Q.Find the anti derivative $F$ of $f$ defined by $f(x) = 4x^3 - 6$, where $F(0) = 3$.

Showing the 12 most recent of 26 on this concept.

1. 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 …

   🔒 Start your 14-day free trial to see the full solution →
2. 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. $2\cos x+4\sin x+1$. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 $f(x) = -(x+1)+C = C-1-x$. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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 …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If $\cos x \dfrac{dy}{dx} - y\sin x = 6x$, $\left(0 < x < \dfrac{\pi}{2}\right)$ and $y\left(\dfrac{\pi}{3}\right) = 0$ then $y\left(\dfrac{\pi}{6}\right) =$ (A) $\dfrac{-\pi^2}{4\sqrt{3}}$ (B) $\dfrac{-\pi^2}{2}$ (C) $\dfrac{-\pi^2}{2\sqrt{3}}$ (D) $\dfrac{\pi^2}{2\sqrt{3}}$

   ›Reveal solutionSolution

   $\cos x\,y' - y\sin x = \dfrac{d}{dx}(y\cos x) = 6x$, so $y\cos x = 3x^2+C$; using $y(\pi/3)=0$ gives $y(\pi/6)=\dfrac{-\pi^2}{2\sqrt3}$.

   Concept and Intuition

   Many first-order linear ODEs are secretly an exact derivative in disguise. Here $\cos x\,y'-y\sin x$ is precisely the product rule expansion of $\dfrac{d}{dx}(y\cos x)$, so no integrating factor is even needed.

   Step-by-Step Solution

   1. Observe $\dfrac{d}{dx}(y\cos x) = y'\cos x - y\sin x$ — exactly the LHS.
   2. So the equation is $\dfrac{d}{dx}(y\cos x)=6x$.
   3. Integrate: $y\cos x = 3x^2+C$.
   4. Apply $y(\pi/3)=0$: $0\cdot\cos(\pi/3) = 3\left(\dfrac{\pi}{3}\right)^2+C \Rightarrow 0=\dfrac{\pi^2}{3}+C \Rightarrow C=-\dfrac{\pi^2}{3}$.
   5. At $x=\pi/6$: $y\cos(\pi/6) = 3\left(\dfrac{\pi}{6}\right)^2 - \dfrac{\pi^2}{3} = \dfrac{\pi^2}{12}-\dfrac{4\pi^2}{12} = -\dfrac{3\pi^2}{12}=-\dfrac{\pi^2}{4}$. …

   🔒 Start your 14-day free trial to see the full solution →
6. 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 $x=0$: $-\dfrac12 e^1 = 2\sin0+C \Rightarrow C=-\dfrac{e}{2}$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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, $x=0$: $g(0)=0+2^0+1=0+1+1=2$. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.If f(x) is a function such that $f'(x)=\sqrt{f^2(x)-1}$ and $f(0)=1$, then $f(1)=$ (A) $\frac{e^{-2}+1}{2e}$ (B) $\frac{e^2+1}{2e}$ (C) $\frac{e^2-1}{2e}$ (D) $\frac{e^{-2}-1}{2e}$

   ›Reveal solutionSolution

   This is a separable differential equation whose natural substitution is the hyperbolic identity $\cosh^2u-\sinh^2u=1$; the answer is $f(1)=\cosh 1=\dfrac{e^2+1}{2e}$.

   Concept and Intuition

   Whenever you see $f'(x)=\sqrt{f^2(x)-1}$, the structure $\sqrt{f^2-1}$ screams hyperbolic substitution, because $\cosh^2u-1=\sinh^2u$. Setting $f=\cosh u$ turns the messy square root into the clean function $\sinh u$, and the chain rule collapses the whole ODE to $u'=1$ — a straight line in disguise.

   Step-by-Step Solution

   1. Separate variables: $\dfrac{df}{\sqrt{f^2-1}}=dx$.
   2. Let $f=\cosh u$, so $df=\sinh u\,du$ and $\sqrt{f^2-1}=\sinh u$ (taking $u\ge 0$ since $f\ge 1$ near $x=0$).
   3. The equation becomes $\dfrac{\sinh u\,du}{\sinh u}=dx\Rightarrow du=dx\Rightarrow u=x+C$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $-\dfrac1v=x-1\Rightarrow v=\dfrac{1}{1-x}$. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.If the slope of the tangent at a point $(x, y)$ on a curve is $\dfrac{y - 4}{x - 3}$ and the curve passes through $(4, 3)$, then the point where it cuts the line $y = x$ is (A) $(1, 1)$ (B) $(3, 3)$ (C) $\left(\dfrac{7}{2}, \dfrac{7}{2}\right)$ (D) $\left(-\dfrac{5}{2}, -\dfrac{5}{2}\right)$

    ›Reveal solutionSolution

    The given slope condition is a separable differential equation whose solution is a straight line through $(3,4)$; using the extra point $(4,3)$ pins down that line exactly, and its intersection with $y=x$ is easy to find.

    Concept and Intuition

    "The slope at every point $(x,y)$ equals $\dfrac{y-4}{x-3}$" is really a differential equation, $\dfrac{dy}{dx}=\dfrac{y-4}{x-3}$. This particular form is separable and integrates to a family of straight lines through the fixed point $(3,4)$ — recognizing that shortcut avoids re-deriving calculus each time, but let's still integrate directly to be rigorous.

    Step-by-Step Solution

    1. Separate variables: $\dfrac{dy}{y-4}=\dfrac{dx}{x-3}$.
    2. Integrate both sides: $\log|y-4|=\log|x-3|+C_1$, so $y-4=K(x-3)$ for some constant $K$.
    3. Use the given point $(4,3)$ (the curve passes through it): $3-4=K(4-3)\Rightarrow -1=K$.
    4. So the curve is $y-4=-(x-3)=3-x$, i.e. $y=7-x$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.If $x\log x\dfrac{dy}{dx} + y = \log x^2$ and $y(e) = 0$, then $y(e^2) =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $\dfrac{3}{2}$

    ›Reveal solutionSolution

    This is a first-order linear ODE in $y$; finding the integrating factor $\log x$ and applying $y(e)=0$ gives $y(e^2)=\frac32$.

    Concept and Intuition

    Dividing the given equation by $x\log x$ puts it in the standard linear form $\dfrac{dy}{dx}+P(x)y=Q(x)$, which is always solvable via an integrating factor $\mu=e^{\int P\,dx}$. Recognizing $\log(x^2)=2\log x$ also simplifies the RHS immediately.

    Step-by-Step Solution

    1. Given: $x\log x\,\dfrac{dy}{dx}+y=\log(x^2)=2\log x$.
    2. Divide throughout by $x\log x$: $\dfrac{dy}{dx}+\dfrac{y}{x\log x}=\dfrac{2}{x}$.
    3. This is linear with $P(x)=\dfrac{1}{x\log x}$. Integrating factor:

    $\mu=e^{\int \frac{1}{x\log x}dx}.$

    Let $u=\log x$, $du=dx/x$, so $\int\frac{dx}{x\log x}=\int\frac{du}{u}=\log|u|=\log|\log x|$. Hence $\mu=e^{\log|\log x|}=\log x$ (positive since $x>1$ in this problem).

    4. Multiply the linear ODE by $\mu=\log x$:

    $\log x\cdot\frac{dy}{dx}+\frac{y}{x}=\frac{2\log x}{x}.$

    5. The LHS is exactly $\dfrac{d}{dx}(y\log x)$ (product rule check: $y'\log x+y\cdot\frac1x$ — matches).
    6. Integrate both sides: $y\log x=\displaystyle\int\frac{2\log x}{x}dx$. With $u=\log x$: $\int 2u\,du=u^2=(\log x)^2$. So $y\log x=(\log x)^2+C$. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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 $(1,2)$: $2 = 1 + 1 + C \Rightarrow C = 0$. …

    🔒 Start your 14-day free trial to see the full solution →