JEE Mathematics 2023 Question with Solution

Given: The mean is $9$ and the variance is $15.08$ for the given frequency distribution.

Find: The value of $\alpha^2 + \beta^2 - \alpha\beta$.

Using the table,

$$N = \sum f_i = 40 + \alpha + \beta$$ $$\sum f_i x_i = 360 + 6\alpha + 12\beta$$ $$\sum f_i x_i^2 = 3904 + 36\alpha + 144\beta$$

Mean $\left(\overline{x}\right)$ is:

$$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = 9$$

So,

$$\frac{360 + 6\alpha + 12\beta}{40 + \alpha + \beta} = 9$$

Hence,

$$360 + 6\alpha + 12\beta = 9(40 + \alpha + \beta)$$ $$360 + 6\alpha + 12\beta = 360 + 9\alpha + 9\beta$$ $$3\beta = 3\alpha \Rightarrow \alpha = \beta$$

Now use the variance formula:

$$\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \left(\frac{\sum f_i x_i}{\sum f_i}\right)^2$$

Substituting the given values,

$$\frac{3904 + 36\alpha + 144\beta}{40 + \alpha + \beta} - 9^2 = 15.08$$

Since $\alpha = \beta$,

$$\frac{3904 + 180\alpha}{40 + 2\alpha} - 81 = 15.08$$

This gives

$$\frac{3904 + 180\alpha}{40 + 2\alpha} = 96.08$$

Solving, we get $\alpha = 5$ and $\beta = 5$. Therefore,

$$\alpha^2 + \beta^2 - \alpha\beta = 5^2 + 5^2 - 5 \cdot 5 = 25$$

So, the required value is $25$.

Given: The frequency distribution has unknown frequencies $\alpha$ and $\beta$, with mean $9$ and variance $15.08$.

Find: $\alpha^2 + \beta^2 - \alpha\beta$.

From the distribution,

• total frequency: $40 + \alpha + \beta$
• sum $\sum f_i x_i$: $360 + 6\alpha + 12\beta$
• sum $\sum f_i x_i^2$: $3904 + 36\alpha + 144\beta$

Using the mean,

$$\frac{360 + 6\alpha + 12\beta}{40 + \alpha + \beta} = 9$$

Cross-multiplying,

$$360 + 6\alpha + 12\beta = 360 + 9\alpha + 9\beta$$ $$12\beta - 9\beta = 9\alpha - 6\alpha$$ $$3\beta = 3\alpha$$ $$\alpha = \beta$$

Now apply variance:

$$15.08 = \frac{3904 + 36\alpha + 144\beta}{40 + \alpha + \beta} - 81$$

So,

$$96.08 = \frac{3904 + 36\alpha + 144\beta}{40 + \alpha + \beta}$$

Using $\alpha = \beta$,

$$96.08 = \frac{3904 + 180\alpha}{40 + 2\alpha}$$

Now solve this linear equation to obtain $\alpha = 5$. Hence $\beta = 5$. Finally,

$$\alpha^2 + \beta^2 - \alpha\beta = 25 + 25 - 25 = 25$$

Therefore, the required numerical value is $25$.