# Algebra - Quadratic Equations - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic Algebra - Quadratic Equations. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 2021 QA Slot 1 | Algebra - Quadratic Equations**

If r is a constant such that |x^{2} – 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, |x^{2} – 4x - 13| = r

∴ x^{2} – 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**: x^{2} – 4x - 13 = r has equal roots, i.e., Discriminant = 0

⇒ x^{2} – 4x – 13 - r = 0 had D = 0

⇒ D = 16 – 4(-13 - r) = 0

⇒ 16 + 52 + 4r = 0

⇒ r = - 17

**Case 2**: x^{2} – 4x - 13 = - r has equal roots, i.e., Discriminant = 0

⇒ x^{2} – 4x – 13 + r = 0 had D = 0

⇒ D = 16 – 4(-13 + r) = 0

⇒ 16 + 52 - 4r = 0

⇒ r = 17

Hence, option (c).

Workspace:

**CAT 2021 QA Slot 2 | Algebra - Quadratic Equations**

Suppose one of the roots of the equation ax^{2} – bx + c = 0 is 2 + √3, where a, b and c are rational numbers and a ≠ 0. If b = c^{3} 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 = c^{3}

⇒ 4a = a^{3}

⇒ a^{2} = 4

⇒ a = ±2

∴ |a| = 2

Hence, option (a).

Workspace:

**CAT 2021 QA Slot 2 | Algebra - Quadratic Equations**

For all real values of x, the range of the function f(x) = $\frac{{x}^{2}+2x+4}{2{x}^{2}+4x+9}$ is

- A.
$\left[\frac{3}{7},\frac{1}{2}\right)$

- B.
$\left[\frac{3}{7},\frac{8}{9}\right)$

- C.
$\left[\frac{4}{9},\frac{1}{2}\right]$

- D.
$\left(\frac{3}{7},\frac{1}{2}\right)$

Answer: Option A

**Explanation** :

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

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

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

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

f(x) will be minimum when (2x^{2} + 4x + 9) is minimum.

Now, 2x^{2} + 4x + 9 will be minimum when x = -(4)/2 × 2 = -1

∴ Minimum value of 2x^{2} + 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 (2x^{2} + 4x + 9) is maximum.

Maximum value of (2x^{2} + 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 (2x^{2} + 4x + 9) will never actually be ∞.

Hence, option (a).

Workspace:

**CAT 2021 QA Slot 3 | Algebra - Quadratic Equations**

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

⇒ 10x^{2} + 42x + 36 = 40x^{2}

⇒ 30x^{2} - 42x - 36 = 0

⇒ 5x^{2} – 7x – 6 = 0

⇒ 5x^{2} – 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:

**CAT 2021 QA Slot 3 | Algebra - Quadratic Equations**

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).

⇒ $\frac{5000}{p+50}$ + $\frac{1800}{p}$ = 64

⇒ 6800p + 90000 = 64p2 + 3200p

⇒ 4p^{2} - 225p - 5625 = 0

⇒ 4p^{2} - 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:

**CAT 2020 QA Slot 1 | Algebra - Quadratic Equations**

How many distinct positive integer-valued solutions exist to the equation ${\left({x}^{2}-7x+11\right)}^{({x}^{2}-13x+42)}$ = 1?

- A.
2

- B.
8

- C.
4

- D.
6

Answer: Option D

**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

⇒ 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:

**CAT 2020 QA Slot 1 | Algebra - Quadratic Equations**

The number of distinct real roots of the equation ${\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)}^{2}-3\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+2$ = 0 equals

Answer: 1

**Explanation** :

Let x + 1/x = y

⇒ y^{2} – 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:

**CAT 2020 QA Slot 2 | Algebra - Quadratic Equations**

In how many ways can a pair of integers (x , a) be chosen such that x^{2} − 2|x| + |a - 2| = 0?

- A.
4

- B.
5

- C.
6

- D.
7

Answer: Option D

**Explanation** :

Given, x^{2} − 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:

**CAT 2020 QA Slot 2 | Algebra - Quadratic Equations**

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:

**CAT 2020 QA Slot 3 | Algebra - Quadratic Equations**

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.

⇒ m^{2} – 8n ≥ 0 …(1)

Also, we have the quadratic equation, x² + 2nx + m = 0

Since its roots are real, that means D ≥ 0.

⇒ (2n)^{2} – 4m ≥ 0

⇒ 4n^{2} – 4m ≥ 0

⇒ n^{2} ≥ m …(2)

Let’s take cases now.

If n = 1,

⇒ m = 1 (from (2))

But this violates eq. (1).

Hence, n ≠ 1.

If n = 2,

According to eq. (1): m ≤ 4.

According to eq. (2): m ≥ 4.

Hence, the only possible value of m is 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:

**CAT 2019 QA Slot 1 | Algebra - Quadratic Equations**

The number of solutions to the equation |x|(6${x}^{2}$ + 1) = 5${x}^{2}$ 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${x}^{2}$ + 1 = 5x ⇒ 6${x}^{2}$ − 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${x}^{2}$ + 1 = −5x ⇒ 6${x}^{2}$ + 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:

**CAT 2019 QA Slot 1 | Algebra - Quadratic Equations**

The product of the distinct roots of ∣x^{2} − x − 6∣ = x + 2 is

- A.
-16

- B.
-4

- C.
-8

- D.
-24

Answer: Option A

**Explanation** :

∣x^{2} − 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 1.

Workspace:

**CAT 2019 QA Slot 2 | Algebra - Quadratic Equations**

The real root of the equation 2^{6x} + 2^{3x+2} - 21 = 0 is

- A.
$\frac{{\mathrm{log}}_{2}\left(3\right)}{3}$

- B.
log

_{2}9 - C.
$\frac{{\mathrm{log}}_{2}\left(7\right)}{3}$

- D.
log

_{2}27

Answer: Option A

**Explanation** :

Given: 2^{6x} + 2^{3x+2} - 21 = 0

Replace 2^{3x} with y

So, 2^{6x} = y^{2}

Now, 2^{6x} + 2^{3x+2} - 21 = 0 can be rewritten as y^{2} + 2^{2} × 2^{3x} - 21 = 0

y^{2} + 4y - 21 = 0

Solving the above quadratic equation,

(y + 7) (y - 3) = 0

So, y = -7 or +3

y = - 7 is rejected (since, y = 2^{3x} which should always be positive)

⇒ 2^{3x} = 3

Taking log on both sides,

log_{2}3 = 3x

x = $\frac{{\mathrm{log}}_{2}\left(3\right)}{3}$

Hence, option (1).

Workspace:

**CAT 2019 QA Slot 2 | Algebra - Quadratic Equations**

Let A be a real number. Then the roots of the equation x^{2} - 4x - log_{2}A = 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 x^{2} − 4x – log_{2}A = 0,

D = (-4)^{2} - (4 × 1 × (-log_{2}A)) > 0

⇒ 16 + 4 × log_{2}A > 0

⇒ log_{2}A > -4

⇒ A > 2^{-4}

⇒ A > 1/16

Hence, option (3).

Workspace:

**CAT 2019 QA Slot 2 | Algebra - Quadratic Equations**

The quadratic equation x^{2} + bx + c = 0 has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of b^{2} + c?

- A.
3721

- B.
549

- C.
361

- D.
427

Answer: Option B

**Explanation** :

Given quadratic equation is x^{2} + bx + c = 0

Sum of roots = -b

⇒ -b = 4a + 3a

⇒ -b = 7a

Product of the roots = c

⇒ c = 4a × 3a

⇒ c = 12a^{2}

Now, b^{2} = 49a^{2} and c =12a^{2}

Hence, b^{2} + c = 49a^{2} + 12a^{2}

⇒ b^{2} + c = 61a^{2}

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 × 3^{2}

Thus, possible value of b^{2} + c = 549.

Hence, option (2).

Workspace:

**CAT 2018 QA Slot 1 | Algebra - Quadratic Equations**

If u^{2} + (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** :

u^{2} + (u – 2v – 1)^{2} = −4v(u + v)

u^{2} + (u – 2v – 1)^{2} + 4v(u + v) = 0

(u – 2v – 1)^{2} + u^{2} + 4uv + 4v^{2} = 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 2.

Workspace:

**CAT 2018 QA Slot 2 | Algebra - Quadratic Equations**

If a and b are integers such that 2x^{2} − ax + 2 > 0 and x^{2} − 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 2x^{2} – ax + 2 > 0 for all value of x, the graph of 2x^{2} – ax + 2 is above the x-axis or the roots of the quadratic equations

2x^{2} – ax + 2 = 0 are imaginary or its discriminant is less than 0. Therefore, we have a^{2} – 16 < 0 or a^{2} < 16 or –4 < a < 4. Therefore the largest possible value of a is 3.

If x^{2} – bx + 8 ≥ 0, the discriminant of the equation is less than or equal to zero.

Therefore, ${b}^{2}-32\le 0or-4\sqrt{2}\le b\le 4\sqrt{2}or-5.64\le b\le 5.64.$

Therefore the smallest possible value of b is –5.

Therefore the maximum possible value of 2a - 6b = 2(3) – 6(–5)=36.

Answer: 36

Workspace:

**CAT 2018 QA Slot 2 | Algebra - Quadratic Equations**

The smallest integer n such that n^{3} – 11n^{2 }+ 32n – 28 > 0 is

Answer: 8

**Explanation** :

The sum of the coefficients of n^{3} -11n^{2} + 32n - 28 is not zero.

Similarly the sum of the odd coefficients of n^{3} -11n^{2} + 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, n^{3} – 11n^{2} + 32n – 28 = 8 – 44 + 64 – 28 = 0.

Therefore (n – 2) is a factor of the polynomial n^{3} - 11n^{2} + 32n – 28.

We have, (n - 2) is a factor of n^{3} – 11n^{2} + 32n – 28

∴ (n^{2} – 9n + 14) = (n – 2)(n – 2)(n - 7) = (n – 2)^{2} (n – 7).

Therefore n = 2 and n = 7 are the two roots of the equation n^{3 }- 11n^{2} + 32n - 28 = 0.

Also, at n = 0, n^{3} – 11n^{2} + 32n - 28 = –28 < 0. Therefore the curve touches the x-axis at 2 and intersects the x-axis at 7.

Therefore the smallest integer for which n^{3 }- 11n^{2} + 32n - 28 > 0 is 8.

Answer: 8

Workspace:

**CAT 2017 QA Slot 1 | Algebra - Quadratic Equations**

If 9^{2x–1} – 81^{x-1} = 1944, then x is

- A.
3

- B.
9/4

- C.
4/9

- D.
1/3

Answer: Option B

**Explanation** :

9^{2x–1} – 81^{x-1} = 1944

⇒ 9^{2x–2 }× 9 – 9^{2x-2} = 1944

⇒ 9^{2x–2} (9 – 1) = 8 × 243 = 8 × 3^{5}

⇒ 3^{4x–4} × 8 = 8 × 3^{5}

⇒ ∴ 4x – 4 = 5

∴ x = 9/4.

Hence, option 2.

Workspace:

**CAT 2017 QA Slot 2 | Algebra - Quadratic Equations**

The minimum possible value of the squares of the roots of the equation

*x*^{2} + (*a* + 3)*x* – (*a* + 5) = 0 is

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

In the given equation, let α and β be the 2 roots of the equation Now *α* + β = - α – 3

Also *α *β= -*a* – 5

(α + β)^{2} = α^{2}+ β^{2 }+2 α β

*a*^{2} + 6*a* + 9 = α^{2} + β^{2} - 2*a* – 10

*a*^{2} + 8*a* + 19 = α^{2} + β^{2}

(*a*^{2} + 8*a* + 16) + 3 = α^{2} + β^{2}

α^{2} + β^{2} = (*a* + 4)^{2} + 3

Now (*a* + 4)^{2} ≥ 0

So minimum value of (*a* + 4)^{2} will be 0 when *a* = -4

At *a* = -4, the of squares of the roots of the equation i.e.,

α^{2} + β^{2} will be 3.

Hence, option 3.

Workspace:

**Answer the next 2 questions based on the information given below.**

Let f(x) = ax^{2} + bx + c, where, a, b and c are certain constants and a ≠ 0. It is known that f(5) = −3f(2) and that 3 is a root of f(x) = 0.

**CAT 2008 QA | Algebra - Quadratic Equations**

What is the correct root of f(x) = 0?

- A.
-7

- B.
-4

- C.
2

- D.
6

- E.
Cannot be determined

Answer: Option B

**Explanation** :

∵ 3 is a root of f(x) = 0,

∴ 9a + 3b + c = 0 …(i)

Also,

f(5) = −3f(2)

∴ 25a + 5b + c = −3(4a + 2b + c)

∴ 37a + 11b + 4c = 0 …(ii)

On solving equations (i) and (ii), we get,

a – b = 0

∴ a = b

We know that sum of the roots of a quadratic equation (ax^{2} + bx + c = 0) is –b/a

∴ 3 + other root = −1

∴ Other root = −4

Hence, option (b).

Workspace:

**CAT 2008 QA | Algebra - Quadratic Equations**

What is the value of a + b + c?

- A.
9

- B.
14

- C.
13

- D.
37

- E.
Cannot be determined

Answer: Option E

**Explanation** :

The roots at f(x) = 0 are 3 and −4

∴ The equation can be written as (x – 3)(x + 4) = 0

Or, x^{2} − x + 12 = 0

The co-efficient of x^{2} is 1 here, but all equations which are multiple of this equation will also have same roots.

For example,

10(x^{2} − x + 12) = 0 will also have same roots

∴ (a + b + c) cannot be determined uniquely.

Hence, option (e).

Workspace:

**CAT 2008 QA | Algebra - Quadratic Equations**

If the roots of the equation x^{3}− ax^{2} + bx – c = 0 are three consecutive integers, then what is the smallest possible value of b?

- A.
$\frac{-1}{\sqrt{3}}$

- B.
-1

- C.
0

- D.
1

- E.
$\frac{1}{\sqrt{3}}$

Answer: Option B

**Explanation** :

If p, q and r are the roots of a cubic equation ax^{3} + bx^{2} + cx + d = 0,

then pq + pr + qr = c/a

If a is 1, then pq + qr + pr = c

Comparing the equation ax^{3} + bx^{2} + cx + d = 0 with the equation in the question x^{3}− ax^{2} + bx – c = 0, we get

pq + qr + pr = b

Let the three roots of the given equation be (n – 1), n and (n + 1).

∴ (n – 1)n + n(n + 1) + (n – 1)(n + 1) = b

∴ n^{2} – n + n^{2 }+ n + n^{2} – 1 = b

∴ 3n^{2} – 1 = b

∵ n^{2} ≥ 0, minimum value of b occurs at n = 0

∴ Minimum value of b = –1

Hence, option (b).

Workspace:

**CAT 2008 QA | Algebra - Quadratic Equations**

Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers?

- A.
1 ≤ m ≤ 3

- B.
4 ≤ m ≤ 6

- C.
7 ≤ m ≤ 9

- D.
10 ≤ m ≤ 12

- E.
13 ≤ m ≤ 15

Answer: Option A

**Explanation** :

Let the three numbers be (a – 2), (a – 1) and a.

∴ (a – 2) + (a – 1)^{2} + a^{3 }= p^{2}

Where p is the sum of the three integers.

Now, a – 2 + a^{2} – 2a + 1 + a^{3 }= p^{2}

∴ a^{3} + a^{2} – a – 1 = p^{2}

∴ a^{2}(a + 1) – 1(a + 1) = p^{2}

∴ (a^{2} – 1)(a + 1) = p^{2}

∴ (a + 1)^{2}(a – 1) = p^{2}

For the above condition to be satisfied, (a – 1) must be a perfect square.

The smallest possible value for a - 1 is 4, since (a – 2) cannot be zero, giving us (a − 2) = 3

The minimum of the three is therefore 3.

Hence, option (a).

Workspace:

**CAT 2007 QA | Algebra - Quadratic Equations**

A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f(x) at x = 10?

- A.
–119

- B.
–159

- C.
–110

- D.
–180

- E.
–105

Answer: Option B

**Explanation** :

Let f(x) = px^{2} + qx + k, where p, q and k are integers, and p ≠ 0

∴ f(0) = k = 1

∴ f(x) = px^{2} + qx + 1

f(x) = px^{2} + qx + k

∴ f'(x) = 2px + q

When f’(x) = 0, x = −q/2p

f(x) attains maximum at x = 1

∴ q = −2p

f(1) = p + q + 1 = 3

∴ 1 – p = 3

∴ p = −2

∴ q = 4

∴ f(x) = −2x^{2 }+ 4x + 1

∴ f(10) = −200 + 40 + 1 = −159

Hence, option (b).

Workspace:

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