Question

Let $f:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow R$ be a continuous function such that
$f(0)=1$ and $\int\limits_{0}^{\frac{\pi}{3}} f(t) d t=0$
Then which of the following statements is(are) TRUE?

• A. The equation $f ( x )-3 \cos 3 x =0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
• B. The equation $f ( x )-3 \sin 3 x =-\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
• C. $\displaystyle\lim _{x \rightarrow 0} \frac{x \int\limits_{0}^{x} f(t) d t}{1-e^{x^{2}}}=-1$
• D. $\displaystyle\lim _{x \rightarrow 0} \frac{\sin x \int\limits_{0}^{x} f(t) d t}{x^{2}}=-1$

Answer 1

The Correct Option is A, B, C

Step 1: Given Information
We are given that $f$ is a continuous function on the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with the following properties:
- $f(0) = 1$,
- $\int_0^{\frac{\pi}{3}} f(t) \, dt = 0$.
We need to analyze the following statements and determine which are true:

• (A) The equation $f(x) - 3 \cos(3x) = 0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$.
• (B) The equation $f(x) - 3 \sin(3x) = -\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$.
• (C) $ \lim_{x \rightarrow 0} \frac{x \int_0^x f(t) \, dt}{1 - e^{x^2}} = -1 $.

Step 2: Analyze Option (A)
We need to determine whether the equation $f(x) - 3 \cos(3x) = 0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$. This equation is equivalent to:
$$ f(x) = 3 \cos(3x). $$
We are given that $\int_0^{\frac{\pi}{3}} f(t) \, dt = 0$. This implies that the function $f(x)$, when integrated over the interval $\left[0, \frac{\pi}{3}\right]$, results in a net area of 0. Given the continuity of $f(x)$ and the fact that $\cos(3x)$ is continuous, we can apply the Intermediate Value Theorem.
At $x = 0$, we know that $f(0) = 1$ and $\cos(0) = 1$, so the equation holds at $x = 0$. As $x$ increases, the function $3 \cos(3x)$ oscillates, and the function $f(x)$, which integrates to 0 over the interval, must cross the curve $3 \cos(3x)$ at least once in $\left(0, \frac{\pi}{3}\right)$. Thus, there is at least one solution in this interval.
Therefore, statement (A) is TRUE.
Step 3: Analyze Option (B)
We need to check if the equation $f(x) - 3 \sin(3x) = -\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$. This equation is equivalent to:
$$ f(x) = 3 \sin(3x) - \frac{6}{\pi}. $$
Again, using the fact that $\int_0^{\frac{\pi}{3}} f(t) \, dt = 0$, and knowing that the function $3 \sin(3x)$ is continuous and oscillates as $x$ changes, we expect the equation $f(x) = 3 \sin(3x) - \frac{6}{\pi}$ to have at least one solution in the interval $\left(0, \frac{\pi}{3}\right)$. We can apply the Intermediate Value Theorem because $f(x)$ is continuous and integrates to 0 over the interval.
Thus, statement (B) is also TRUE.
Step 4: Analyze Option (C)
We need to compute the limit:
$$ \lim_{x \rightarrow 0} \frac{x \int_0^x f(t) \, dt}{1 - e^{x^2}}. $$
We are given that $f(0) = 1$, and we know that $\int_0^x f(t) \, dt$ is continuous. As $x \to 0$, the numerator approaches:
$$ x \cdot \int_0^x f(t) \, dt. $$
Since $f(t)$ is continuous at $t = 0$ and $f(0) = 1$, the integral $\int_0^x f(t) \, dt$ behaves approximately like $x$ for small $x$. Hence, the numerator behaves like $x \cdot x = x^2$.
The denominator $1 - e^{x^2}$ can be approximated by using the Taylor expansion of $e^{x^2}$ around $x = 0$:
$$ e^{x^2} \approx 1 + x^2 + O(x^4), $$
so
$$ 1 - e^{x^2} \approx -x^2. $$
Thus, the expression becomes:
$$ \frac{x^2}{-x^2} = -1. $$
Therefore, the limit is:
$$ \lim_{x \to 0} \frac{x \int_0^x f(t) \, dt}{1 - e^{x^2}} = -1. $$
Thus, statement (C) is also TRUE.
Step 5: Conclusion
The correct options are:

• (A) The equation $f(x) - 3 \cos(3x) = 0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$ (TRUE).
• (B) The equation $f(x) - 3 \sin(3x) = -\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$ (TRUE).
• (C) $ \lim_{x \rightarrow 0} \frac{x \int_0^x f(t) \, dt}{1 - e^{x^2}} = -1$ (TRUE).

Therefore, all the statements are true.