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

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.The inverse of the function $y = \dfrac{10^x - 10^{-x}}{10^x + 10^{-x}}$ is (A) $\frac{1}{2}\log_{10}\left(\frac{1+x}{1-x}\right)$ (B) $\frac{1}{2}\log_{10}\left(\frac{2+x}{2-x}\right)$ (C) $\frac{1}{2}\log_{10}\left(\frac{1-x}{1+x}\right)$ (D) $\frac{1}{2}\log_{10}\left(\frac{2-x}{2+x}\right)$

   ›Reveal solutionSolution

   Substituting $t=10^x$ turns the given function into a rational expression in $t^2$; solving for $t$ (hence $x$) in terms of $y$ gives the inverse.

   Concept and Intuition

   This function has the same algebraic shape as the hyperbolic tangent, just with base 10 instead of base $e$. The standard trick to invert it is to substitute $t=10^x$, reducing the expression to a simple rational function of $t^2$, solve that algebraically for $t^2$, then take a logarithm to recover $x$ in terms of $y$.

   Step-by-Step Solution

   1. Let $t = 10^x$. Then $y = \dfrac{t - t^{-1}}{t+t^{-1}} = \dfrac{t^2-1}{t^2+1}$ (multiplying numerator and denominator by $t$).
   2. Cross-multiply: $y(t^2+1) = t^2 - 1 \;\Rightarrow\; yt^2+y = t^2-1 \;\Rightarrow\; t^2(y-1) = -(1+y)$.
   3. So $t^2 = \dfrac{1+y}{1-y}$.
   4. Since $t=10^x$: $x = \log_{10}(t) = \dfrac12\log_{10}(t^2) = \dfrac12\log_{10}\!\left(\dfrac{1+y}{1-y}\right)$.
   5. This expresses $x$ in terms of $y$ — i.e., it IS the inverse function; renaming the input variable back to $x$: $f^{-1}(x)=\dfrac12\log_{10}\!\left(\dfrac{1+x}{1-x}\right)$.

   Common Mistakes

   • Sign errors when cross-multiplying and isolating $t^2$, which flip $(1+y)/(1-y)$ into $(1-y)/(1+y)$ — leading to the wrong-looking but similar option (C).

   ✓Final answer

   The correct option is (A) — $\frac{1}{2}\log_{10}\left(\frac{1+x}{1-x}\right)$.

   ANSWER: A

2. 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 $y\ge0$).
   4. So $y=\log(2+\sqrt3)$.

   Common Mistakes

   • Using the formula for $\sinh^{-1}$ (which has a $+$ inside differently) instead of $\cosh^{-1}$.
   • Picking the smaller root $2-\sqrt3$ (which would give a negative $y$, not the principal value).

   ✓Final answer

   The correct option is (A) — $\log(2+\sqrt3)$.

   ANSWER: A

3. 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 $1/x=\sinh u=-1/7$.
   5. $\tanh^{-1}(1/x)=\tanh^{-1}\!\left(-\dfrac17\right)=\dfrac12\log\dfrac{1+(-1/7)}{1-(-1/7)}=\dfrac12\log\dfrac{6/7}{8/7}=\dfrac12\log\dfrac34=\log\sqrt{3/4}=\log\dfrac{\sqrt3}{2}$.

   Common Mistakes

   • Forgetting $\text{Cosech}^{-1}x=\sinh^{-1}(1/x)$ (mixing it up with $\sinh^{-1}x$ directly).
   • Sign slips when rationalising $e^{-u}$.

   ✓Final answer

   The correct option is (B) — $\log\dfrac{\sqrt{3}}{2}$.

   ANSWER: B

4. 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. $x-y=\dfrac43-\dfrac{13}{12}=\dfrac{16}{12}-\dfrac{13}{12}=\dfrac{3}{12}=\dfrac14$.
   4. $\operatorname{Tanh}^{-1}\!\left(\dfrac14\right)=\dfrac12\log\!\left(\dfrac{1+\frac14}{1-\frac14}\right)=\dfrac12\log\!\left(\dfrac{5/4}{3/4}\right)=\dfrac12\log\dfrac53$.
   5. $\dfrac12\log\dfrac53=\log\sqrt{\dfrac53}=\log\sqrt{\dfrac53}$.

   Common Mistakes

   • Confusing $\operatorname{Cosh}^{-1}$ with $\operatorname{Sinh}^{-1}$ formulas — $\cosh(\log k)=\frac{k+1/k}2$ has a $+$, not $-$.
   • Arithmetic slip subtracting the fractions $\frac43-\frac{13}{12}$ without a common denominator.

   ✓Final answer

   The correct option is (A) — $\log\sqrt{\dfrac53}$.

   ANSWER: A

5. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If $\cos^{-1}x = \sin^{-1}(3x)$ then $x =$ (A) $\frac{\sqrt{10}}{10}$ (B) $\frac{\sqrt{5}}{5}$ (C) $\frac{5}{2\sqrt{6}}$ (D) $\frac{-\sqrt{10}}{10}$

   ›Reveal solutionSolution

   Equating the two inverse-trig expressions to a common angle $\theta$ and using $\sin^2\theta+\cos^2\theta=1$ gives $x=\frac{\sqrt{10}}{10}$ (the positive root, forced by the range of $\sin^{-1}$).

   Concept and Intuition

   $\cos^{-1}x$ has range $[0,\pi]$ while $\sin^{-1}(3x)$ has range $[-\pi/2,\pi/2]$. For the two to be equal, the common value $\theta$ must lie in the overlap $[0,\pi/2]$, which restricts the sign of $x$.

   Step-by-Step Solution

   1. Let $\theta = \cos^{-1}x = \sin^{-1}(3x)$. Then $\cos\theta=x$ and $\sin\theta=3x$.
   2. Since $\theta=\sin^{-1}(3x)$, $\theta\in[-\pi/2,\pi/2]$; but also $\theta=\cos^{-1}x\in[0,\pi]$. The intersection forces $\theta\in[0,\pi/2]$.
   3. Use $\sin^2\theta+\cos^2\theta=1$: $(3x)^2+x^2=1 \Rightarrow 10x^2=1 \Rightarrow x^2=\frac{1}{10}$.
   4. Since $\theta\in[0,\pi/2]$, $\cos\theta=x\ge0$, so we take the positive root: $x=\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}$.

   Common Mistakes

   • Taking $x=\pm\frac{\sqrt{10}}{10}$ without using the range argument to fix the sign.

   ✓Final answer

   The correct option is (A) — $\frac{\sqrt{10}}{10}$.

   ANSWER: A

6. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $g(x)$ is the inverse of the function $f(x)$ and $f'(x) = \dfrac{1}{h(x)}$, then $g'(x) =$ (A) $h(g(x))$ (B) $g(h(x))$ (C) $h'(f(x))$ (D) $f(h(x))$

   ›Reveal solutionSolution

   The inverse function derivative rule directly gives $g'(x)=h(g(x))$.

   Concept and Intuition

   If $g$ is the inverse of $f$, then $g'(x)=\dfrac{1}{f'(g(x))}$ — this is the standard inverse-function derivative formula.

   Step-by-Step Solution

   1. $g'(x) = \dfrac{1}{f'(g(x))}$.
   2. Given $f'(x)=\dfrac{1}{h(x)}$, so $f'(g(x))=\dfrac{1}{h(g(x))}$.
   3. $g'(x)=\dfrac{1}{1/h(g(x))}=h(g(x))$.

   Common Mistakes

   • Writing $g'(x)=1/f'(x)$ instead of $1/f'(g(x))$ (forgetting to compose with $g$).

   ✓Final answer

   The correct option is (A) — $h(g(x))$.

   ANSWER: A

7. 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: $y+1=5y-5\Rightarrow6=4y\Rightarrow y=\dfrac32$.
   5. $xy=\dfrac23\cdot\dfrac32=1$.
   6. $\tan^{-1}(xy)=\tan^{-1}(1)=\dfrac{\pi}{4}$.

   Common Mistakes

   • Confusing $\coth^{-1}$'s log form with $\tanh^{-1}$'s (they're reciprocal-flavored, not identical) — mixing them up gives the wrong $y$.
   • Forgetting $\log\sqrt5=\tfrac12\log5$ and mishandling the exponent when clearing logs.

   ✓Final answer

   The correct option is (A) — $\dfrac{\pi}{4}$.

   ANSWER: A

8. 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.
   6. Check directly: $f(-1)=0-1=-1$ and $f^{-1}(-1)=-1$ (since $f(-1)=-1$) — matches. $f(0)=1-1=0$ and $f^{-1}(0)=0$ — matches.
   7. So the solution set, restricted to the real domain $x\ge -1$, is exactly $\{0,-1\}$.

   Common Mistakes

   • Including the complex roots of $u^3=1$ (option (C)) — these don't correspond to any real $x$, and the domain here is real numbers only.
   • Assuming $f(x)=f^{-1}(x)$ only ever happens at points where $f(x)=x$ without checking whether $f(f(x))=x$ has other real solutions (here it doesn't, but that must be verified, not assumed).

   ✓Final answer

   The correct option is (A) — $\{0,-1\}$.

   ANSWER: A

9. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.Let $f(x) = (x + 2)^2 - 2, x \ge -2$. Then $f^{-1}(x) =$ (A) $-\sqrt{2 + x} - 2$ (B) $\sqrt{2 + x} + 2$ (C) $\sqrt{2 + x} - 2$ (D) $-\sqrt{2 + x} + 2$

   ›Reveal solutionSolution

   Solve $y=(x+2)^2-2$ for $x$, keeping only the branch consistent with the given domain

   $x\ge-2$; this gives $f^{-1}(x)=\sqrt{2+x}-2$.

   Concept and Intuition

   $f$ is only invertible because its domain is restricted to $x\ge-2$, which makes it one-to-one (a

   parabola shifted so its vertex is exactly at the left edge of the domain). Inverting means solving

   $y=f(x)$ for $x$ in terms of $y$ and choosing the square-root sign that matches the given domain

   restriction.

   Step-by-Step Solution

   1. Start with $y = (x+2)^2 - 2$.
   2. Add 2: $y + 2 = (x+2)^2$.
   3. Take square roots: $x + 2 = \pm\sqrt{y+2}$.
   4. Since $x \ge -2$, we have $x + 2 \ge 0$, so we must choose the positive root: $x + 2 = \sqrt{y+2}$.
   5. Solve for $x$: $x = \sqrt{y+2} - 2$.
   6. Relabel $y \to x$: $f^{-1}(x) = \sqrt{2+x} - 2$.

   Common Mistakes

   • Picking the negative square-root branch, which would correspond to $x \le -2$, the wrong half of the parabola relative to the stated domain.

   ✓Final answer

   The correct option is (C) — $\sqrt{2+x}-2$.

   ANSWER: C

10. 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 $t=g(x)$).
    5. Rewrite $(\star)$ using $t$: $f'(t) = \dfrac{1}{1+\sec^2 x} = \dfrac{1}{1+\sec^2 f(t)}$ (substituting $x=f(t)$).
    6. Renaming the dummy variable $t$ back to $x$:

    $f'(x) = \frac{1}{1+\sec^2 f(x)}$

    Common Mistakes

    • Stopping at $f'(g(x))=\dfrac1{1+\sec^2x}$ and mistakenly picking option (C) ($\sec^2g(x)$) — that expression is $f'$ evaluated at $g(x)$, not $f'(x)$ itself as a function of $x$.
    • Forgetting that $f$ and $g$ being mutual inverses means $x=f(g(x))=f(t)$ when $t=g(x)$, which is the key substitution step.
    • Sign or reciprocal errors in the inverse-derivative formula.

    ✓Final answer

    The correct option is (B) — $\dfrac{1}{1+\sec^2 f(x)}$.

    ANSWER: B

11. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.If $a, b, c, d \ne 0$ and $f\left(\dfrac{at + b}{ct + d}\right) = t$, then $\displaystyle\int f(x)\,dx =$ (A) $\dfrac{-d}{c}x + \dfrac{bc - ad}{c^2}\log(cx - a) + K$ (B) $\dfrac{-d}{c}x + \dfrac{bc - ad}{c^2}\log(cx + d) + K$ (C) $\dfrac{a}{c}x + \dfrac{ad - bc}{c^2}\log(cx + a) + K$ (D) $\dfrac{a}{c}x - \dfrac{ad - bc}{c}\log(cx + a) + K$

    ›Reveal solutionSolution

    The functional equation secretly defines $f$ as the inverse of a Möbius (linear-fractional) map; solving for $t$ in terms of $x$ gives an explicit formula for $f(x)$, which is then a routine "linear plus $1/(\text{linear})$" integral.

    Concept and Intuition

    Being told $f\!\left(\dfrac{at+b}{ct+d}\right)=t$ means: whatever expression is plugged into $f$, the output is the parameter $t$ that produced it. So if we let $x=\dfrac{at+b}{ct+d}$ and solve this equation for $t$ in terms of $x$, we get an explicit closed form for $f(x)$ itself.

    Step-by-Step Solution

    1. Set $x=\dfrac{at+b}{ct+d}$. Cross-multiplying: $x(ct+d)=at+b\Rightarrow xct+xd=at+b$.
    2. Collect $t$ terms: $t(xc-a)=b-xd\Rightarrow t=\dfrac{b-xd}{xc-a}$.
    3. So $f(x)=\dfrac{-dx+b}{cx-a}$.
    4. Do polynomial-style division: write $-dx+b = -\dfrac{d}{c}(cx-a) + \left(b-\dfrac{ad}{c}\right)$ (check: $-\frac{d}{c}(cx-a)=-dx+\frac{ad}{c}$, and adding $b-\frac{ad}{c}$ recovers $-dx+b$ ✓).
    5. So $f(x)=-\dfrac{d}{c}+\dfrac{b-ad/c}{cx-a}=-\dfrac{d}{c}+\dfrac{bc-ad}{c}\cdot\dfrac{1}{cx-a}$.
    6. Integrate term by term:

    $\int f(x)\,dx = -\frac{d}{c}x + \frac{bc-ad}{c}\cdot\frac{1}{c}\log|cx-a|+K = -\frac{d}{c}x+\frac{bc-ad}{c^2}\log(cx-a)+K.$

    7. This matches option (A) exactly.

    Common Mistakes

    • Sign errors while solving the cross-multiplied equation for $t$ — easy to flip a sign and end up with $(ad-bc)$ instead of $(bc-ad)$.
    • Forgetting the extra factor of $1/c$ that appears from $\int\frac{1}{cx-a}dx=\frac1c\log|cx-a|$.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

12. 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

    • Trying to explicitly write $g(y)=\tan y$ and differentiate at $f(2)=\tan^{-1}2$ directly (correct but needlessly harder) instead of using the inverse-derivative shortcut.
    • Computing $f'(x)$ incorrectly as $\frac{1}{1-x^2}$ (confusing with $\tanh^{-1}$).

    ✓Final answer

    The correct option is (D) — $5$.

    ANSWER: D