Q.The maximum distance of a point on the graph of the function $y = \sqrt{3}\,\sin x + \cos x$ from $x$-axis is ______.

The key insight is that the exponents $\sin^2 x$ and $\cos^2 x$ are not independent — they sum to 1. That means if we set $u = 81^{\sin^2 x}$, then $81^{\cos^2 x} = 81^{1 - \sin^2 x} = \frac{81}{81^{\sin^2 x}} = \frac{81}{u}$. This turns the original equation into an equation in one variable, which we can solve.

1. Rewrite using the identity

   Since $\sin^2 x + \cos^2 x = 1$, we have $\cos^2 x = 1 - \sin^2 x$.

   Then $81^{\cos^2 x} = 81^{1 - \sin^2 x} = 81 \cdot 81^{-\sin^2 x} = \frac{81}{81^{\sin^2 x}}$.

2. Substitute

   Let $t = 81^{\sin^2 x} > 0$. The equation becomes

3. Clear the denominator Multiply through by $t$:

4. Solve the quadratic Discriminant: $30^2 - 4 \cdot 81 = 900 - 324 = 576 = 24^2$. So

5. Back-substitute

   • If $t = 27$, then $81^{\sin^2 x} = 27$.

     Write $81 = 3^4$ and $27 = 3^3$, so $(3^4)^{\sin^2 x} = 3^{4\sin^2 x} = 3^3$.

     Hence $4\sin^2 x = 3$ $\Rightarrow$ $\sin^2 x = \frac{3}{4}$ $\Rightarrow$ $\sin x = \frac{\sqrt{3}}{2}$ (positive in $[0,\pi]$).

     So $x = \frac{\pi}{3}$ or $\frac{2\pi}{3}$. Neither is among the options.

   • If $t = 3$, then $81^{\sin^2 x} = 3$. …

We are given:

We need $\sin^2 x + \cosh^2 y$.

Concept & Intuition

These equations look like the real and imaginary parts of a complex number expressed in two ways. Recall Euler’s formula and the hyperbolic-trigonometric identity:

but here we have $\sin x \cosh y$ and $\cos x \sinh y$ — that’s actually the expansion of $\sin(x + iy)$:

So the given pair says:

Thus the problem reduces to a single complex equation, and we can use the modulus to find the required expression.

Step-by-step

1. Recognize the complex sine identity For any real $x, y$,

The given equations match exactly:

Hence

2. Take the modulus squared The modulus of $e^{i\theta}$ is $1$, so

For a complex number $z = a + ib$, $|z|^2 = a^2 + b^2$. Here $a = \sin x \cosh y$, $b = \cos x \sinh y$, so

3. Expand and simplify

Use $\cos^2 x = 1 - \sin^2 x$ and $\sinh^2 y = \cosh^2 y - 1$ (since $\cosh^2 y - \sinh^2 y = 1$):

Expand the second term:

$\sin^2 x \cosh^2 y + (\cosh^2 y - 1) - \sin^2 x (\cosh^2 y - 1) = 1.$ …

Translating the perimeter condition

The perimeter equals $6$ times the arithmetic mean of the sines of the angles:

Applying the law of sines

With circumradius $R$, $a = 2R\sin A$, $b=2R\sin B$, $c=2R\sin C$, so

Equating the two expressions: …

The concept first

For $f(x) = ax^2 + bx + c$, completing the square gives

The squared bracket is always $\ge 0$.

• If $a < 0$, the term $a(\cdot)^2 \le 0$, so $f$ is largest when the bracket is zero, i.e. at $x = -\dfrac{b}{2a}$: a maximum exists there. (If $a>0$ the same point is a minimum.) So the Reason (R) is a correct statement.

Step-by-step: test the Assertion

1. Here $f(x) = -x^2 + 3x + 1$, so $a = -1,\ b = 3,\ c = 1$. Since $a<0$, a maximum exists.
2. Locate it. $x = -\dfrac{b}{2a} = -\dfrac{3}{2(-1)} = \dfrac{3}{2}$. (Calculus check: $f'(x) = -2x + 3 = 0 \Rightarrow x = \tfrac32$; $f''(x) = -2 < 0$, confirming a maximum.)
3. Evaluate it. …