Question

Consider a function $f:(-1,1) \rightarrow A$ defined as $f(x)=\int \frac{dx}{(1-x)^{\frac{1}{2}}(1+x)^{\frac{3}{2}}}$, $f(0)=-1$. Then which of the following option(s) is/are correct?

• A. If f(x) is bijective function, then $A = (0, \infty) \cup \{-1\}$
• B. If f(x) is bijective function, then $A = (-\infty, 0)$
• C. $\lim_{x \rightarrow 1} f(x) = \infty$
• D. $f(\frac{1}{2}) = -\frac{1}{\sqrt{3}}$

Answer 1

Answer: (B), (D)

B, D

Answer 2

1. Evaluate the integral $f(x) = \int \frac{dx}{(1-x)^{\frac{1}{2}}(1+x)^{\frac{3}{2}}}$ using the substitution $x = \cos 2\theta$. This yields $f(x) = -\tan\theta + C = -\sqrt{\frac{1-x}{1+x}} + C$.

2. Apply the condition $f(0)=-1$ to find $C$: $f(0) = -\sqrt{\frac{1-0}{1+0}} + C = -1+C$. Since $f(0)=-1$, $C=0$. So, $f(x) = -\sqrt{\frac{1-x}{1+x}}$.

3. Determine the range of $f(x)$ for $x \in (-1, 1)$. As $x \rightarrow -1^+$, $f(x) \rightarrow -\infty$. As $x \rightarrow 1^-$, $f(x) \rightarrow 0$. Thus, the range is $(-\infty, 0)$.

4. Check injectivity by finding $f'(x)$. $f'(x) = \frac{1}{(1-x)^{1/2}(1+x)^{3/2}}$. For $x \in (-1, 1)$, $f'(x) > 0$, so $f(x)$ is strictly increasing and thus injective.

5. For $f(x)$ to be bijective, its codomain $A$ must be equal to its range. Therefore, $A = (-\infty, 0)$. This confirms Option B is correct.

6. Evaluate $\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1^-} -\sqrt{\frac{1-x}{1+x}} = 0$. This makes Option C incorrect.

7. Evaluate $f(\frac{1}{2}) = -\sqrt{\frac{1-\frac{1}{2}}{1+\frac{1}{2}}} = -\sqrt{\frac{1/2}{3/2}} = -\sqrt{\frac{1}{3}} = -\frac{1}{\sqrt{3}}$. This confirms Option D is correct.

8. Option A is incorrect as the range is $(-\infty, 0)$.