Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
The function $f(x) = \begin{cases} \frac{x^2}{a}, & 0 \le x < 1 \\ b, & 1 \le x < \sqrt{2} \\ \frac{2b^2 - 4b}{x^2}, & \sqrt{2} \le x < \infty \end{cases}$ is continuous for $0 \le x < \infty$. Then which of the following statements is correct?
The number of all possible ordered pairs $(a, b)$ is 3
The number of all possible order pairs $(a, b)$ is 4
The product of all possible values of $b$ is $-1$
The product of all possible values of $b$ is 1.
Step-by-Step Solution
Key Concept: Difference of squares formula allows quick computation of products of conjugate-like expressions.
We need to find $(a, b)$ such that the product $(1.1 + \sqrt{2})(1.1 - \sqrt{2})$ equals some expression. Using the difference of squares formula $(x+y)(x-y) = x^2 - y^2$, we get $(1.1)^2 - (\sqrt{2})^2 = 1.21 - 2 = -0.79$. Therefore $(a, b) = (-1, 1)$ with the given factors.
Correct Answer: 1,3