∞.
 
Hence, option (a).

Answer: 34

Video Explanation

Text 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.

Answer: Option B

Video Explanation

Text Explanation :

Let the price of a smaller shirt be Rs. p. Hence, price of a larger shirt will be Rs. (p + 50).

⇒ $\frac{5000}{p+50}$ + $\frac{1800}{p}$ = 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).

Answer: Option D

Video Explanation

Text Explanation :

Given, ${\left(x^2-7x+11\right)}^{(x^2-13x+42)}$ = 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).

Answer: 1

Video Explanation

Text 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.

Answer: Option D

Video Explanation

Text 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).

Answer: Option D

Video Explanation

Text 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).

Answer: Option D

Video Explanation

Text 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)
⇒ n^3/2 ≥ 2^3/2
⇒ 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).