Question

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 :

• A. \(x^2-35x+306=0\)
• B. \(x^2-41x+420=0\)
• C. \(x^2-45x+506=0\)
• D. \(x^2-37x+342=0\)

Answer 1

The Correct Option is D

Concept:

• Mean \(= \dfrac{\text{Sum of observations}}{n}\)
• Variance \(= \dfrac{\sum x_i^2}{n}-\mu^2\)

Step 1: {Use the mean condition.} Mean \(=8\), number of observations \(=7\) \[ \text{Sum} = 7 \times 8 = 56 \] Sum of given numbers: \[ 2+4+8+12+14 = 40 \] Thus \[ \alpha+\beta+40 = 56 \] \[ \alpha+\beta = 16 \] 

Step 2: {Use the variance condition.} Variance \(=16\) \[ 16=\frac{\sum x^2}{7}-8^2 \] \[ \frac{\sum x^2}{7}=80 \] \[ \sum x^2 = 560 \] Squares of known numbers: \[ 2^2+4^2+8^2+12^2+14^2 \] \[ =4+16+64+144+196 \] \[ =424 \] Thus \[ \alpha^2+\beta^2+424=560 \] \[ \alpha^2+\beta^2=136 \] 

Step 3: {Find \( \alpha\beta \).} \[ (\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta \] \[ 256=136+2\alpha\beta \] \[ 2\alpha\beta=120 \] \[ \alpha\beta=60 \] 

Step 4: {Find \( \alpha \) and \( \beta \).} \[ x^2-16x+60=0 \] \[ (x-6)(x-10)=0 \] Since \( \alpha < \beta \): \[ \alpha=6,\quad \beta=10 \] 

Step 5: {Form the required quadratic equation.} Roots: \[ 3\alpha+2 = 20 \] \[ 2\beta+1 = 21 \] Sum of roots: \[ 41 \] Product: \[ 420 \] Thus equation: \[ x^2-41x+420=0 \]