Q.State True or False: The value of the expression $(\cos^{-1}x)^2$ is equal to $\sec^2 x$.

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $f(x) = \dfrac{4x}{3} + \cos x$ and $g(x)$ is the inverse of $f(x)$, then $g'(2\pi) =$ (A) $\dfrac{6}{11}$ (B) $\dfrac{1}{\pi}$ (C) $\dfrac{3}{7}$ (D) $\dfrac{2}{\pi}$

   ›Reveal solutionSolution

   Finding the point $x=3\pi/2$ where $f(x)=2\pi$ and applying the inverse-function derivative rule gives $g'(2\pi)=\dfrac37$.

   Concept and Intuition

   For an invertible function $f$ with inverse $g$, the key identity is $g'(y_0)=\dfrac{1}{f'(x_0)}$ where $f(x_0)=y_0$. So instead of trying to write $g$ explicitly (often impossible here, since $f$ mixes a linear term with $\cos x$), we just need to (a) find the $x_0$ with $f(x_0)=2\pi$, and (b) evaluate $f'$ there.

   Step-by-Step Solution

   1. $f(x)=\dfrac{4x}{3}+\cos x$. We want $x_0$ with $f(x_0)=2\pi$.
   2. Try $x_0=\dfrac{3\pi}{2}$: $f\left(\dfrac{3\pi}{2}\right)=\dfrac{4}{3}\cdot\dfrac{3\pi}{2}+\cos\dfrac{3\pi}{2} = 2\pi + 0 = 2\pi$. ✓ So $g(2\pi)=\dfrac{3\pi}{2}$.
   3. $f'(x)=\dfrac{4}{3}-\sin x$. …

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2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $\text{Cosech}^{-1}x = \log\left(\dfrac{5\sqrt{2}-1}{7}\right)$ then, $\text{Tanh}^{-1}\left(\dfrac{1}{x}\right) =$ (A) $\log\dfrac{1}{2}$ (B) $\log\dfrac{\sqrt{3}}{2}$ (C) $\log\sqrt{3}$ (D) $\log\dfrac{1}{\sqrt{3}}$

   ›Reveal solutionSolution

   This tests converting an inverse hyperbolic given as a log expression into $e^{u}$ and $e^{-u}$, extracting $\sinh u$, and then plugging into the $\tanh^{-1}$ log formula.

   Concept and Intuition

   $\text{csch}^{-1}x=\sinh^{-1}(1/x)$, and $\sinh^{-1}(t)=\log(t+\sqrt{t^2+1})$. If we are told $\sinh^{-1}(1/x)$ equals a specific $\log$ expression, that expression IS $e^{u}$ where $u=\sinh^{-1}(1/x)$. From $e^u$ we can recover $\sinh u = 1/x$ directly using $e^{-u}=1/e^{u}$, without ever solving for $u$ itself.

   Step-by-Step Solution

   1. Let $u=\text{Cosech}^{-1}x=\log\dfrac{5\sqrt2-1}{7}$, so $e^{u}=\dfrac{5\sqrt2-1}{7}$.
   2. $e^{-u}=\dfrac{7}{5\sqrt2-1}=\dfrac{7(5\sqrt2+1)}{(5\sqrt2)^2-1^2}=\dfrac{7(5\sqrt2+1)}{49}=\dfrac{5\sqrt2+1}{7}$.
   3. $\sinh u=\dfrac{e^u-e^{-u}}2=\dfrac{(5\sqrt2-1)-(5\sqrt2+1)}{14}=\dfrac{-2}{14}=-\dfrac17$.
   4. Since $u=\sinh^{-1}(1/x)$, we get $1/x=\sinh u=-1/7$. …

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3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $f(x) = \mathrm{Tan}^{-1}x$ and $g$ is the inverse of 'f', then $g'(f(2)) =$ (A) $1$ (B) $2$ (C) $4$ (D) $5$

   ›Reveal solutionSolution

   This tests the derivative-of-inverse-function rule. Since $g$ undoes $f$, $g'(f(2))=\frac1{f'(2)}=5$.

   Concept and Intuition

   If $g$ is the inverse of $f$, then for any $x$, $g'(f(x))=\frac{1}{f'(x)}$ — this is the standard inverse function derivative rule, and it avoids ever needing an explicit formula for $g$.

   Step-by-Step Solution

   1. $f(x)=\tan^{-1}x\Rightarrow f'(x)=\frac1{1+x^2}$.
   2. Since $g=f^{-1}$, differentiating $g(f(x))=x$ gives $g'(f(x))\cdot f'(x)=1$, i.e. $g'(f(x))=\frac1{f'(x)}$.
   3. Set $x=2$: $g'(f(2))=\frac1{f'(2)}$.
   4. $f'(2)=\frac1{1+4}=\frac15$.
   5. So $g'(f(2))=\frac1{1/5}=5$.

   Common Mistakes …

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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.f is differentiable function such that $f(1) = 8$ and $f'(1) = \dfrac{1}{8}$. If f is invertible and $g = f^{-1}$, then (A) $g'(1) = 8$ (B) $g'(1) = \dfrac{1}{8}$ (C) $g'(8) = 8$ (D) $g'(8) = \dfrac{1}{8}$

   ›Reveal solutionSolution

   Using the inverse-function derivative rule $g'(y)=1/f'(g(y))$ with $g(8)=1$ (since

   $f(1)=8$), we get $g'(8)=1/f'(1)=8$.

   Concept and Intuition

   If $g=f^{-1}$, then differentiating $f(g(y))=y$ using the chain rule gives

   $f'(g(y))\cdot g'(y)=1$, i.e. $g'(y)=\dfrac{1}{f'(g(y))}$. The key bookkeeping step is

   correctly matching which $y$-value corresponds to which $x$-value under $f$.

   Step-by-Step Solution

   1. Given $f(1)=8$ and $f'(1)=\dfrac18$, and $g=f^{-1}$.
   2. Since $f(1)=8$, we have $g(8)=1$.
   3. Inverse derivative rule: $g'(8)=\dfrac{1}{f'(g(8))}=\dfrac{1}{f'(1)}=\dfrac{1}{1/8}=8$. …

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5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If g is the inverse of the function f(x) and $g(x) = x + \tan x$ then, $f'(x) =$ (A) $1+\sec^2 x$ (B) $\dfrac{1}{1+\sec^2 f(x)}$ (C) $\dfrac{1}{1+\sec^2 g(x)}$ (D) $1+\sec^2 f(x)$

   ›Reveal solutionSolution

   Using the inverse-function derivative rule $f'(g(x))\cdot g'(x)=1$ and re-expressing the result purely in terms of $x$ (using $x=f(g(x))$) gives $f'(x)=\dfrac1{1+\sec^2 f(x)}$.

   Concept and Intuition

   If $g$ is the inverse of $f$, then applying one after the other gives back the input: $f(g(x))=x$. Differentiating this identity via the chain rule directly links $f'$ at the point $g(x)$ to $g'(x)$. The subtlety in this problem is that the answer must be expressed as a function purely of $x$ (i.e. $f'(x)$, not $f'(g(x))$) — this requires a careful relabelling step using the inverse relationship again.

   Step-by-Step Solution

   1. Since $g=f^{-1}$, we have $f(g(x))=x$ for all $x$ in the domain.
   2. Differentiate both sides with respect to $x$ using the chain rule:

   $f'(g(x))\cdot g'(x) = 1 \implies f'(g(x)) = \frac{1}{g'(x)}$

   3. Given $g(x)=x+\tan x$, so $g'(x)=1+\sec^2x$. Thus:

   $f'(g(x)) = \frac1{1+\sec^2x} \qquad (\star)$

   4. Equation $(\star)$ gives the value of $f'$ at the point $g(x)$, in terms of $x$. To express $f'$ as a function of its own argument, substitute $t=g(x)$. Since $f$ and $g$ are inverses, $x=f(t)$ (because $f(g(x))=x$ means $f(t)=x$ when $t=g(x)$). …

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6. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $f(x)=(x+1)^2-1, x \ge -1$, then $\{x \mid f(x)=f^{-1}(x)\}$ is (A) $\{0, -1\}$ (B) $\{-1, 0, 1\}$ (C) $\left\{-1, 0, \dfrac{-3+\sqrt{3}i}{2}, \dfrac{-3-\sqrt{3}i}{2}\right\}$ (D) an empty set

   ›Reveal solutionSolution

   For an increasing invertible function, $f(x)=f^{-1}(x)$ forces $f(f(x))=x$; solving that quartic-in-disguise on the restricted domain $x\ge-1$ gives exactly $x=-1,0$.

   Concept and Intuition

   $f(x)=(x+1)^2-1$ is a rightward-shifted-up parabola restricted to $x\ge -1$, where it is strictly increasing and hence one-to-one — so $f^{-1}$ exists on this domain. If $y=f^{-1}(x)$ then by definition $f(y)=x$. If additionally $f(x)=y$, substituting gives $f(f(x))=f(y)=x$. So every solution of $f(x)=f^{-1}(x)$ must satisfy $f(f(x))=x$, and we can solve that equation instead.

   Step-by-Step Solution

   1. $f(x)=(x+1)^2-1$, domain $x\ge -1$.
   2. Let $y=f(x)$, so $y+1=(x+1)^2$. Requiring $f(y)=x$ means $(y+1)^2-1=x$, i.e. $(y+1)^2=x+1$.
   3. Substitute $y+1=(x+1)^2$: $\big((x+1)^2\big)^2=x+1$.
   4. Let $u=x+1\ge 0$: $u^4=u \implies u(u^3-1)=0 \implies u=0$ or $u^3=1$.
   5. Since $u\ge 0$ real, the only real solutions are $u=0$ (giving $x=-1$) and $u=1$ (giving $x=0$) — the other two cube roots of $1$ are complex and don't correspond to any real $x$ in the domain. …

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7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.$\operatorname{Tanh}^{-1}(\sin\theta)=$ (A) $\operatorname{Sinh}^{-1}(\operatorname{cosec}\theta)$ (B) $\operatorname{Sinh}^{-1}(\sec\theta)$ (C) $\operatorname{Cosh}^{-1}(\operatorname{cosec}\theta)$ (D) $\operatorname{Cosh}^{-1}(\sec\theta)$

   ›Reveal solutionSolution

   This tests converting between inverse hyperbolic functions and their logarithmic definitions, plus a classical trig-log identity. Answer: $\operatorname{Cosh}^{-1}(\sec\theta)$.

   Concept and Intuition

   Inverse hyperbolic functions have explicit log forms: $\operatorname{Tanh}^{-1}y=\frac12\log\frac{1+y}{1-y}$ and $\operatorname{Cosh}^{-1}x=\log\left(x+\sqrt{x^2-1}\right)$ for $x\ge1$. Separately, $\log(\sec\theta+\tan\theta)$ is the classical "integral of secant" expression, which turns out to equal $\frac12\log\frac{1+\sin\theta}{1-\sin\theta}$. Recognizing both sides collapse to the same log expression proves the identity.

   Step-by-Step Solution

   1. By definition, $\operatorname{Tanh}^{-1}(\sin\theta)=\dfrac12\log\dfrac{1+\sin\theta}{1-\sin\theta}$.
   2. Multiply numerator and denominator inside the log by $(1+\sin\theta)$: $\dfrac{1+\sin\theta}{1-\sin\theta}=\dfrac{(1+\sin\theta)^2}{1-\sin^2\theta}=\dfrac{(1+\sin\theta)^2}{\cos^2\theta}=\left(\dfrac{1+\sin\theta}{\cos\theta}\right)^2=(\sec\theta+\tan\theta)^2$.
   3. So $\operatorname{Tanh}^{-1}(\sin\theta)=\dfrac12\log(\sec\theta+\tan\theta)^2 = \log(\sec\theta+\tan\theta)$ (taking $\sec\theta+\tan\theta>0$).
   4. Now compute $\operatorname{Cosh}^{-1}(\sec\theta)=\log\left(\sec\theta+\sqrt{\sec^2\theta-1}\right)$. Since $\sec^2\theta-1=\tan^2\theta$, this is $\log(\sec\theta+\tan\theta)$ (taking $\tan\theta\ge0$ in the relevant range). …

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8. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If $\cot(\operatorname{Cos}^{-1}x) = \sec\left\{\operatorname{Tan}^{-1}\left(\dfrac{a}{\sqrt{b^2-a^2}}\right)\right\}$, $b>a$, then $x =$ (A) $\dfrac{b}{\sqrt{2b^2-a^2}}$ (B) $\dfrac{a}{\sqrt{2b^2-a^2}}$ (C) $\dfrac{\sqrt{b^2-a^2}}{a}$ (D) $\dfrac{\sqrt{b^2-a^2}}{b}$

   ›Reveal solutionSolution

   Converting both inverse-trig expressions to algebraic ratios and equating $\cot(\cos^{-1}x)=\sec(\tan^{-1}(\cdots))$ leads to a quadratic in $x$ solved as $x=\dfrac b{\sqrt{2b^2-a^2}}$.

   Concept and Intuition

   The standard technique for equations mixing different inverse trig functions is to convert each side to a right-triangle ratio: if $\theta=\cos^{-1}x$, build a right triangle with adjacent $=x$, hypotenuse $=1$, opposite $=\sqrt{1-x^2}$, and read off $\cot\theta$ directly. Similarly for $\phi=\tan^{-1}(\text{something})$, build a triangle with opposite/adjacent given by that "something" and read off $\sec\phi$ via $\sqrt{1+\tan^2\phi}$.

   Step-by-Step Solution

   1. Let $\theta=\operatorname{Cos}^{-1}x$ (principal range $[0,\pi]$), so $\cos\theta=x$ and $\sin\theta=\sqrt{1-x^2}\ (\ge0)$.
   2. Then $\cot\theta=\dfrac{\cos\theta}{\sin\theta}=\dfrac{x}{\sqrt{1-x^2}}$.
   3. Let $\phi=\operatorname{Tan}^{-1}\!\left(\dfrac{a}{\sqrt{b^2-a^2}}\right)$ (principal range $(-\pi/2,\pi/2)$, so $\sec\phi>0$), so $\tan\phi=\dfrac{a}{\sqrt{b^2-a^2}}$.
   4. $\sec\phi=\sqrt{1+\tan^2\phi}=\sqrt{1+\dfrac{a^2}{b^2-a^2}}=\sqrt{\dfrac{b^2}{b^2-a^2}}=\dfrac b{\sqrt{b^2-a^2}}$.
   5. Equation becomes $\dfrac{x}{\sqrt{1-x^2}}=\dfrac{b}{\sqrt{b^2-a^2}}$. …

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9. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If $\operatorname{Sinh}^{-1}x = \log 3$ and $\operatorname{Cosh}^{-1}y = \log\dfrac{3}{2}$, then $\operatorname{Tanh}^{-1}(x-y) =$ (A) $\log\sqrt{\dfrac{5}{3}}$ (B) $\log\dfrac{5}{3}$ (C) $\log\dfrac{4}{3}$ (D) $\log\dfrac{2}{\sqrt{3}}$

   ›Reveal solutionSolution

   Directly computing $x=\sinh(\log3)$ and $y=\cosh(\log\frac32)$ gives $x-y=\frac14$, and $\operatorname{Tanh}^{-1}(\tfrac14)$ simplifies to $\log\sqrt{5/3}$.

   Concept and Intuition

   Hyperbolic functions of a logarithm collapse nicely because $e^{\log k}=k$: $\sinh(\log k)=\dfrac{k-1/k}2$ and $\cosh(\log k)=\dfrac{k+1/k}2$. Once $x$ and $y$ are plain numbers, $\operatorname{Tanh}^{-1}z=\dfrac12\log\dfrac{1+z}{1-z}$ finishes the problem as ordinary logarithm algebra.

   Step-by-Step Solution

   1. $\operatorname{Sinh}^{-1}x=\log3\Rightarrow x=\sinh(\log3)=\dfrac{e^{\log3}-e^{-\log3}}2=\dfrac{3-\frac13}2=\dfrac{8/3}2=\dfrac43$.
   2. $\operatorname{Cosh}^{-1}y=\log\frac32\Rightarrow y=\cosh\!\left(\log\tfrac32\right)=\dfrac{\frac32+\frac23}2=\dfrac{\frac{9}{6}+\frac{4}{6}}2=\dfrac{13/6}2=\dfrac{13}{12}$.
   3. $x-y=\dfrac43-\dfrac{13}{12}=\dfrac{16}{12}-\dfrac{13}{12}=\dfrac{3}{12}=\dfrac14$. …

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10. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If $\text{Tanh}^{-1}x=\text{Coth}^{-1}y=\log\sqrt5$, then $\text{Tan}^{-1}(xy)=$ (A) $\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{\pi}{6}$ (D) $\dfrac{3\pi}{4}$

    ›Reveal solutionSolution

    Convert the inverse hyperbolic definitions to logarithmic form, solve for $x$ and $y$ individually, and the product collapses to exactly $1$. Answer: $\tan^{-1}(xy)=\pi/4$.

    Concept and Intuition

    $\tanh^{-1}x=\tfrac12\log\dfrac{1+x}{1-x}$ and $\coth^{-1}y=\tfrac12\log\dfrac{y+1}{y-1}$ are standard logarithmic definitions of the inverse hyperbolic functions; setting both equal to the same value $\log\sqrt5=\tfrac12\ln5$ gives two independent linear equations.

    Step-by-Step Solution

    1. $\tanh^{-1}x=\tfrac12\log\dfrac{1+x}{1-x}=\log\sqrt5=\tfrac12\ln5$, so $\log\dfrac{1+x}{1-x}=\ln5\Rightarrow\dfrac{1+x}{1-x}=5$.
    2. Solve: $1+x=5-5x\Rightarrow6x=4\Rightarrow x=\dfrac23$.
    3. $\coth^{-1}y=\tfrac12\log\dfrac{y+1}{y-1}=\tfrac12\ln5\Rightarrow\dfrac{y+1}{y-1}=5$.
    4. Solve: $y+1=5y-5\Rightarrow6=4y\Rightarrow y=\dfrac32$. …

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11. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.$\mathrm{Cosh}^{-1} 2 =$ (A) $\log(2 + \sqrt{3})$ (B) $\log(2 + \sqrt{5})$ (C) $\log(2 - \sqrt{5})$ (D) $\log(2 + \sqrt{2})$

    ›Reveal solutionSolution

    Using the logarithmic form of the inverse hyperbolic cosine, $\cosh^{-1}2=\log(2+\sqrt3)$.

    Concept and Intuition

    The inverse hyperbolic cosine has the closed form $\cosh^{-1}x = \log\left(x+\sqrt{x^2-1}\right)$ for $x\ge1$, derived by solving $x=\cosh y=\dfrac{e^y+e^{-y}}{2}$ for $y$ using the quadratic formula in $e^y$.

    Step-by-Step Solution

    1. Let $y=\cosh^{-1}2$, so $\cosh y=2 \Rightarrow \dfrac{e^y+e^{-y}}{2}=2 \Rightarrow e^y+e^{-y}=4$.
    2. Multiply by $e^y$: $e^{2y}-4e^y+1=0$. Solve as a quadratic in $e^y$: $e^y=\dfrac{4\pm\sqrt{16-4}}{2}=2\pm\sqrt3$.
    3. Since $\cosh^{-1}$ is defined as the non-negative branch, take $e^y=2+\sqrt3$ (the larger root, giving $y\ge0$). …

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12. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.If $y=f(x)$ is a thrice differentiable function and a bijection then $\dfrac{d^2x}{dy^2}\left(\dfrac{dy}{dx}\right)^3+\dfrac{d^2y}{dx^2}=$ (A) $y$ (B) $-y$ (C) $x$ (D) $0$

    ›Reveal solutionSolution

    This is a direct application of the well-known second-derivative inverse-function relation; the two terms exactly cancel to give $0$.

    Concept and Intuition

    When $y=f(x)$ is a differentiable bijection, $x$ can be viewed as a function of $y$ (via the inverse function), and there's a standard relation connecting the second derivatives in the two directions. Differentiating $\dfrac{dx}{dy}=\left(\dfrac{dy}{dx}\right)^{-1}$ with respect to $y$ (using the chain rule) produces exactly the identity used here.

    Step-by-Step Solution

    1. Start from $\dfrac{dx}{dy}=\dfrac{1}{dy/dx}$.
    2. Differentiate both sides with respect to $y$: $\dfrac{d^2x}{dy^2}=\dfrac{d}{dy}\left(\dfrac{1}{dy/dx}\right) = -\dfrac{1}{(dy/dx)^2}\cdot\dfrac{d}{dy}\left(\dfrac{dy}{dx}\right)$.
    3. Convert the inner derivative w.r.t. $y$ to w.r.t. $x$ via chain rule: $\dfrac{d}{dy}\left(\dfrac{dy}{dx}\right) = \dfrac{d^2y}{dx^2}\cdot\dfrac{dx}{dy} = \dfrac{d^2y/dx^2}{dy/dx}$.
    4. So $\dfrac{d^2x}{dy^2} = -\dfrac{1}{(dy/dx)^2}\cdot\dfrac{d^2y/dx^2}{dy/dx} = -\dfrac{d^2y/dx^2}{(dy/dx)^3}$. …

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