Q.If $\frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}$ such that $f(2) = 0$. Then $f(x)$ is (A) $x^4 + \frac{1}{x^3} + \frac{129}{8}$ (B) $x^3 + \frac{1}{x^4} + \frac{129}{8}$ (C) $x^4 + \frac{1}{x^3} - \frac{129}{8}$ (D) $x^3 + \frac{1}{x^4} - \frac{129}{8}$

Showing the 12 most recent of 26 on this concept.

1. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.If $f'(x) = \tan^2(x) + \cot^2(x)$ and $f\left(\dfrac{\pi}{4}\right) = 0$, then $f(x) =$ ______ (A) $\tan(x) - \cot(x) - x + \dfrac{\pi}{2}$ (B) $\tan(x) - \cot(x) - 2x + \dfrac{\pi}{2}$ (C) $\tan(x) + \cot(x) - 2x + \dfrac{\pi}{2}$ (D) $\sec(x) - \text{cosec}(x) - 2x + \dfrac{\pi}{2}$

   ›Reveal solutionSolution

   Rewrite $\tan^2x+\cot^2x$ using Pythagorean identities, integrate term by term, then fix the constant using $f(\pi/4)=0$.

   Concept and Intuition

   Whenever you see $\tan^2x$ or $\cot^2x$ in something you must integrate, immediately convert using $\tan^2x=\sec^2x-1$ and $\cot^2x=\csc^2x-1$ — these have known antiderivatives ($\tan x$, $-\cot x$), unlike the squared trig functions themselves.

   Step-by-Step Solution

   1. Rewrite: $f'(x) = (\sec^2x-1)+(\csc^2x-1) = \sec^2x+\csc^2x-2$.
   2. Integrate: $f(x) = \int(\sec^2x+\csc^2x-2)\,dx = \tan x - \cot x - 2x + C$.
   3. Apply the condition $f(\pi/4)=0$: $\tan(\pi/4)=1$, $\cot(\pi/4)=1$, so …

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2. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.If $f(x) = \dfrac{1}{(\cos^2 x)\sqrt{1 + \tan x}}$, then its anti-derivative $F(x) =$ ______, given $F(0) = 4$ (A) $\sqrt{1 + \tan x} + 4$ (B) $\dfrac{2}{3}(1 + \tan x)^{3/2}$ (C) $2\left(\sqrt{1 + \tan x} + 1\right)$ (D) $\sqrt{1 + \tan x} + 2$

   ›Reveal solutionSolution

   A substitution-based integration followed by using the initial condition to fix the constant. The answer is $2(\sqrt{1+\tan x}+1)$.

   Concept and Intuition

   Recognizing that $\sec^2x\,dx$ is exactly the derivative of $1+\tan x$ lets a substitution reduce the integrand to a simple power of $u$, which integrates immediately.

   Step-by-Step Solution

   1. $f(x)=\dfrac{1}{\cos^2x\sqrt{1+\tan x}}=\dfrac{\sec^2x}{\sqrt{1+\tan x}}$.
   2. Let $u=1+\tan x$, so $du=\sec^2x\,dx$.
   3. $F(x)=\displaystyle\int\dfrac{du}{\sqrt u}=2\sqrt u+C=2\sqrt{1+\tan x}+C$.
   4. Apply $F(0)=4$: $2\sqrt{1+\tan0}+C=2\sqrt1+C=2+C=4\Rightarrow C=2$.
   5. So $F(x)=2\sqrt{1+\tan x}+2=2\left(\sqrt{1+\tan x}+1\right)$.

   Common Mistakes …

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3. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.If $f'(x)=a\cos x+b\sin x$ and $f'(0)=4,\ f(0)=3,\ f\left(\dfrac{\pi}{2}\right)=5$, then $f(x)=$ (A) $2\cos x+4\sin x+1$ (B) $4\cos x+2\sin x+1$ (C) $2\cos x+3\sin x+1$ (D) $4\cos x+\sin x+1$

   ›Reveal solutionSolution

   $a$ is pinned by $f'(0)$, then integrating and using the two function values $f(0)=3,f(\pi/2)=5$ gives $b$ and the constant of integration.

   Concept and Intuition

   Given $f'(x)$ explicitly (up to unknown constants) and enough boundary values of $f$ itself, integrate first (introducing a constant $C$), then use every given condition (there are exactly as many conditions as unknowns: $a,b,C$).

   Step-by-Step Solution

   1. $f'(x)=a\cos x+b\sin x$. At $x=0$: $f'(0)=a\cos0+b\sin0=a$. Given $f'(0)=4\Rightarrow a=4$.
   2. Integrate: $f(x)=\int(4\cos x+b\sin x)dx=4\sin x-b\cos x+C$.
   3. $f(0)=4(0)-b(1)+C=-b+C$. Given $=3$: $-b+C=3$.
   4. $f(\pi/2)=4(1)-b(0)+C=4+C$. Given $=5$: $C=1$.
   5. From $-b+1=3\Rightarrow b=-2$.
   6. $f(x)=4\sin x-(-2)\cos x+1=4\sin x+2\cos x+1$, i.e. $2\cos x+4\sin x+1$. …

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4. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.If the solution of $\dfrac{dy}{dx} = \dfrac{y^3 \cos\sqrt{x}}{\sqrt{x}\, e^{1/y^2}}$, $y(0) = 1$ is $\dfrac{1}{y^2} = \log_e(f(x))$, then $f(x) =$ (A) $4 + 4\sin\sqrt{x}$ (B) $e\sin\sqrt{x}$ (C) $1 - 4\sin\sqrt{x}$ (D) $e - 4\sin\sqrt{x}$

   ›Reveal solutionSolution

   This is a separable ODE; separating variables and using the initial condition $y(0)=1$ gives $f(x)=e-4\sin\sqrt x$.

   Concept and Intuition

   The equation separates cleanly into a function of $y$ times $dy$ equal to a function of $x$ times $dx$. Both sides then need simple substitutions ($u=1/y^2$ and $v=\sqrt x$) to become directly integrable exponential/trig forms.

   Step-by-Step Solution

   1. $\dfrac{dy}{dx} = \dfrac{y^3\cos\sqrt x}{\sqrt x\,e^{1/y^2}} \Rightarrow e^{1/y^2}\,y^{-3}\,dy = \dfrac{\cos\sqrt x}{\sqrt x}\,dx$.
   2. LHS: let $u=1/y^2 \Rightarrow du = -2y^{-3}dy \Rightarrow y^{-3}dy = -\tfrac12 du$. So $\displaystyle\int e^u\left(-\tfrac12\right)du = -\tfrac12 e^u + C_1 = -\tfrac12 e^{1/y^2}+C_1$.
   3. RHS: let $v=\sqrt x \Rightarrow dv=\dfrac{dx}{2\sqrt x}\Rightarrow \dfrac{dx}{\sqrt x}=2\,dv$. So $\displaystyle\int\cos v\cdot2\,dv = 2\sin v + C_2 = 2\sin\sqrt x + C_2$.
   4. Combine: $-\dfrac12e^{1/y^2} = 2\sin\sqrt x + C$.
   5. Apply $y(0)=1 \Rightarrow 1/y^2=1$ at $x=0$: $-\dfrac12 e^1 = 2\sin0+C \Rightarrow C=-\dfrac{e}{2}$. …

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5. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If the solution of $\dfrac{dy}{dx} = \log(x + 1)$ when $y(0) = 3$ is $y = (x + 1)\log(x + 1) + f(x)$, then $f(x) =$ (A) $3 - x$ (B) $x - 3$ (C) $1 - x$ (D) $x - 1$

   ›Reveal solutionSolution

   This tests integrating $\log(x+1)$ by substitution and then pinning the constant with the initial condition. The answer is $f(x)=3-x$.

   Concept and Intuition

   The differential equation $\dfrac{dy}{dx}=\log(x+1)$ is already separated — $y$ is just the antiderivative of the right-hand side. The initial condition $y(0)=3$ fixes the single constant of integration, and comparing the resulting expression to the given form of the solution lets us read off $f(x)$ without guessing.

   Step-by-Step Solution

   1. Integrate: $y=\displaystyle\int \log(x+1)\,dx + C$.
   2. Substitute $u=x+1$, $du=dx$: $\displaystyle\int \log u\,du = u\log u - u + C$ (standard result, by parts with $dv=du$).
   3. Back-substitute: $y=(x+1)\log(x+1)-(x+1)+C$.
   4. Compare with the given form $y=(x+1)\log(x+1)+f(x)$: so $f(x) = -(x+1)+C = C-1-x$. …

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6. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.g(x) is an anti derivative of $f(x)=1+2^x\log 2$ and the graph of $y=g(x)$ passes through $\left(-1,\dfrac{1}{2}\right)$. Then the curve meets the Y-axis at (A) (0, 1) (B) (0, 2) (C) (0, -2) (D) (1, 1)

   ›Reveal solutionSolution

   Integrating $f(x)=1+2^x\log2$ gives $g(x)=x+2^x+C$; the given point fixes $C=1$, and evaluating at $x=0$ gives the Y-intercept $(0,2)$.

   Concept and Intuition

   Since $\dfrac{d}{dx}2^x=2^x\log 2$, the antiderivative of $2^x\log2$ is simply $2^x$ — recognizing this avoids unnecessary substitution work.

   Step-by-Step Solution

   1. $g(x)=\int(1+2^x\log2)\,dx=x+2^x+C$.
   2. Use the given point $\left(-1,\tfrac12\right)$: $g(-1)=-1+2^{-1}+C=-1+\tfrac12+C=-\tfrac12+C=\tfrac12\Rightarrow C=1$.
   3. So $g(x)=x+2^x+1$.
   4. Where the curve meets the Y-axis, $x=0$: $g(0)=0+2^0+1=0+1+1=2$. …

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7. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If a curve passes through $(1, 2)$ and has the slope of its tangent $1 - \dfrac{1}{x^2}$ at a point $(x, y)$, then the equation of that curve is (A) $y = 3x - \dfrac{1}{x}$ (B) $y = x + \dfrac{1}{x}$ (C) $y = 2x + \dfrac{1}{x} - 1$ (D) $y = x + \dfrac{2}{x} - 1$

   ›Reveal solutionSolution

   Direct integration of the given slope expression, followed by using the point $(1,2)$ to fix the constant, gives $y=x+\dfrac{1}{x}$.

   Concept and Intuition

   When the slope $\dfrac{dy}{dx}$ is given purely as a function of $x$, the curve is found by straightforward integration (no need for separation of variables in a more complex sense) — then a known point on the curve pins down the constant of integration.

   Step-by-Step Solution

   1. $\dfrac{dy}{dx} = 1 - \dfrac{1}{x^2}$.
   2. Integrate both sides: $y = \displaystyle\int\left(1-\dfrac{1}{x^2}\right)dx = x + \dfrac{1}{x} + C$.
   3. Apply the point $(1,2)$: $2 = 1 + 1 + C \Rightarrow C = 0$. …

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8. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.Let $y = Y(x)$ be the solution of the differential equation $\frac{dy}{dx} + y\tan x = 2x + x^2\tan x$, $x \in \left( \frac{-\pi}{2}, \frac{\pi}{2} \right)$, such that $Y(0) = 1$, then ________ (A) $y\left(\frac{\pi}{4}\right) + Y\left(\frac{-\pi}{4}\right) = \frac{\pi^2}{2} + 2$ (B) $y'\left(\frac{\pi}{4}\right) + Y'\left(\frac{-\pi}{4}\right) = -\sqrt{2}$ (C) $y\left(\frac{\pi}{4}\right) - Y\left(\frac{-\pi}{4}\right) = \sqrt{2}$ (D) $y'\left(\frac{\pi}{4}\right) - Y'\left(\frac{-\pi}{4}\right) = \pi - \sqrt{2}$

   ›Reveal solutionSolution

   Solving the first-order linear ODE explicitly gives $Y(x)=x^2+\cos x$; checking each option against this closed form picks out (D). Answer: $\pi-\sqrt2$.

   Concept and Intuition

   This is a standard first-order linear ODE $\frac{dy}{dx}+P(x)y=Q(x)$, solved with an integrating factor. Once we have the explicit closed form for $Y(x)$, we can just plug in and test every option directly instead of guessing.

   Step-by-Step Solution

   1. ODE: $\dfrac{dy}{dx}+y\tan x = 2x+x^2\tan x$. Integrating factor: $\mu=e^{\int\tan x\,dx}=e^{-\log|\cos x|}=\sec x$.
   2. Multiply through by $\sec x$: $\sec x\,\dfrac{dy}{dx}+y\sec x\tan x = 2x\sec x + x^2\sec x\tan x$.
   3. LHS $=\dfrac{d}{dx}(y\sec x)$. Check RHS: $\dfrac{d}{dx}(x^2\sec x)=2x\sec x+x^2\sec x\tan x$ — matches exactly!
   4. So $\dfrac{d}{dx}(y\sec x)=\dfrac{d}{dx}(x^2\sec x)\Rightarrow y\sec x = x^2\sec x + C \Rightarrow y = x^2+C\cos x$.
   5. Apply $Y(0)=1$: $0+C(1)=1\Rightarrow C=1$. So $Y(x)=x^2+\cos x$.
   6. $Y'(x)=2x-\sin x$. Compute $Y'(\pi/4)=\dfrac\pi2-\dfrac{\sqrt2}2$ and $Y'(-\pi/4)=-\dfrac\pi2+\dfrac{\sqrt2}2$. …

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9. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.If y=y(x) is the solution of $\dfrac{dy}{dx}=\dfrac{x-y\cos x}{1+\sin x}$, $y\left(\dfrac{\pi}{2}\right)=\dfrac{\pi^2}{8}$, then $y(\pi)=$ (A) $\dfrac{5\pi^2}{8}$ (B) $\dfrac{7\pi^2}{8}$ (C) $\dfrac{9\pi^2}{8}$ (D) $\dfrac{12\pi^2}{7}$

   ›Reveal solutionSolution

   Recognise the left side of the rearranged ODE as an exact derivative $\dfrac{d}{dx}[y(1+\sin x)]$, integrate directly, use the initial condition to find the constant, then evaluate at $x=\pi$.

   Concept and Intuition

   Many "linear-looking" first-order ODEs are secretly exact derivatives in disguise — recognising the pattern $y'g(x)+y\,g'(x)=\dfrac{d}{dx}[y\,g(x)]$ turns a substitution-heavy linear-ODE problem into direct integration.

   Step-by-Step Solution

   1. Given: $\dfrac{dy}{dx}=\dfrac{x-y\cos x}{1+\sin x}$. Multiply through by $(1+\sin x)$: $(1+\sin x)\dfrac{dy}{dx}+y\cos x=x$.
   2. Note $\dfrac{d}{dx}\left[y(1+\sin x)\right]=y'(1+\sin x)+y\cos x$ — exactly the left side above.
   3. So $\dfrac{d}{dx}\left[y(1+\sin x)\right]=x$. Integrate: $y(1+\sin x)=\dfrac{x^2}{2}+C$.
   4. Apply $y(\pi/2)=\dfrac{\pi^2}{8}$: $1+\sin(\pi/2)=2$, so LHS $=\dfrac{\pi^2}{8}\times 2=\dfrac{\pi^2}{4}$.
   5. RHS at $x=\pi/2$: $\dfrac{(\pi/2)^2}{2}+C=\dfrac{\pi^2}{8}+C$.
   6. Equate: $\dfrac{\pi^2}{4}=\dfrac{\pi^2}{8}+C \Rightarrow C=\dfrac{\pi^2}{8}$.
   7. General solution: $y(1+\sin x)=\dfrac{x^2}{2}+\dfrac{\pi^2}{8}$. …

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10. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.If $\lim_{x\to\infty} y(x) = \dfrac{\pi}{2}$, then the solution of $x^3\sin y\dfrac{dy}{dx}=2$ is $\cos y=$ (A) $\dfrac{3}{x^2}$ (B) $\dfrac{1}{x}$ (C) $\dfrac{1}{x^2}$ (D) $\dfrac{2}{x^3}$

    ›Reveal solutionSolution

    This is a separable ODE; integrate directly and use the given limiting condition at infinity to pin down the constant of integration.

    Concept and Intuition

    $x^3\sin y\,dy/dx=2$ separates cleanly into a function of $y$ times $dy$ equals a function of $x$ times $dx$. The unusual boundary condition (a limit as $x\to\infty$, rather than a value at a finite point) still fixes the constant because both sides of the solution tend to definite limits.

    Step-by-Step Solution

    1. Rewrite: $x^3\sin y\,\dfrac{dy}{dx}=2 \Rightarrow \sin y\,dy=\dfrac{2}{x^3}dx$.
    2. Integrate both sides: $\displaystyle\int \sin y\,dy=\int 2x^{-3}dx \Rightarrow -\cos y = -x^{-2}+C_0$.
    3. Rearranged: $\cos y = x^{-2}-C_0 = \dfrac{1}{x^2}-C$ (renaming the constant). …

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11. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.The solution of the differential equation $x^2(y+1)\dfrac{dy}{dx} + y^2(x+1)^2 = 0$, when $y(1) = 2$, is (A) $\log|x^2y| = \dfrac{2}{x} + \dfrac{1}{y} + x - 1$ (B) $\log\left|\dfrac{1}{4}x^2y\right| = \dfrac{1}{x} + \dfrac{2}{y} + x - 1$ (C) $\log\left|\dfrac{1}{2}x^2y\right| = \dfrac{1}{x} + \dfrac{1}{y} - x - \dfrac{1}{2}$ (D) $\log\left|\dfrac{1}{3}x^2y\right| = \dfrac{1}{x} + \dfrac{1}{y} - x + \dfrac{1}{2}$

    ›Reveal solutionSolution

    This is a separable ODE; separating and integrating gives an implicit relation, and the initial condition $y(1)=2$ pins the constant to match option (C).

    Concept and Intuition

    $x^2(y+1)\,dy = -y^2(x+1)^2\,dx$ separates cleanly because all the $y$-terms can be gathered on one side (as $\frac{y+1}{y^2}$) and all $x$-terms on the other (as $\frac{(x+1)^2}{x^2}$). Both sides then reduce to standard integrals: $\frac1y+\frac{1}{y^2}$ integrates via power rule and $\log$, and $\frac{(x+1)^2}{x^2}=1+\frac2x+\frac1{x^2}$ likewise.

    Step-by-Step Solution

    1. Rearranging: $x^2(y+1)\,dy=-y^2(x+1)^2\,dx \Rightarrow \dfrac{y+1}{y^2}\,dy=-\dfrac{(x+1)^2}{x^2}\,dx$.
    2. LHS: $\dfrac{y+1}{y^2}=\dfrac1y+\dfrac{1}{y^2}$, so $\displaystyle\int\left(\frac1y+\frac1{y^2}\right)dy=\log|y|-\frac1y+C_1$.
    3. RHS: $\dfrac{(x+1)^2}{x^2}=\dfrac{x^2+2x+1}{x^2}=1+\dfrac2x+\dfrac1{x^2}$, so $\displaystyle-\int\left(1+\frac2x+\frac1{x^2}\right)dx=-x-2\log|x|+\frac1x+C_2$.
    4. Equate: $\log|y|-\dfrac1y=-x-2\log|x|+\dfrac1x+C$.
    5. Rearranged: $\log|y|+2\log|x|=\dfrac1x+\dfrac1y-x+C \Rightarrow \log(x^2y)=\dfrac1x+\dfrac1y-x+C$.
    6. Apply $y(1)=2$: $\log(1\cdot2)=1+\dfrac12-1+C \Rightarrow \ln2=\dfrac12+C \Rightarrow C=\ln2-\dfrac12$.
    7. Substitute back: $\log(x^2y)=\dfrac1x+\dfrac1y-x+\ln2-\dfrac12$, i.e. $\log(x^2y)-\ln2=\dfrac1x+\dfrac1y-x-\dfrac12$.
    8. $\log(x^2y)-\ln2=\log\!\left(\dfrac{x^2y}{2}\right)=\log\left|\dfrac12x^2y\right|$, so finally …

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12. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
    Q.If $y = f(x)$ is a solution of $\dfrac{dy}{dx} = (y - Kx)^2 + K$ when $x = 0$ and $y = 1$, then $f(2) =$ (A) $2K + 1$ (B) $2K - 1$ (C) $2K + 5$ (D) $2K - 5$

    ›Reveal solutionSolution

    A shift substitution $v=y-Kx$ turns the equation into the separable $dv/dx=v^2$; solving and evaluating at $x=2$ gives $f(2)=2K-1$.

    Concept and Intuition

    The right side $(y-Kx)^2+K$ is not separable in $y$ directly, but the combination $y-Kx$ appears squared — this is the signal to substitute $v=y-Kx$, which typically removes the extra additive constant $K$ from the derivative and leaves a clean separable equation in $v$.

    Step-by-Step Solution

    1. Let $v=y-Kx$. Then $\dfrac{dv}{dx}=\dfrac{dy}{dx}-K$.
    2. Given $\dfrac{dy}{dx}=(y-Kx)^2+K=v^2+K$, so $\dfrac{dv}{dx}=v^2+K-K=v^2$.
    3. Separate variables: $\dfrac{dv}{v^2}=dx\Rightarrow -\dfrac{1}{v}=x+c$.
    4. Initial condition: at $x=0$, $y=1\Rightarrow v=y-K\cdot0=1$. So $-1=0+c\Rightarrow c=-1$.
    5. Thus $-\dfrac1v=x-1\Rightarrow v=\dfrac{1}{1-x}$. …

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