JEE Main 2026 · 6 Apr · Shift 2 · Mathematics Q8

← JEE Main 2026 · 6 Apr · Shift 2

MathematicsSingle correct+4 / -1Question 8 of 75

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 $3\alpha+2$ and $2\beta+1$ is :

x^2-35x+306=0

x^2-41x+420=0

x^2-45x+506=0

x^2-37x+342=0

Answer and worked solution

Official answer: B

Mean. $40+\alpha+\beta=56$ , so $\alpha+\beta=16$ .

Variance. $\dfrac{\sum x^2}7-64=16$ , so $\sum x^2=560$ . The known values give $424$ , so $\alpha^2+\beta^2=136$ and $\alpha\beta=60$ .

Values. $\alpha=6$ , $\beta=10$ . The roots are $20$ and $21$ .

Equation: $x^2-41x+420=0$ .

(B)

Topics:Variance and standard deviation Roots and coefficients

← Previous question Next question →

JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 8