Search Previous Year Questions

Given:

p(x) leaves a remainder of 2 when divided by (x - 1).

p(x) leaves a remainder of 1 when divided by (x - 2).

We need the remainder when p(x) is divided by (x - 1)(x - 2).

Formula used:

The remainder when p(x) is divided by (x - 1)(x - 2) will be a linear polynomial of the form ax + b.

From the given conditions: p(1) = 2 and p(2) = 1.

Calculations:

Assume remainder = ax + b.

From p(1) = 2:

⇒ a(1) + b = 2

⇒ a + b = 2

From p(2) = 1:

⇒ a(2) + b = 1

⇒ 2a + b = 1

Solving the two equations:

1. a + b = 2
2. 2a + b = 1

⇒ Subtract equation 1 from equation 2:

⇒ (2a + b) - (a + b) = 1 - 2

⇒ a = -1

Substitute a = -1 in equation 1:

⇒ -1 + b = 2

⇒ b = 3

Final Remainder:

Remainder = ax + b = -x + 3 = 3 - x

∴ The correct answer is option (3).

Given:

I. x + (1/x) > 1

II. (x + (1/x))² > 2

III. (x + (1/x))⁴ > 9

Formula used:

For positive real numbers, x + (1/x) ≥ 2 when x > 0.

Calculation:

Check I: x + (1/x) > 1

⇒ x > 0 ensures x + (1/x) ≥ 2 > 1

⇒ I is correct.

Check II: (x + (1/x))² > 2

⇒ x + (1/x) ≥ 2

⇒ (x + (1/x))² ≥ 2² = 4

⇒ II is correct.

Check III: (x + (1/x))⁴ > 9

⇒ (x + (1/x)) ≥ 2

⇒ (x + (1/x))⁴ ≥ 2⁴ = 16 > 9

⇒ III is correct.

∴ The correct answer is option (4).

Given:

Let q² - 5p - 4 be negative, where q > p and p, q are natural numbers.

Formula used:

For q2 - 5p - 4 < 0 , solve for p, considering q > p.

Calculation:

To find the largest p:

Let q = p + 1 (since q > p ).

⇒ (p + 1)² - 5p - 4 < 0

⇒ p² + 2p + 1 - 5p - 4 < 0

⇒ p² - 3p - 3 < 0

Factorize or test values for p:

For p = 3:

⇒ 3² - 3×3 - 3 = 9 - 9 - 3 = -3 (negative).

For p = 4:

⇒ 4² - 3×4 - 3 = 16 - 12 - 3 = 1 (positive).

Largest p is 3 where q² - 5p - 4 < 0 .

∴ The correct answer is option (1).

Given:

x, y are natural numbers < 20

x, y, (x + y), (x − y) are prime numbers

Formula used:

Prime rule: Except 2, all primes are odd

Calculation:

x + y is prime

⇒ Sum of two odd primes is even (not prime)

⇒ One number must be 2

Since (x − y) is prime

⇒ x > y

⇒ y = 2

Now x is prime < 20 and x > 2

Check values:

x = 3 ⇒ x − y = 1 (not prime) 

x = 5 ⇒ x − y = 3, x + y = 7 

x = 7 ⇒ x + y = 9 

x = 11 ⇒ x − y = 9 

x = 13 ⇒ x + y = 15 

x = 17 ⇒ x − y = 15 

x = 19 ⇒ x + y = 21 

Valid set:

(x, y, x + y, x − y) = (5, 2, 7, 3)

∴ Number of possible combinations = One (Option 1).

Given:

(x + 1)(x + p)(x² + p²) = x⁴ − 1

Formula used:

x⁴ − 1 = (x − 1)(x + 1)(x² + 1)

Calculation:

x⁴ − 1 = (x − 1)(x + 1)(x² + 1)

Given expression:

(x + 1)(x + p)(x² + p²)

Common factor: (x + 1)

⇒ Remaining factors must match

⇒ (x + p) = (x − 1)

⇒ p = −1

Check:

p² = 1

⇒ (x² + p²) = (x² + 1)

Hence LHS = (x + 1)(x − 1)(x² + 1)

= x⁴ − 1

∴ The value of p is −1.

Given:

(2 + √3)^x + (2 - √3)^x = 2

Formula used:

Let a = 2 + √3, b = 2 - √3. Then a × b = 1, so b^x = a^-x.

Calculations:

Let y = (2 + √3)^x. Then given ⇒ y + 1/y = 2.

⇒ y + 1/y = 2

⇒ y² + 1 = 2y

⇒ y² - 2y + 1 = 0

⇒ (y - 1)² = 0

⇒ y = 1 ⇒ (2 + √3)^x = 1 ⇒ x = 0

Therefore, (2 + √3)^x - (2 - √3)^x = 1 - 1 = 0

∴ The required value is 0.

Given:

$\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{5}{6}$

$\dfrac{1}{a^2}+\dfrac{1}{b^2}=\dfrac{13}{36}$

Formula used:

$\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2 = \dfrac{1}{a^2}+\dfrac{1}{b^2}+2\cdot\dfrac{1}{ab}$

$\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3 = \dfrac{1}{a^3}+\dfrac{1}{b^3}+3\cdot\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\cdot\dfrac{1}{ab}$

Calculation:

⇒ S = $\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{5}{6}$

⇒ S² = $\dfrac{25}{36}$

⇒ Use S² = (1/a²+1/b²) + 2×(1/ab)

⇒ $\dfrac{25}{36} = \dfrac{13}{36} + 2\times\dfrac{1}{ab}$

⇒ 2×(1/ab) = $\dfrac{12}{36}=\dfrac{1}{3}$

⇒ $\dfrac{1}{ab}=\dfrac{1}{6}$

⇒ Now S³ = $\dfrac{125}{216}$

⇒ Use S³ = (1/a³+1/b³) + 3×S×(1/ab)

⇒ $\dfrac{125}{216} = \left(\dfrac{1}{a^3}+\dfrac{1}{b^3}\right) + 3\times\dfrac{5}{6}\times\dfrac{1}{6}$

⇒ 3×S×(1/ab) = 3×(5/6)×(1/6) = $\dfrac{15}{36}=\dfrac{5}{12}=\dfrac{90}{216}$

⇒ $\dfrac{1}{a^3}+\dfrac{1}{b^3} = \dfrac{125}{216}-\dfrac{90}{216}=\dfrac{35}{216}$

∴ The correct answer is $\dfrac{35}{216}$.

Shortcut Trick

To find the remainder when a polynomial is divided by (x² + 1), substitute x² = −1 into the polynomial.

Given polynomial: x⁶

Rewrite as: (x²)³

Substitute x² = −1 ⇒ (−1)³ = −1

∴ The correct answer is −1.

Alternate Method

Given: Dividend = x⁶, Divisor = x² + 1

Formula Used: Remainder Theorem for polynomials.

⇒ Let f(x) = x⁶

⇒ We use the identity: a³ + b³ = (a + b)(a² − ab + b²)

⇒ Consider x⁶ + 1 = (x²)³ + 1³

⇒ x⁶ + 1 = (x² + 1)(x⁴ − x² + 1)

⇒ x⁶ = (x² + 1)(x⁴ − x² + 1) − 1

⇒ This is in the form: Dividend = (Divisor × Quotient) + Remainder

⇒ Comparing the terms, Remainder = −1

∴ The correct answer is −1.

Additional Information

Remainder Theorem

If a polynomial f(x) is divided by (x − a), the remainder is f(a).

Imaginary Unit Substitution

When dividing by (x² + 1), the roots are x = ±i. Substituting x² = −1 is a direct application of this property.

Degree of Remainder

The degree of the remainder is always strictly less than the degree of the divisor. For a quadratic divisor, the remainder is at most linear (ax + b).

Given:

(x + 2) is a factor of which one of the following polynomials:

Formula used:

If (x + c) is a factor of a polynomial f(x), then f(-c) = 0.

Calculation:

Here, (x + 2) is the factor, so c = -2. Substitute x = -2 into the given polynomials:

Option 1:

f(x) = x⁵ - 4x⁴ - 3x³ + 8x² - 14x + 12

f(-2) = (-2)⁵ - 4(-2)⁴ - 3(-2)³ + 8(-2)² - 14(-2) + 12

= -32 - 64 + 24 + 32 + 28 + 12

= 0

Since f(-2) = 0, (x + 2) is a factor of Option 1.

Option 2:

f(x) = x⁵ + 4x⁴ - 3x³ + 8x² - 14x + 12

f(-2) = (-2)⁵ + 4(-2)⁴ - 3(-2)³ + 8(-2)² - 14(-2) + 12

= -32 + 64 + 24 + 32 + 28 + 12

= 128 (Not 0)

Option 3:

f(x) = x⁵ - 4x⁴ + 3x³ + 8x² - 14x + 12

f(-2) = (-2)⁵ - 4(-2)⁴ + 3(-2)³ + 8(-2)² - 14(-2) + 12

= -32 - 64 - 24 + 32 + 28 + 12

= -48 (Not 0)

Option 4:

f(x) = x⁵ - 4x⁴ - 3x³ + 8x² + 14x + 12

f(-2) = (-2)⁵ - 4(-2)⁴ - 3(-2)³ + 8(-2)² + 14(-2) + 12

= -32 - 64 + 24 + 32 - 28 + 12

= -56 (Not 0)

∴ The correct answer is Option 1.

Given:

log₁₀2 = 0.301

log₁₀3 = 0.477

Find number of digits in 60⁶⁰

Formula used:

The number of digits in any number ‘n’ is equal to + 1, where X is a floor function and it gives the largest integer less than or equal to x.

Calculations:

log₁₀60 = log₁₀(6 × 10) = log₁₀6 + 1

log₁₀6 = log₁₀2 + log₁₀3 = 0.301 + 0.477 = 0.778

⇒ log₁₀60 = 0.778 + 1 = 1.778

⇒ log₁₀60⁶⁰ = 60 × 1.778 = 106.68

⇒ No. of digits in (60)⁶⁰ = 106.68 + 1 = 106 + 1 = 107

∴ Number of digits = 107.