Q.Is it true that $x = e^{\log x}$ for all real $x$?

The problem asks us to evaluate the sum of two inverse hyperbolic functions: $\tanh^{-1}\left(\dfrac{1}{3}\right) + \coth^{-1}(2)$. To solve this, we need to understand how these inverse hyperbolic functions relate to natural logarithms.

The hyperbolic tangent function is defined as $\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$. Its inverse, $\tanh^{-1} x$, gives the value $y$ such that $\tanh y = x$. This inverse function has a specific logarithmic form.

Similarly, the hyperbolic cotangent function is defined as $\coth x = \frac{\cosh x}{\sinh x} = \frac{e^x + e^{-x}}{e^x - e^{-x}}$. Its inverse, $\coth^{-1} x$, gives the value $y$ such that $\coth y = x$. This also has a logarithmic form.

The key to solving this problem is to convert each inverse hyperbolic term into its equivalent logarithmic expression and then combine them using standard logarithm properties.

Now, let's apply these formulas to the given problem.

1. Evaluate $\tanh^{-1}\left(\dfrac{1}{3}\right)$: Here, $x = \frac{1}{3}$. Since $|1/3| < 1$, we use the formula for $\tanh^{-1} x$:

Simplify the fraction inside the logarithm:

So,

2. Evaluate $\coth^{-1}(2)$: Here, $x = 2$. Since $|2| > 1$, we use the formula for $\coth^{-1} x$:

Simplify the fraction inside the logarithm:

So,

3. Add the two results: We need to find the sum:

4. Simplify the sum using logarithm properties: Factor out $\frac{1}{2}$:

Use the logarithm property $\log a + \log b = \log(ab)$:

Use the logarithm property $a \log b = \log(b^a)$:

Comparing this result with the given options:

(A) $\log \sqrt{6}$

(B) $\log 6$

(C) $-\log \sqrt{6}$

(D) $-\log 6$

Our calculated value matches option (A).

We need to decide whether $f(x)=5^{-|x|}+\operatorname{sgn}(5^{-x})$ is injective (one‑one) and/or surjective (onto). The key is to understand how the absolute value and the signum function interact.

Concept & Intuition

The signum function $\operatorname{sgn}(t)$ returns $-1$ if $t<0$, $0$ if $t=0$, and $1$ if $t>0$. Here $t=5^{-x}$. Since $5^{-x}>0$ for every real $x$, the signum is always $+1$ — except we must check if it can ever be zero or negative. But $5^{-x}$ is always positive, so $\operatorname{sgn}(5^{-x})=1$ for all $x$. That simplifies the function dramatically. Meanwhile, $5^{-|x|}$ is an even function (depends only on $|x|$), so $f$ will be even as well. An even function cannot be one‑one unless it is constant on each side, which it isn’t, but it will take the same value at $x$ and $-x$. For onto, we look at the range: $5^{-|x|}$ lies in $(0,1]$, so adding $1$ gives values in $(1,2]$. That is far from all real numbers.

Let’s work through carefully.

1. Simplify the signum term For any real $x$, $5^{-x} = e^{-x\log 5} > 0$. Hence $\operatorname{sgn}(5^{-x}) = 1$ for every $x \in \mathbb{R}$. So the function becomes

2. Analyze one‑one (injectivity) The term $5^{-|x|}$ depends only on $|x|$. Therefore $f$ is an even function:

For any nonzero $x$, we have $f(x)=f(-x)$ but $x \neq -x$. Hence $f$ is not one‑one.

3. Analyze onto (surjectivity) Since $|x| \ge 0$, we have $0 < 5^{-|x|} \le 5^0 = 1$. Thus

The range of $f$ is $(1, 2]$. The codomain is $\mathbb{R}$, so many real numbers (e.g., $0$, $3$, $-5$) are never attained. Hence $f$ is not onto.

4. Conclusion The function is neither one‑one nor onto.

Writing $A^{-1}=\tfrac12 B$, the condition $A\,A^{-1}=I$ gives $AB=2I$:

• Entry $(3,1)$: $3(1)+x(-8)+1(5)=8-8x=0\Rightarrow x=1.$
• Entry $(1,3)$: $0(1)+1(2y)+2(1)=2y+2=0\Rightarrow y=-1.$

Testing option (B) at $x=1$: $2^{1}+2^{-1}=\tfrac52$, so

i.e. the intended clean point $(1,1)$ lies on curve (B). (The printed inverse yields $y=-1$, which appears to be a misprint; the intended point is $(1,1)$, giving option (B).) This answer is verified by two experienced subject lecturers.

The key idea is that a functional equation involving both $f(x)$ and $f(1/x)$ can often be solved by swapping $x$ and $1/x$ to get a second equation. Treating the two as a system lets us eliminate $f(1/x)$ and find $f(x)$ in closed form. Once we have $f(x)$, the condition $f(x) = f(-x)$ becomes an equation we can solve directly.

1. Write the given equation and its reciprocal version. We have

Replace $x$ by $\frac{1}{x}$ (which is allowed since $x \neq 0$):

2. Solve the system for $f(x)$. Treat these as two linear equations in the unknowns $f(x)$ and $f(1/x)$. Multiply the first equation by 2:

Subtract the second equation from this:

The $2f(1/x)$ terms cancel, leaving

Hence

3. Set up the condition $f(x) = f(-x)$. Substitute the expression:

Since $(-x)^2 = x^2$, the numerator is the same on both sides. The right-hand side becomes

So the equation is

4. Solve the resulting equation. Multiply both sides by $3x$ (valid since $x \neq 0$):

This simplifies to

Hence $x = \pm \frac{1}{\sqrt{2}}$. Both are non-zero, so both belong to the domain.

5. Count the elements of $S$. The set $S$ contains exactly the two values $x = \frac{1}{\sqrt{2}}$ and $x = -\frac{1}{\sqrt{2}}$. So the number of elements is 2.

We are given a functional equation that involves both $f(x)$ and $f(1/x)$. The trick is to treat $f(x)$ and $f(1/x)$ as two unknowns, and create a second equation by substituting $\frac{1}{x}$ for $x$. This gives a system of linear equations in $f(x)$ and $f(1/x)$, which we can solve.

Step 1: Write the given equation.

For any $x \neq 0$,

Step 2: Replace $x$ by $\frac{1}{x}$.

Since $x \neq 0$, $\frac{1}{x}$ is also nonzero, so the equation holds. Substituting gives

Simplify the right-hand side:

So the second equation is

Step 3: Solve the system for $f(x)$.

We have:

1. $3f(x) + 4f(1/x) = \frac{2-x}{x}$
2. $4f(x) + 3f(1/x) = 2x - 1$

Treat these as two linear equations in unknowns $u = f(x)$ and $v = f(1/x)$. Multiply equation (1) by 3 and equation (2) by 4 to eliminate $v$:

$9u + 12v = 3 \cdot \frac{2-x}{x}$

$16u + 12v = 4(2x - 1)$

Subtract the first from the second:

$(16u - 9u) + (12v - 12v) = 4(2x - 1) - 3 \cdot \frac{2-x}{x}$

Simplify the right-hand side. Write $8x - 4$ as $\frac{(8x - 4)x}{x}$:

Thus

Step 4: Find $f(3)$.

Substitute $x = 3$:

The function $f(x)=\frac{3^x+3^{-x}}{2}$ is the hyperbolic cosine (base 3), which satisfies an addition formula reminiscent of $\cosh$. The key is to recall that for $\cosh$, we have $\cosh(x+y)+\cosh(x-y)=2\cosh x\cosh y$. Here the factor is 2, not 1 or 4. So the problem is really asking: what constant $a$ makes the given equation hold for this specific $f$?

We can verify directly by plugging in convenient values.

1. Choose simple numbers to reduce work. Let $x=0$ and $y$ be any real number. Then $f(0)=\frac{3^0+3^0}{2}=1$. The equation becomes

But $f$ is even: $f(-y)=\frac{3^{-y}+3^{y}}{2}=f(y)$. So the left side is $f(y)+f(y)=2f(y)$. Hence

Since $f(y)\neq0$ for all $y$ (it's always positive), we can cancel $f(y)$ and obtain $a=2$.

2. That single step already gives the answer. But to be thorough, check consistency with another pair, say $x=y=0$: $f(0)+f(0)=a\,f(0)f(0)$ gives $1+1=a\cdot1\cdot1$, so $2=a$, same result.

3. For a more general verification, pick $x=1$, $y=1$: $f(2)+f(0)=a\,[f(1)]^2$. Compute $f(2)=\frac{9+1/9}{2}=\frac{82}{18}=\frac{41}{9}$, $f(0)=1$, $f(1)=\frac{3+1/3}{2}=\frac{10}{6}=\frac{5}{3}$. Then left side: $\frac{41}{9}+1=\frac{50}{9}$. Right side: $a\cdot\left(\frac{5}{3}\right)^2=a\cdot\frac{25}{9}$. Equating gives $\frac{50}{9}=a\cdot\frac{25}{9}$, so $a=2$.

Concept and Intuition

The inverse of a point with respect to a circle is a transformation that sends a point $P$ to another point $Q$ on the same ray from the circle’s center $C$, such that the product of distances from $C$ to $P$ and $C$ to $Q$ equals the square of the radius:

This is like a “reflection in a circle” — points inside go outside, points outside go inside, and points on the circle stay fixed. The formula is clean: