Question

Statement-1: If $\sin \frac{3x}{2} \cos \frac{5y}{3}=k^8-4k^4+5$, where $x,y \in R$ then exactly four distinct real are possible.

because

Statement-2: $\sin \frac{3x}{2}$ and $\cos \frac{5y}{3}$ both are less than or equal to one and greater than or equal

• A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-
• B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for state
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true.

Answer 1

Answer: (D)

Statement-1 is false, statement-2 is true.

Answer 2

Statement-1 Analysis: The given equation is $\sin \frac{3x}{2} \cos \frac{5y}{3}=k^8-4k^4+5$. Let $LHS = \sin \frac{3x}{2} \cos \frac{5y}{3}$. Since $x, y \in R$, we know that $-1 \le \sin \frac{3x}{2} \le 1$ and $-1 \le \cos \frac{5y}{3} \le 1$. The product of two numbers, each in $[-1, 1]$, will also be in $[-1, 1]$. Thus, $-1 \le LHS \le 1$.

Let $RHS = k^8-4k^4+5$. We can rewrite the RHS by completing the square for $k^4$: $RHS = (k^4)^2 - 4(k^4) + 5 = (k^4 - 2)^2 - 4 + 5 = (k^4 - 2)^2 + 1$. Since $(k^4 - 2)^2 \ge 0$ for any real $k$, the minimum value of $RHS$ is $1$ (when $k^4 - 2 = 0$, i.e., $k^4 = 2$). So, $RHS \ge 1$.

For the equality $LHS = RHS$ to hold, both sides must be equal to $1$, because $LHS \le 1$ and $RHS \ge 1$. Therefore, $\sin \frac{3x}{2} \cos \frac{5y}{3} = 1$ and $k^8-4k^4+5 = 1$.

From $k^8-4k^4+5 = 1$: $(k^4 - 2)^2 + 1 = 1$ $(k^4 - 2)^2 = 0$ $k^4 - 2 = 0$ $k^4 = 2$ This implies $k = \pm \sqrt[4]{2}$. These are two distinct real values for $k$. Statement-1 says "exactly four distinct real are possible". Assuming this refers to values of $k$, it is false, as we found only two distinct real values of $k$.

Statement-2 Analysis: Statement-2 says: $\sin \frac{3x}{2}$ and $\cos \frac{5y}{3}$ both are less than or equal to one and greater than or equal to -1. This is a fundamental property of sine and cosine functions. The range of $\sin(\theta)$ and $\cos(\theta)$ for any real $\theta$ is indeed $[-1, 1]$. So, Statement-2 is true.

Conclusion for Question 23: Statement-1 is false. Statement-2 is true. This corresponds to option (D).