Let the mean and the variance of seven observations 2, 4, alpha, 8, beta, 12, 14, alpha <= beta, be 8 and 16 respectively. Then the quadratic equation whose roots are 3alpha + 2 and 2beta + 1 is :

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q8 – JEE-MAIN 2026

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Home Maths StatisticsQuestion #429

Let the mean and the variance of seven observations

2, 4, α, 8, β, 12, 14

α \leq β

, be

8

and

16

respectively. Then the quadratic equation whose roots are

3 α + 2

and

2 β + 1

is :

Clue 1 of 3: Statistical Definitions

Recall the formulas for mean and variance of a set of observations.

🧠 Full Solution Path

Step 1: Calculate

α + β

using the mean✦ Active

The sum of observations is $2 + 4 + α + 8 + β + 12 + 14 = 40 + α + β$ . Given the mean is $8$ for $7$ observations, we have:

\frac{40 + α + β}{7} = 8 ⟹ 40 + α + β = 56 ⟹ α + β = 16

Step 2: Calculate

α^{2} + β^{2}

using the variance○ Expand

The sum of squares of observations is $2^{2} + 4^{2} + α^{2} + 8^{2} + β^{2} + 12^{2} + 14^{2} = 4 + 16 + α^{2} + 64 + β^{2} + 144 + 196 = 424 + α^{2} + β^{2}$ . Given variance is $16$ and mean is $8$ , using the formula $σ^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2}$ :

16 = \frac{424 + α^{2} + β^{2}}{7} - 8^{2}

16 = \frac{424 + α^{2} + β^{2}}{7} - 64

80 = \frac{424 + α^{2} + β^{2}}{7}

560 = 424 + α^{2} + β^{2} ⟹ α^{2} + β^{2} = 136

Step 3: Determine

α

and

β

and form the final quadratic equation○ Expand

We have $α + β = 16$ and $α^{2} + β^{2} = 136$ . Using the identity $(α + β)^{2} = α^{2} + β^{2} + 2 α β$ :

16^{2} = 136 + 2 α β ⟹ 256 = 136 + 2 α β ⟹ 2 α β = 120 ⟹ α β = 60

$α$ and $β$ are the roots of the quadratic equation $t^{2} - (α + β) t + α β = 0$ , which is $t^{2} - 16 t + 60 = 0$ . Factoring this equation gives $(t - 6) (t - 10) = 0$ . So, the roots are $t = 6$ or $t = 10$ . Given $α \leq β$ , we have $α = 6$ and $β = 10$ . The new roots for the required quadratic equation are:

r_{1} = 3 α + 2 = 3 (6) + 2 = 18 + 2 = 20

r_{2} = 2 β + 1 = 2 (10) + 1 = 20 + 1 = 21

The sum of the new roots is $S = r_{1} + r_{2} = 20 + 21 = 41$ . The product of the new roots is $P = r_{1} r_{2} = 20 \times 21 = 420$ . The quadratic equation is $x^{2} - S x + P = 0$ :

x^{2} - 41 x + 420 = 0

This matches option 2.

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Maths Question 8 – JEE-MAIN 2026

Clue 1 of 3: Statistical Definitions

🧠 Full Solution Path