Let an input x(t) = 2 sin(10πt)+ 5 cos(15πt) +7 sin(42πt) + 4cos(45πt) is passed through an LTI system having an impulse response,

h(t)=2(sin(10πt)/πt) cos(40πt)

The output of the system is

The correct option is: 7 sin(42πt) + 4 cos(45πt)

Here is the step-by-step mathematical explanation of why this is the correct output:

Step 1: Understand the Input Signal
The input signal is given as:
$$x\left(t\right)=2sin\left(10\pi t\right)+5cos\left(15\pi t\right)+7sin\left(42\pi t\right)+4cos\left(45\pi t\right)$$
This signal consists of four distinct frequency components with angular frequencies:
1. ${\omega }_1=10\pi \text{rad/s}$
2. ${\omega }_2=15\pi \text{rad/s}$
3. ${\omega }_3=42\pi \text{rad/s}$
4. ${\omega }_4=45\pi \text{rad/s}$

Step 2: Analyze the Impulse Response of the System
The impulse response of the LTI system is:
$$h\left(t\right)=2\left(\frac{sin\left(10\pi t\right)}{\pi t}\right)cos\left(40\pi t\right)$$
Let us define a low-pass filter impulse response $g\left(t\right)$ as:
$$g\left(t\right)=\frac{sin\left(10\pi t\right)}{\pi t}$$
The Fourier transform of $g\left(t\right)$, denoted as $G\left(\omega \right)$, represents an ideal brick-wall low-pass filter:
$$G\left(\omega \right)=\{\begin{array}{ll}1,& \left|\omega \right|<10\pi \\ 0,& \left|\omega \right|>10\pi \end{array}$$

Step 3: Apply the Modulation Property
Now, the impulse response can be written as:
$$h\left(t\right)=2g\left(t\right)cos\left(40\pi t\right)$$
Using the modulation property of the Fourier transform:
$$h\left(t\right)=2g\left(t\right)cos\left({\omega }_{0}t\right)\Rightarrow H\left(\omega \right)=G(\omega -{\omega }_0)+G(\omega +{\omega }_0)$$
Here, ${\omega }_0=40\pi \text{rad/s}$. Thus, the frequency response of the system is:
$$H\left(\omega \right)=G(\omega -40\pi )+G(\omega +40\pi )$$
Since $G\left(\omega \right)$ is a rectangular pulse from $-10\pi$ to $10\pi$, shifting it by $\pm 40\pi$ creates a band-pass filter. The passband of this system is:
For positive frequencies: $40\pi -10\pi <\omega <40\pi +10\pi \Rightarrow 30\pi <\omega <50\pi$
For negative frequencies: $-50\pi <\omega <-30\pi$
So, the filter passes any frequency components whose angular frequency satisfies:
$$30\pi <\left|\omega \right|<50\pi$$
with a gain of 1, and blocks all other frequencies (gain is 0).

Step 4: Evaluate Each Component of the Input Signal
Let us check which components fall inside the passband range $(30\pi ,50\pi )$:
1. For $2sin\left(10\pi t\right)$, $\omega =10\pi$. Since $10\pi <30\pi$, this component is blocked.
2. For $5cos\left(15\pi t\right)$, $\omega =15\pi$. Since $15\pi <30\pi$, this component is blocked.
3. For $7sin\left(42\pi t\right)$, $\omega =42\pi$. Since $30\pi <42\pi <50\pi$, this component is passed with a gain of 1.
4. For $4cos\left(45\pi t\right)$, $\omega =45\pi$. Since $30\pi <45\pi <50\pi$, this component is passed with a gain of 1.

Step 5: Write the Final Output Signal
Combining only the passed components, the output of the LTI system is:
$$y\left(t\right)=7sin\left(42\pi t\right)+4cos\left(45\pi t\right)$$