Q.(C) cos−1(3𝑥) (D) 3 cos−1 𝑥 1 h

We start with the function $y = \sec^{-1} x$. The key idea is to recall the derivative of the inverse secant and then differentiate again, being meticulous about the domain and the absolute value that appears. The absolute value is not just a decoration—it affects the sign of the derivative and must be handled algebraically.

Step-by-step derivation:

1. Recall the first derivative. For $y = \sec^{-1} x$, the standard formula is

The absolute value arises because the range of $\sec^{-1} x$ is $[0, \pi]$ excluding $\pi/2$, and the derivative must be positive for $x > 1$ and also positive for $x < -1$ (since the function is decreasing there, but the derivative formula with $|x|$ corrects the sign).

2. Rewrite the derivative without the absolute value using a piecewise approach or a clever identity. A useful trick: $|x| = \sqrt{x^2}$, but that can be messy for differentiation. Instead, note that

but a cleaner method is to write

and

However, we can unify these by observing that

and for $x < -1$, $x$ is negative, so $1/x$ is negative, but we need the derivative to be positive. Actually, check: For $x < -1$, $\sec^{-1} x$ is in $(\pi/2, \pi]$, and its derivative is positive. If we used $\frac{1}{x\sqrt{x^2-1}}$, then $x$ negative gives a negative value—wrong. So the absolute value is essential.

The unified form $\frac{1}{|x|\sqrt{x^2-1}}$ is correct for all $|x|>1$.

3. Differentiate again using the chain rule. Let

But differentiating $|x|$ directly is tricky. Instead, write $|x| = \sqrt{x^2}$, so

Now differentiate using the product rule:

4. Compute each derivative.
   • First term:

 But $(x^2)^{3/2} = |x|^3$, so this is $-\frac{x}{|x|^3}$.  

• Second term:

5. Assemble the second derivative.

Factor common terms:

$\frac{d^2 y}{dx^2} = -\frac{x}{|x|} \left[ \frac{1}{|x|^2 \sqrt{x^2 - 1}} + \frac{1}{(x^2 - 1)^{3/2}} \right].$ …

Step 1 — Evaluate $\text{Sech}^{-1}\!\left(\tfrac{3}{5}\right)$.

For $0<x\le 1$, $\ \text{Sech}^{-1}(x)=\ln\!\left(\dfrac{1+\sqrt{1-x^{2}}}{x}\right)$. With $x=\tfrac{3}{5}$:

Step 2 — Evaluate $\text{Tanh}^{-1}\!\left(\tfrac{3}{5}\right)$.

For $|x|<1$, $\ \text{Tanh}^{-1}(x)=\tfrac{1}{2}\ln\!\left(\dfrac{1+x}{1-x}\right)$. With $x=\tfrac{3}{5}$: …

The problem gives $\sinh x = \tan A$ and asks for $|\tanh x|$ in terms of $A$. The core idea is to relate hyperbolic functions through their fundamental identity: $\cosh^2 x - \sinh^2 x = 1$. This is the hyperbolic analogue of $\cos^2 \theta + \sin^2 \theta = 1$, and it lets us express $\tanh x$ purely in terms of $\sinh x$ (or $\cosh x$). Once we do that, substituting $\sinh x = \tan A$ turns the expression into something involving only trigonometric functions of $A$, which simplifies neatly.

1. Recall the definition of $\tanh x$

   $\tanh x = \frac{\sinh x}{\cosh x}$.

   So to find $|\tanh x|$, we need $\cosh x$ in terms of $\sinh x$.

2. Use the hyperbolic identity

   $\cosh^2 x - \sinh^2 x = 1 \quad \Rightarrow \quad \cosh^2 x = 1 + \sinh^2 x$.

   Since $\cosh x \ge 1$ for all real $x$, we take the positive square root:

   $\cosh x = \sqrt{1 + \sinh^2 x}$.

3. Express $\tanh x$

   $\tanh x = \frac{\sinh x}{\sqrt{1 + \sinh^2 x}}$.

4. Substitute the given $\sinh x = \tan A$

   $\tanh x = \frac{\tan A}{\sqrt{1 + \tan^2 A}}$.

5. Simplify the denominator using a trigonometric identity

   Recall: $1 + \tan^2 A = \sec^2 A$.

   So $\sqrt{1 + \tan^2 A} = \sqrt{\sec^2 A} = |\sec A|$ (the absolute value is needed because $\sec A$ can be negative, but the square root gives a non-negative result).

6. Write $\tanh x$ in terms of $A$

   $\tanh x = \frac{\tan A}{|\sec A|}$.

7. Simplify further …

The key idea here is that $f$ is a function from $A$ to $A$, meaning every $x$ in $A$ pairs with some $y$ in $A$ such that $(x,y) \in f$. The given equation $a^x + a^y = a$ must hold for these pairs. Since $a>1$, the exponential $a^t$ is strictly increasing, and its range is $(0, \infty)$. We need to find the largest set $A$ such that for every $x$ in $A$, there exists a $y$ in $A$ satisfying the equation — and vice versa, because $f$ is a relation on $A \times A$.

Let’s work through the constraints step by step.

1. Rewrite the condition.

   From $a^x + a^y = a$, we get $a^y = a - a^x$. Since $a^y > 0$ for any real $y$, we require $a - a^x > 0$, i.e., $a^x < a$. Because $a>1$, the exponential function is increasing, so $a^x < a$ implies $x < 1$. Similarly, by symmetry, $y < 1$ as well. So both coordinates must be strictly less than $1$.

2. What about lower bounds?

   There is no lower bound from positivity alone: $a^x$ can be arbitrarily close to $0$ as $x \to -\infty$, and then $a^y = a - a^x$ approaches $a$, so $y$ approaches $1$ from below. Conversely, if $x$ is very close to $1$ from below, $a^x$ is just under $a$, making $a^y$ very small, so $y \to -\infty$. So $x$ and $y$ can each be any real number less than $1$, with no minimum.

3. Check the endpoints.

   Can $x = 1$? Then $a^x = a$, so $a^y = 0$, which is impossible because $a^y > 0$ for all real $y$. So $1$ is not allowed. Can $x$ be exactly $0$? Yes, $a^0 = 1$, then $a^y = a-1 > 0$, so $y = \log_a(a-1)$, which is a real number less than $1$. So $0$ is fine. Similarly, any negative number works.

4. What about $x$ being greater than $1$? …