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 values of k 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 to -1.

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

Analysis of Statement-2:

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 trigonometric sine and cosine functions. For any real angle $\theta$, the range of $\sin\theta$ and $\cos\theta$ is $[-1, 1]$. Therefore, Statement-2 is True.

Analysis of Statement-1:

The given equation is $\sin\frac{3x}{2}\cos\frac{5y}{3}=k^8-4k^4+5$.

Let's analyze the Left Hand Side (LHS): $\text{LHS} = \sin\frac{3x}{2}\cos\frac{5y}{3}$. From Statement-2 (which is true), 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 the range $[-1, 1]$, will also be in the range $[-1, 1]$. The maximum value of LHS is $1 \times 1 = 1$. The minimum value of LHS is $1 \times (-1) = -1$ or $(-1) \times 1 = -1$. So, the range of the LHS is $[-1, 1]$. This means $-1 \le \sin\frac{3x}{2}\cos\frac{5y}{3} \le 1$.

Now, let's analyze the Right Hand Side (RHS): $\text{RHS} = k^8-4k^4+5$. Let $u = k^4$. Since $k$ is a real number, $k^4 \ge 0$, so $u \ge 0$. Substituting $u$ into the RHS expression, we get: $\text{RHS} = u^2 - 4u + 5$. To find the minimum value of this quadratic expression, we can complete the square: $u^2 - 4u + 5 = (u^2 - 4u + 4) + 1 = (u-2)^2 + 1$. Since $(u-2)^2 \ge 0$ for all real $u$, the minimum value of $(u-2)^2 + 1$ is $1$. This minimum occurs when $u-2=0$, i.e., $u=2$. Since $u=k^4$, $k^4=2$ is a valid non-negative value for $u$. Therefore, the range of the RHS is $[1, \infty)$. This means $k^8-4k^4+5 \ge 1$.

For the given equation $\text{LHS} = \text{RHS}$ to hold, we must have: $-1 \le \text{LHS} \le 1$ AND $\text{RHS} \ge 1$

The only way for these two conditions to be simultaneously satisfied is if both sides are equal to 1. So, we must have:

1. $\sin\frac{3x}{2}\cos\frac{5y}{3} = 1$
2. $k^8-4k^4+5 = 1$

Let's solve the second equation for $k$: $k^8-4k^4+5 = 1$ $k^8-4k^4+4 = 0$ This is a perfect square: $(k^4)^2 - 4(k^4) + 4 = 0$ $(k^4-2)^2 = 0$ $k^4-2 = 0$ $k^4 = 2$ Taking the fourth root, $k = \pm \sqrt[4]{2}$. These are two distinct real values for $k$: $\sqrt[4]{2}$ and $-\sqrt[4]{2}$.

Now, let's check if the first condition ($\sin\frac{3x}{2}\cos\frac{5y}{3} = 1$) is possible. Yes, it is possible. For example, if $\sin\frac{3x}{2} = 1$ and $\cos\frac{5y}{3} = 1$. This can happen for specific real values of $x$ and $y$ (e.g., $\frac{3x}{2} = \frac{\pi}{2} \implies x=\frac{\pi}{3}$ and $\frac{5y}{3} = 0 \implies y=0$). It is also possible if $\sin\frac{3x}{2} = -1$ and $\cos\frac{5y}{3} = -1$. (e.g., $\frac{3x}{2} = \frac{3\pi}{2} \implies x=\pi$ and $\frac{5y}{3} = \pi \implies y=\frac{3\pi}{5}$). Since $x, y \in \mathbb{R}$, such values exist.

Therefore, the equation holds if and only if $k^4=2$, which gives exactly two distinct real values for $k$. Statement-1 says "exactly four distinct real values of k are possible." This contradicts our finding. Therefore, Statement-1 is False.

Conclusion: Statement-1 is False. Statement-2 is True.