Search Previous Year Questions

Given:

If x² - 5x + 4 is a factor of x⁴ - px² + q.

Formula used:

If (x² - 5x + 4) is a factor of (x⁴ - px² + q), then the roots of x² - 5x + 4 (i.e., 1 and 4) must satisfy x⁴ - px² + q = 0.

Calculation:

Roots of x² - 5x + 4 are x = 1 and x = 4.

Substitute x = 1 in x⁴ - px² + q = 0

⇒ 1⁴ - p(1²) + q = 0

⇒ 1 - p + q = 0

⇒ q = p - 1 ----- (1)

Substitute x = 4 in x⁴ - px² + q = 0

⇒ 4⁴ - p(4²) + q = 0

⇒ 256 - 16p + q = 0

⇒ q = 16p - 256 ------ (2)

Equate q values from 1 and 2

⇒ p - 1 = 16p - 256

⇒ 255 = 15p

⇒ p = 17

Substitute p = 17 into q = p - 1

⇒ q = 17 - 1

⇒ q = 16

∴ The values of p and q are 17 and 16 respectively.

The correct answer is option (1).

Given:

Two quadratic equations: $px^2 + px + 4 = 0$ and $x^2 + qx + q = 0$

Common root = 2

Formula used:

For a quadratic equation $ax^2 + bx + c = 0$, if a root is given, substitute it into the equation to find the unknowns.

Calculation:

Substituting the root 2 into the first equation:

$p(2)^2 + p(2) + 4 = 0$

⇒ $4p + 2p + 4 = 0$

⇒ $6p + 4 = 0$

⇒ $6p = -4$

⇒ $p = -\frac{2}{3}$

Substituting the root 2 into the second equation:

$(2)^2 + q(2) + q = 0$

⇒ $4 + 2q + q = 0$

⇒ $4 + 3q = 0$

⇒ $3q = -4$

⇒ $q = -\frac{4}{3}$

Now, calculate $p + q$:

$p + q = -\frac{2}{3} - \frac{4}{3}$

⇒ $p + q = -\frac{6}{3}$

⇒ $p + q = -2$

∴ The correct answer is option (2).

Given:

Find the HCF of: f(x) = x⁸ + x⁴ + 1 and g(x) = x⁴ + x² + 1

Formula used:

a² + ab + b² = (a + b)² − ab

f(x) = x⁸ + x⁴ + 1 = (x⁴)² + x⁴ × 1 + 1²

So f(x) = x⁸ + x⁴ + 1 = a² + ab + b² where a = x⁴, b = 1

⇒ f(x) = (x⁴)² + x⁴ × 1 + 1² = x⁸ + x⁴ + 1

Similarly, g(x) = x⁴ + x² + 1 = a² + ab + b² where a = x², b = 1

So: f(x) = a² + ab + b² where a = x⁴, b = 1

g(x) = a² + ab + b² where a = x², b = 1

Now, we know that: a² + ab + b² = (a + b)² − ab

So: f(x) = (x⁴ + 1)² − x⁴

f(x) = (x⁴ + x² + 1)(x⁴ − x² + 1)

f(x) = x⁸ + x⁴ + 1 = (x⁴ + x² + 1)(x⁴ − x² + 1)

g(x) = x⁴ + x² + 1

Since x4 + x2 + 1 is common in both terms.

∴ HCF is: x⁴ + x² + 1

Given:

An arc AB of a circle subtends an angle x radian at the center O.

The area of sector AOB = Half of the square of the length of arc AB.

Formula used:

Area of sector = (1/2) × r² × x

Length of arc = r × x

Given condition: (1/2) × r² × x = (1/2) × (r × x)²

Calculations:

(1/2) × r² × x = (1/2) × r² × x²

⇒ r² × x = r² × x²

⇒ x = x²

⇒ x² - x = 0

⇒ x(x - 1) = 0

⇒ x = 0 or x = 1

Since x cannot be 0 (as it represents an angle),

∴ x = 1 radian.

Correct answer is option (3).

Given:

p and q are prime numbers

Statement I: Their LCM is always an odd number

Formula used:

LCM of two numbers is odd only if both numbers are odd

Among prime numbers, 2 is the only even prime

Calculation:

If p = 2, q = 3 ⇒ LCM = 6 (even)

So, LCM is not always odd

Statement I is incorrect

Statement II: Sum of their LCM and HCF is always even

HCF of two distinct primes = 1

LCM of p and q = p × q

Sum = p × q + 1

Check:

p = 2, q = 3 ⇒ LCM = 6, HCF = 1 ⇒ Sum = 6 + 1 = 7  odd

p = 3, q = 5 ⇒ LCM = 15, HCF = 1 ⇒ Sum = 15 + 1 = 16  even

p = 3, q = 7 ⇒ 21 + 1 = 22  even

But counter example: p = 2, q = 3 ⇒ 6 + 1 = 7 

Statement II is incorrect

∴ The correct answer is: Neither I nor II.

Given:

Frequency distribution of X with total observations = 205

x = 3, 5, 6, 7

Frequencies = f, f + 2, f - 3, f + 6

Formula used:

Sum of all frequencies = Total observations

Calculations:

f + (f + 2) + (f - 3) + (f + 6) = 205

⇒ 4f + 5 = 205

⇒ 4f = 200

⇒ f = 50

∴ The value of f is 50.

Given:

Frequency distribution of X:

x = 3, 5, 6, 7

Frequencies = 50, 52, 47, 56 (from previous calculation f = 50)

Total observations = 205

Formula used:

Median for grouped data = L + [(N/2 - CF)/f] × h

Where, L = lower boundary of median class, N = total observations, CF = cumulative frequency before median class, f = frequency of median class, h = class width

Calculations:

Cumulative frequencies:

3 → 50

5 → 50 + 52 = 102

6 → 102 + 47 = 149

7 → 149 + 56 = 205

N/2 = 205/2 = 102.5 → lies in class x = 6 (since CF before = 102)

Assuming class width h = 1, L = 6

Median = 6 + [(102.5 - 102)/47] × 1

⇒ Median = 6 + [0.5/47]

⇒ Median ≈ 6 + 0.0106 ≈ 6

∴ The median of the frequency distribution is approximately 6.

Given:

Frequency distribution of X:

x = 3, 5, 6, 7

Frequencies = 50, 52, 47, 56

Total observations = 205

Formula used:

Mode for grouped data = L + [(f1 - f0) / (2f1 - f0 - f2)] × h

Where, L = lower boundary of modal class, f1 = frequency of modal class, f0 = frequency of previous class, f2 = frequency of next class, h = class width

Calculations:

Highest frequency = 56 → modal class x = 7

f1 = 56, f0 = 47 (previous), f2 = 0 (no class after 7), h = 1, L = 7

Mode = 7 + [(56 - 47)/(2×56 - 47 - 0)] × 1

⇒ Mode = 7 + [9 / (112 - 47)]

⇒ Mode = 7 + 9 / 65 ≈ 7 + 0.138 ≈ 7

∴ The mode of the frequency distribution is approximately 7.

Given:

Frequency distribution of X:

x = 3, 5, 6, 7

Frequencies = 50, 52, 47, 56

Formula used:

For discrete data, suitable graphs are bar graph or histogram (for continuous intervals); here X is discrete.

Calculations:

Plot x-values on x-axis (3, 5, 6, 7)

Plot frequencies on y-axis (50, 52, 47, 56)

Draw bars corresponding to each x-value with height equal to its frequency

∴ The most appropriate graphical representation is a bar graph.

Given:

Frequency distribution of X:

x = 3, 5, 6, 7

Frequencies = 50, 52, 47, 56

Total observations = 205

Formula used:

Mean = Σ(f × x) ÷ Σf

Calculations:

f × x:

3 × 50 = 150

5 × 52 = 260

6 × 47 = 282

7 × 56 = 392

Σ(f × x) = 150 + 260 + 282 + 392 = 1084

Mean = 1084 ÷ 205 ≈ 5.29

∴ The mean of the frequency distribution is approximately 5.29.