JEE MAIN 2026 Previous Year Questions (PYQs) – Page 25 of 25
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Let $\alpha$ and $\beta$ be the roots of equation $x^2 + 2ax + (3a + 10) = 0$ such that $\alpha < 1 < \beta$. Then the set of all possible values of $a$ is:
1
2
3
4
Solution
$\alpha < 1 < \beta$
$f(1) < 0$
$\Rightarrow 1 + 2a + (3a + 10) < 0$
$\Rightarrow 5a + 11 < 0$
$\Rightarrow a < -\frac{11}{5}$
$\therefore a \in (-\infty, -\frac{11}{5})$
A random variable $X$ takes values $0, 1, 2, 3$ with probabilities
$\frac{2a+1}{30}, \frac{8a-1}{30}, \frac{4a+1}{30},b$ respectively, where $a, b \in R$. Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of $X$ such that $\sigma^2 + \mu^2 = 2$. Then $\frac{1}{b}$ is equal to:
1
2
3
4
Solution
$x \quad 0 \quad 1 \quad 2 \quad 3$
$p(x) \quad \frac{2a+1}{30} \quad \frac{8a-1}{30} \quad \frac{4a+1}{30} \quad b$
$\sigma^2 = \sum x^2 p(x) - \mu^2$
$\sigma^2 + \mu^2 = \sum x^2 p(x)$
$= 0 + 1\cdot \frac{8a-1}{30} + 4\cdot \frac{4a+1}{30} + 9b$
$= \frac{24a + 270b + 3}{30} = 2$
$\Rightarrow 24a + 270b = 57$
$\Rightarrow 8a + 90b = 19 \quad ...(1)$
Also $\sum p(x) = 1$
$\Rightarrow \frac{2a+1}{30} + \frac{8a-1}{30} + \frac{4a+1}{30} + b = 1$
$\Rightarrow 14a + 30b = 29 \quad ...(2)$
Solving (1) & (2):
$a = 2,; b = \frac{1}{60}$
$\therefore \frac{1}{b} = 60$
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