JEE Mathematics 2026 Question with Solution

Given: The class mid-points are $6, 10, 14, 18$ with frequencies $3, \lambda, 4, 7$. The variance is $19$.

Find: $\lambda + \mu$.

Using step-deviation, take assumed mean $A = 14$ and class width $h = 4$. Define

$$u_i = \frac{x_i - 14}{4}$$

So the values are $-2, -1, 0, 1$.

The total frequency is

$$N = 3 + \lambda + 4 + 7 = 14 + \lambda$$

Now

$$\bar{u} = \frac{\sum f_i u_i}{N} = \frac{3(-2) + \lambda(-1) + 4(0) + 7(1)}{14+\lambda} = \frac{1-\lambda}{14+\lambda}$$

Also,

$$\sum f_i u_i^2 = 3(4) + \lambda(1) + 4(0) + 7(1) = 19 + \lambda$$

Using variance under scale change,

$$\sigma_x^2 = h^2 \sigma_u^2 = 16 \left[ \frac{\sum f_i u_i^2}{N} - (\bar{u})^2 \right] = 19$$

Hence,

$$19 = 16 \left[ \frac{19+\lambda}{14+\lambda} - \left(\frac{1-\lambda}{14+\lambda}\right)^2 \right]$$

So,

$$\frac{19}{16} = \frac{(19+\lambda)(14+\lambda) - (1-\lambda)^2}{(14+\lambda)^2}$$

That gives

$$19(14+\lambda)^2 = 16 \left[ (266 + 33\lambda + \lambda^2) - (1 - 2\lambda + \lambda^2) \right]$$ $$19(196 + 28\lambda + \lambda^2) = 16(265 + 35\lambda)$$ $$3724 + 532\lambda + 19\lambda^2 = 4240 + 560\lambda$$ $$19\lambda^2 - 28\lambda - 516 = 0$$

Solving,

$$\lambda = \frac{28 \pm 200}{38}$$

Since frequency must be positive,

$$\lambda = 6$$

Now,

$$\bar{u} = \frac{1-6}{20} = -\frac{1}{4}$$

Therefore,

$$\mu = 14 + 4\left(-\frac{1}{4}\right) = 13$$

So,

$$\lambda + \mu = 6 + 13 = 19$$

Therefore, the correct option is C.

Given: A grouped frequency distribution with unknown frequency $\lambda$ and mean $\mu$.

Find: $\lambda + \mu$.

The key idea is that variance is unaffected by change of origin but changes by the square of the scale factor. With

$$u = \frac{x-14}{4}$$

we use

$$\operatorname{Var}(x) = 16\operatorname{Var}(u)$$

The frequency table in terms of $u$ becomes:

• for $x = 6$, $u = -2$ with frequency $3$
• for $x = 10$, $u = -1$ with frequency $\lambda$
• for $x = 14$, $u = 0$ with frequency $4$
• for $x = 18$, $u = 1$ with frequency $7$

Now compute the required sums:

$$\sum f_i u_i = 3(-2) + \lambda(-1) + 4(0) + 7(1) = 1 - \lambda$$ $$N = 14 + \lambda$$ $$\bar{u} = \frac{1-\lambda}{14+\lambda}$$

Also,

$$\sum f_i u_i^2 = 3(4) + \lambda(1) + 4(0) + 7(1) = 19 + \lambda$$

So,

$$\sigma_u^2 = \frac{19+\lambda}{14+\lambda} - \left(\frac{1-\lambda}{14+\lambda}\right)^2$$

Since $\sigma_x^2 = 19$,

$$19 = 16\sigma_u^2$$

Substituting,

$$19 = 16 \left[ \frac{19+\lambda}{14+\lambda} - \left(\frac{1-\lambda}{14+\lambda}\right)^2 \right]$$

After simplification,

$$19\lambda^2 - 28\lambda - 516 = 0$$

This gives

$$\lambda = 6$$

Now the mean is recovered from

$$\mu = A + h\bar{u} = 14 + 4\left(\frac{1-6}{20}\right) = 13$$

Hence,

$$\lambda + \mu = 19$$

Thus, the correct option is C.