Q.A function $f : \mathbb{R} \to \mathbb{R}$ satisfies the equation $f(x + y) = f(x)\,f(y)$ for all $x, y \in \mathbb{R}$, $f(x) \ne 0$. Suppose that the function is differentiable at $x = 0$ and $f'(0) = 2$. Prove that $f'(x) = 2f(x)$.

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $f$ is real valued function such that $f(0) = 1$, $f(x + 2y) = f(x)(f(y))^2$ for all $x, y \in \mathbb{R}$ and 'f' is derivable at $x = 0$, then $f'(x) =$ (A) $f(x)$ (B) $f(x)f'(0)$ (C) $\frac{f(x)}{f'(0)}$ (D) $f'(0) + f(x)$

   ›Reveal solutionSolution

   This tests deriving a differential relation from a Cauchy-type functional equation. Differentiating with respect to $y$ and setting $y=0$ gives $f'(x)=f(x)f'(0)$.

   Concept and Intuition

   When a functional equation like $f(x+2y)=f(x)f(y)^2$ holds for all $x,y$ and $f$ is differentiable at a single point, we can differentiate both sides with respect to one variable (holding the other fixed) and then plug in a convenient value — usually the point where differentiability is known — to get a genuine differential equation for $f$.

   Step-by-Step Solution

   1. Given: $f(x+2y)=f(x)\,f(y)^2$ for all $x,y\in\mathbb R$, $f(0)=1$, $f$ derivable at $0$.
   2. Differentiate both sides with respect to $y$ (treating $x$ as fixed): $\frac{d}{dy}f(x+2y)=2f'(x+2y)$, and $\frac{d}{dy}\left[f(x)f(y)^2\right]=f(x)\cdot2f(y)f'(y)$.
   3. So $2f'(x+2y)=2f(x)f(y)f'(y)$, i.e. $f'(x+2y)=f(x)f(y)f'(y)$.
   4. Set $y=0$: $f'(x)=f(x)f(0)f'(0)$.
   5. Since $f(0)=1$: $f'(x)=f(x)f'(0)$.

   Common Mistakes …

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2. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.A function $f: R \to R$ satisfies the relation $f(x+y) = f(x) \cdot f(y)$, $\forall\, x,y \in R$ and $f(x) \ne 0$, $\forall\, x \in R$. If 'f' is differentiable at $x=0$, $f'(0) = 4$ and $f(6) = 3$, then $f'(6) =$ ____ (A) 0 (B) 12 (C) 3 (D) 6

   ›Reveal solutionSolution

   This tests the classic "Cauchy exponential" functional equation trick: differentiability at one point forces differentiability everywhere, giving $f'(x)=f(x)f'(0)$.

   Concept and Intuition

   $f(x+y)=f(x)f(y)$ with $f$ never zero is the defining property of exponential-type functions ($f(x)=a^x$ satisfies this). Such functions need only be "nice" (differentiable) at a single point to be differentiable everywhere, and their derivative is always proportional to themselves.

   Step-by-Step Solution

   1. Put $x=y=0$: $f(0)=f(0)^2$, and since $f(0)\ne 0$, we get $f(0)=1$.
   2. By definition, $f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\dfrac{f(x)f(h)-f(x)}{h}=f(x)\lim_{h\to 0}\dfrac{f(h)-1}{h}$. …

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3. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.Let $f(x)$ be a differentiable function such that $f(1)=2,\ f(2)=6$ and $f(x+y)=f(x)+kxy+\dfrac{4}{3}y^2\ \forall x,y\in\mathbb{R}$, then $f(x)=$ (A) $4x-2$ (B) $y-4x^2+2x-4$ (C) $\dfrac{8}{3}x^2+\dfrac{4}{3}$ (D) $\dfrac{4}{3}x^2+\dfrac{2}{3}$

   ›Reveal solutionSolution

   The functional equation is a disguised statement about the derivative of $f$; differentiating with respect to the "$y$" slot (by taking $h\to0$) directly gives $f'(x)=kx$, and $k$ is found from the two given values.

   Concept and Intuition

   Functional equations of the form $f(x+y)=f(x)+(\text{stuff involving }x,y)$ are often best attacked by treating $y=h$ as a small increment and forming the difference quotient $\dfrac{f(x+h)-f(x)}{h}$, which as $h\to0$ gives $f'(x)$ directly — turning a functional equation into an ODE.

   Step-by-Step Solution

   1. Set $x=1,y=1$: $f(2)=f(1)+k(1)(1)+\dfrac43(1)^2=2+k+\dfrac43$. Given $f(2)=6$: $k+\dfrac43=4\Rightarrow k=\dfrac83$.
   2. From $f(x+h)=f(x)+kxh+\dfrac43h^2$: $\dfrac{f(x+h)-f(x)}h = kx+\dfrac43h \xrightarrow{h\to0} kx$. So $f'(x)=kx=\dfrac83x$.
   3. Integrate: $f(x)=\dfrac83\cdot\dfrac{x^2}2+C=\dfrac43x^2+C$.
   4. Use $f(1)=2$: $\dfrac43+C=2\Rightarrow C=\dfrac23$. …

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4. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If $3f(x) - 2f\left(\dfrac{1}{x}\right) = x$ then $f'(2) =$ (A) $1$ (B) $\dfrac{1}{2}$ (C) $2$ (D) $\dfrac{7}{2}$

   ›Reveal solutionSolution

   Substituting $x\to1/x$ produces a second linear equation in $f(x),f(1/x)$; solving the pair recovers $f(x)$ explicitly, then differentiate.

   Concept and Intuition

   Functional equations relating $f(x)$ and $f(1/x)$ are typically solved by substituting $x\to1/x$ to get a second equation, then eliminating $f(1/x)$ by linear combination — turning a functional equation into an explicit formula for $f(x)$.

   Step-by-Step Solution

   1. Given: $3f(x)-2f\!\left(\dfrac1x\right)=x$. — (1)
   2. Replace $x\to\dfrac1x$: $3f\!\left(\dfrac1x\right)-2f(x)=\dfrac1x$. — (2)
   3. $3\times(1)$: $9f(x)-6f(1/x)=3x$. $2\times(2)$: $-4f(x)+6f(1/x)=\dfrac2x$.
   4. Add: $5f(x) = 3x+\dfrac2x \Rightarrow f(x)=\dfrac{3x^2+2}{5x} = \dfrac{3x}5+\dfrac{2}{5x}$.
   5. Differentiate: $f'(x) = \dfrac35 - \dfrac{2}{5x^2}$. …

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5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If f is a derivable function and $2f(\sin x)+f(\cos x)=x$ for all $x\in R$, then $f'(x)=$ (A) $\sin x+\cos x$ (B) $\sin x-\cos x$ (C) $\sqrt{1-x^2}$ (D) $\dfrac{1}{\sqrt{1-x^2}}$

   ›Reveal solutionSolution

   Substituting $x\to\pi/2-x$ creates a second equation that, combined with the original, isolates $f(\sin x)$ explicitly as $\arcsin x - \pi/6$, giving $f'(x)=1/\sqrt{1-x^2}$.

   Concept and Intuition

   A functional equation linking $f(\sin x)$ and $f(\cos x)$ can be "solved" by generating a second, independent equation via the substitution $x\mapsto \pi/2-x$ (which swaps $\sin$ and $\cos$, since $\sin(\pi/2-x)=\cos x$ and $\cos(\pi/2-x)=\sin x$). Two linear equations in the two unknowns $f(\sin x)$ and $f(\cos x)$ can then be solved like a simultaneous system.

   Step-by-Step Solution

   1. Original: $2f(\sin x)+f(\cos x)=x$. — (1)
   2. Replace $x\to \dfrac{\pi}{2}-x$: $2f\big(\sin(\tfrac{\pi}{2}-x)\big)+f\big(\cos(\tfrac{\pi}{2}-x)\big)=\dfrac{\pi}{2}-x$, i.e. $2f(\cos x)+f(\sin x)=\dfrac{\pi}{2}-x$. — (2)
   3. Treat $u=f(\sin x),\ v=f(\cos x)$: (1) is $2u+v=x$; (2) is $u+2v=\pi/2-x$.
   4. Multiply (1) by 2: $4u+2v=2x$. Subtract (2): $4u+2v-(u+2v) = 2x-(\pi/2-x)$, i.e. $3u = 3x-\pi/2$.
   5. So $f(\sin x) = u = x - \dfrac{\pi}{6}$. …

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6. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.If a function f satisfies $f(x+1) + f(x-1) = \sqrt{2} f(x)$, then $f(x+2) + f(x-2)=$ (A) $2f(x)$ (B) $f(x+1) - f(x-1)$ (C) $4f(x)$ (D) $0$

   ›Reveal solutionSolution

   The recurrence $f(x+1)+f(x-1)=\sqrt2 f(x)$ is a linear (constant-coefficient) recursion in the shift index; applying it twice shows $f(x+2)+f(x-2)=0$.

   Concept and Intuition

   This functional equation behaves exactly like a linear recurrence relation in an integer shift $n$: define $g(n)=f(x+n)$ for a fixed $x$. The given equation says $g(1)+g(-1)=\sqrt2\,g(0)$; because the original functional relation holds for every real argument, it also holds with $x$ replaced by $x+n$, giving the general recursion $g(n+1)+g(n-1)=\sqrt2\,g(n)$ for every integer $n$. This is structurally identical to $\cos((n+1)\theta)+\cos((n-1)\theta)=2\cos\theta\cos(n\theta)$ with $2\cos\theta=\sqrt2$, i.e. $\theta=\pi/4$ — so $f$ behaves like a cosine of period 8 in the shift, and by the recursion, values 4 apart in the shift are related in a way that makes the sum $f(x+2)+f(x-2)$ vanish.

   Step-by-Step Solution

   1. Let $g(n)=f(x+n)$. Given: $g(1)+g(-1)=\sqrt2\,g(0)$, and by shifting $x$, generally $g(n+1)+g(n-1)=\sqrt2\,g(n)$.
   2. Apply with $n=1$: $g(2)+g(0)=\sqrt2\,g(1) \Rightarrow g(2)=\sqrt2\,g(1)-g(0)$.
   3. Apply with $n=-1$: $g(0)+g(-2)=\sqrt2\,g(-1) \Rightarrow g(-2)=\sqrt2\,g(-1)-g(0)$.
   4. Add: $g(2)+g(-2) = \sqrt2\big(g(1)+g(-1)\big) - 2g(0)$. …

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7. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.If $3f(x) - 2f\left(\frac{1}{x}\right) = x$, then $f'(2) =$ ________ (A) $\frac{1}{2}$ (B) $\frac{-1}{2}$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   Substituting $x\to1/x$ gives a second equation that, solved together with the first, isolates $f(x)$ explicitly. Answer: $f'(2)=1/2$.

   Concept and Intuition

   Functional equations relating $f(x)$ and $f(1/x)$ are solved by generating a second independent equation via the substitution $x\to1/x$, then treating $f(x)$ and $f(1/x)$ as two unknowns in linear equations.

   Step-by-Step Solution

   1. Given: $3f(x)-2f(1/x)=x$ — (i).
   2. Replace $x$ by $1/x$: $3f(1/x)-2f(x)=1/x$ — (ii).
   3. Multiply (i) by 3 and (ii) by 2: $9f(x)-6f(1/x)=3x$; $-4f(x)+6f(1/x)=2/x$.
   4. Add: $5f(x)=3x+\dfrac2x \Rightarrow f(x)=\dfrac{3x}{5}+\dfrac{2}{5x}$.
   5. Differentiate: $f'(x)=\dfrac35-\dfrac{2}{5x^2}$. …

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8. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If $f(x+ay) + g(x-ay) = 0$, then $a\dfrac{dy}{dx} =$ (A) $\dfrac{f'(x-ay)+g'(x+ay)}{g'(x+ay)-f'(x-ay)}$ (B) $\dfrac{f'(x+ay)+g'(x-ay)}{g'(x-ay)-f'(x+ay)}$ (C) $\dfrac{f'(x+ay)g'(x-ay)}{f'(x+ay)+g'(x-ay)}$ (D) $\dfrac{f'(x+ay)+g'(x-ay)}{f'(x+ay)g'(x-ay)}$

   ›Reveal solutionSolution

   Implicit differentiation of the given functional equation, treating $y$ as a function of $x$, directly yields the requested expression for $a\,dy/dx$.

   Concept and Intuition

   This is a chain-rule / implicit-differentiation exercise: differentiate both composite terms w.r.t. $x$, remembering that the argument $x\pm ay$ itself depends on $x$ through $y(x)$, then solve algebraically for $y'$.

   Step-by-Step Solution

   1. Differentiate $f(x+ay)+g(x-ay)=0$ w.r.t. $x$: $f'(x+ay)\cdot(1+ay') + g'(x-ay)\cdot(1-ay') = 0$.
   2. Expand: $f'(x+ay) + ay'f'(x+ay) + g'(x-ay) - ay'g'(x-ay) = 0$.
   3. Group the $y'$ terms: $ay'\big[f'(x+ay)-g'(x-ay)\big] = -\big[f'(x+ay)+g'(x-ay)\big]$. …

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9. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.If $f(x)$ is differentiable on $\mathbb{R}$, $f(x)f'(-x) - f(-x)f'(x) = 0$, $f(0) = 3$ and $f(3) = 9$, then $\left(1 + f(-3)\right)^3 + 1 =$ (A) $2$ (B) $9$ (C) $28$ (D) $0$

   ›Reveal solutionSolution

   The given identity says $f(x)f(-x)$ has zero derivative, so it's a constant, pinned by $f(0)=3$.

   Concept and Intuition

   Whenever we're given a relation like $f(x)f'(-x)-f(-x)f'(x)=0$, it's worth checking whether it's the derivative of a simple combination of $f(x)$ and $f(-x)$ — here it turns out to be (up to sign) the derivative of the product $f(x)f(-x)$.

   Step-by-Step Solution

   1. Define $\psi(x)=f(x)f(-x)$.
   2. $\psi'(x)=f'(x)f(-x)+f(x)\cdot(-f'(-x))=f'(x)f(-x)-f(x)f'(-x)$.
   3. The given condition $f(x)f'(-x)-f(-x)f'(x)=0$ rearranges to $f'(x)f(-x)-f(x)f'(-x)=0$, which is exactly $\psi'(x)$.
   4. So $\psi'(x)=0$ for all $x$, meaning $\psi(x)$ is constant.
   5. Evaluate at $x=0$: $\psi(0)=f(0)f(0)=3\times3=9$. …

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10. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.If $5f(x) + 3f\left(\dfrac{1}{x}\right) = x + 2$ and $y = xf(x)$, then $\dfrac{dy}{dx}$ at $x=1$ is equal to (A) $14$ (B) $\dfrac{7}{8}$ (C) $1$ (D) $7$

    ›Reveal solutionSolution

    A classic "functional equation with $x$ and $1/x$" — write the equation twice (once with $x$, once with $1/x$) and solve the resulting linear system for $f(x)$. Answer: $\dfrac{7}{8}$.

    Concept and Intuition

    Whenever a functional equation relates $f(x)$ and $f(1/x)$ linearly, substituting $x\to 1/x$ produces a second independent linear equation in the same two unknowns $f(x), f(1/x)$ — exactly like a $2\times2$ system, solvable by elimination.

    Step-by-Step Solution

    1. Given: $5f(x)+3f\!\left(\dfrac1x\right)=x+2$. — (i)
    2. Substitute $x\to\dfrac1x$: $5f\!\left(\dfrac1x\right)+3f(x)=\dfrac1x+2$. — (ii)
    3. Eliminate $f(1/x)$: $5\times$(i) $-3\times$(ii):

    $25f(x)+15f(1/x) - \left[9f(x)+15f(1/x)\right] = (5x+10)-\left(\frac3x+6\right)$

    $16f(x) = 5x+4-\frac3x.$

    4. So $f(x)=\dfrac{1}{16}\left(5x+4-\dfrac3x\right)$.
    5. Then $y=xf(x)=\dfrac{1}{16}\left(5x^2+4x-3\right)$.
    6. Differentiate: $\dfrac{dy}{dx}=\dfrac{1}{16}(10x+4)=\dfrac{5x+2}{8}$. …

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11. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.If $x \ne 0$ and $f(x)$ satisfies $8f(x) + 6f\left(\dfrac{1}{x}\right) = x + 5$, then $\dfrac{d}{dx}\left(x^2 f(x)\right)$ at $x = 1$ is (A) $-\dfrac{1}{14}$ (B) $\dfrac{25}{14}$ (C) $\dfrac{9}{14}$ (D) $\dfrac{19}{14}$

    ›Reveal solutionSolution

    Substituting $x\to 1/x$ in the given functional equation creates a second linear equation in $f(x)$ and $f(1/x)$; solving the pair gives an explicit formula for $f(x)$, from which the required derivative at $x=1$ is $\frac{19}{14}$.

    Concept and Intuition

    Functional equations relating $f(x)$ and $f(1/x)$ are classically solved by writing the equation once as given, then again with $x$ replaced by $1/x$ — producing two independent linear equations in the two unknowns $f(x)$ and $f(1/x)$, which can be solved like a simultaneous system.

    Step-by-Step Solution

    1. Given: $8f(x)+6f\!\left(\dfrac1x\right) = x+5$. ... (1)
    2. Replace $x\to\dfrac1x$: $8f\!\left(\dfrac1x\right)+6f(x) = \dfrac1x+5$. ... (2)
    3. Multiply (1) by 8: $64f(x)+48f(1/x)=8x+40$.
    4. Multiply (2) by 6: $36f(x)+48f(1/x)=\dfrac6x+30$.
    5. Subtract: $28f(x) = 8x+10-\dfrac6x \Rightarrow f(x)=\dfrac{4x+5-3/x}{14}$.
    6. Compute $x^2f(x) = \dfrac{x^2(4x+5)-3x}{14} = \dfrac{4x^3+5x^2-3x}{14}$.
    7. Differentiate: $\dfrac{d}{dx}\left(x^2f(x)\right) = \dfrac{12x^2+10x-3}{14}$.
    8. Evaluate at $x=1$: $\dfrac{12+10-3}{14}=\dfrac{19}{14}$. …

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12. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.If $3f(\cos x) + 2f(\sin x) = 5x$, then $f'(\cos x) + f'(\sin x) =$ (A) $-5(\sin x + \cos x)$ (B) $-5\sin x \cos x$ (C) $\dfrac{-5}{\sin x} - \dfrac{5}{\cos x}$ (D) $\dfrac{5}{\sin x} + \dfrac{5}{\cos x}$

    ›Reveal solutionSolution

    Substituting $x\to\frac\pi2-x$ turns the single functional equation into a solvable pair, from which $f'(\cos x)+f'(\sin x) = -\frac5{\sin x}-\frac5{\cos x}$.

    Concept and Intuition

    A single equation $3f(\cos x)+2f(\sin x)=5x$ mixes $f(\cos x)$ and $f(\sin x)$, so we can't isolate either directly. But swapping $x\to\frac\pi2-x$ swaps $\sin$ and $\cos$ (since $\cos(\frac\pi2-x)=\sin x$, $\sin(\frac\pi2-x)=\cos x$), giving a second independent linear equation — enough to solve for both unknowns.

    Step-by-Step Solution

    1. Original: $3f(\cos x)+2f(\sin x)=5x$ … (I)
    2. Replace $x\to \frac\pi2-x$: $3f(\sin x)+2f(\cos x) = 5\left(\frac\pi2-x\right)$ … (II)
    3. Treat $f(\cos x)=u$, $f(\sin x)=v$: (I) $3u+2v=5x$; (II) $2u+3v=\frac{5\pi}2-5x$.
    4. Solve: $3\times$(I) $-2\times$(II): $5u = 15x - 5\pi+10x=25x-5\pi \Rightarrow u=f(\cos x)=5x-\pi$. $3\times$(II) $-2\times$(I): $5v = \frac{15\pi}2-15x-10x = \frac{15\pi}2-25x \Rightarrow v=f(\sin x)=\frac{3\pi}2-5x$.
    5. Differentiate $f(\cos x)=5x-\pi$ w.r.t. $x$ using the chain rule: $f'(\cos x)\cdot(-\sin x)=5 \Rightarrow f'(\cos x)=-\dfrac{5}{\sin x}$. …

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