Q.If 𝑓(𝑥 + 𝑦) = 𝑓(𝑥)𝑓(𝑦) for all 𝑥 , 𝑦 ∈ R and 𝑓(5)= 2, 𝑓′(0) = 3, then using the definition of derivatives, find 𝑓′(5). 2

Showing the 12 most recent of 31 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $f(x) = (\log x)^{\sin x},\ x > e$, then $f'(\pi) =$ (A) $\log \pi$ (B) $-\log \pi$ (C) $\log(\log \pi)$ (D) $-\log(\log \pi)$

   ›Reveal solutionSolution

   Use logarithmic differentiation to handle a variable exponent. The derivative at $x = \pi$ simplifies to $-\log(\log \pi)$, so the correct option is (D).

   When you see a function of the form $[g(x)]^{h(x)}$ — where both the base and the exponent are functions of $x$ — the standard power rule or exponential rule alone won't work. The cleanest method is to take the natural logarithm of both sides, differentiate implicitly, and then solve for $f'(x)$. This is called logarithmic differentiation, and it turns the messy exponent into a product you can handle with the product rule.

   Here, $f(x) = (\log x)^{\sin x}$ with $x > e$ (so $\log x > 1$, and the function is well-defined and positive). We want $f'(\pi)$.

   1. Take logs. Let $y = f(x) = (\log x)^{\sin x}$. Then

   $\log y = \sin x \cdot \log(\log x).$

   The right side is now a product of two functions of $x$, which is much easier to differentiate.

   2. Differentiate both sides with respect to $x$. On the left, by the chain rule: $\frac{d}{dx}(\log y) = \frac{1}{y} \cdot y' = \frac{y'}{y}$. On the right, use the product rule:

   $\frac{d}{dx}\bigl[\sin x \cdot \log(\log x)\bigr] = \cos x \cdot \log(\log x) + \sin x \cdot \frac{1}{\log x} \cdot \frac{1}{x}.$

   (The derivative of $\log(\log x)$ is $\frac{1}{\log x} \cdot \frac{1}{x}$ by the chain rule.)

   So we have:

   $\frac{y'}{y} = \cos x \cdot \log(\log x) + \frac{\sin x}{x \log x}.$

   3. Solve for $y'$. Multiply through by $y = (\log x)^{\sin x}$:

   $f'(x) = (\log x)^{\sin x} \left[ \cos x \cdot \log(\log x) + \frac{\sin x}{x \log x} \right].$

   4. Evaluate at $x = \pi$. At $x = \pi$, we have $\sin \pi = 0$ and $\cos \pi = -1$. The second term inside the brackets vanishes because $\sin \pi = 0$. So:

   $f'(\pi) = (\log \pi)^{\sin \pi} \left[ (-1) \cdot \log(\log \pi) + 0 \right] = (\log \pi)^{0} \cdot \bigl[ -\log(\log \pi) \bigr].$

   Since $(\log \pi)^0 = 1$, we get:

   $f'(\pi) = -\log(\log \pi).$

   In the notation of the problem, $\log$ means natural log, so $\log(\log \pi) = \log(\log \pi)$. Thus:

   $f'(\pi) = -\log(\log \pi).$

   Watch out

   A common mistake is to treat $(\log x)^{\sin x}$ as if it were $e^{\sin x \cdot \log(\log x)}$ and then forget to differentiate the exponent properly — or to try the power rule as if the exponent were constant. Logarithmic differentiation avoids both traps.

   ✓Final answer

   The correct option is (D): $f'(\pi) = -\log(\log \pi)$.

2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $(3y)^{2x} = 5(2^{3x})$, then $\left(\dfrac{dy}{dx}\right)_{x=1} =$ (A) $\dfrac{\sqrt{10} \log 5}{3}$ (B) $-\dfrac{\sqrt{10}}{3}$ (C) $-\dfrac{\sqrt{10} \log 5}{3}$ (D) $\dfrac{\sqrt{10}}{3}$

   ›Reveal solutionSolution

   Logarithmic differentiation of $(3y)^{2x} = 5\cdot 2^{3x}$, then evaluating at $x=1$ (where $3y = \sqrt{40}$), gives $\left(\frac{dy}{dx}\right)_{x=1} = -\frac{\sqrt{10}\,\log 5}{3}$.

   Concept — logarithmic differentiation. When the variable appears in both base and exponent, take natural logs first.

   Step 1 — take logarithms.

   $(3y)^{2x} = 5\cdot 2^{3x} \;\Rightarrow\; 2x\,\log(3y) = \log 5 + 3x\log 2.$

   Step 2 — differentiate implicitly w.r.t. $x$.

   $2\log(3y) + 2x\cdot\frac{3y'}{3y} = 3\log 2 \;\Rightarrow\; 2\log(3y) + \frac{2x\,y'}{y} = 3\log 2.$

   Step 3 — find $y$ at $x=1$. The original relation at $x=1$: $(3y)^2 = 5\cdot 2^3 = 40$, so $3y = \sqrt{40} = 2\sqrt{10}$ (positive root for the log to exist), i.e. $y = \dfrac{2\sqrt{10}}{3}$, and $\log(3y) = \tfrac{1}{2}\log 40$.

   Step 4 — substitute $x=1$.

   $2\cdot\tfrac{1}{2}\log 40 + \frac{2y'}{y} = 3\log 2 \;\Rightarrow\; \frac{2y'}{y} = \log 8 - \log 40 = \log\frac{1}{5} = -\log 5.$

   Step 5 — solve for $y'$.

   $y' = -\frac{y\log 5}{2} = -\frac{1}{2}\cdot\frac{2\sqrt{10}}{3}\,\log 5 = -\frac{\sqrt{10}\,\log 5}{3}.$

   ✓Final answer

   The correct option is (C): $\left(\dfrac{dy}{dx}\right)_{x=1} = -\dfrac{\sqrt{10}\,\log 5}{3}$.

   ANSWER: C

3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.$f(x)$ is a linear polynomial such that $f(ax+by) = af(x) + bf(y)$ for all $x, y \in \mathbb{R}$. If $f(0) = 7$ and $f'(0) = 5$, then $\dfrac{f(1) + f(-1)}{a+b} =$ (A) 10 (B) 12 (C) 2 (D) 14

   ›Reveal solutionSolution

   The functional equation forces $f$ to be linear, and the given conditions determine $f(x)=5x+7$. Substituting $x=1$ and $x=-1$ gives $f(1)+f(-1)=14$, and the denominator $a+b$ must be $1$ for the equation to hold for all $x,y$, so the quotient is $14$.

   We are told $f$ is a linear polynomial, so $f(x)=mx+c$ for constants $m$ and $c$. The functional equation $f(ax+by)=af(x)+bf(y)$ must hold for all real $x,y$. This is a strong condition — it essentially says $f$ is both additive and homogeneous in a very specific way. The trick is to realize that such an equation forces $a+b=1$ (otherwise the left and right sides can't match for all $x,y$), and then the rest is just plugging in.

   Let’s work through it step by step.

   1. Write the general form of $f$. Since $f$ is linear, $f(x)=mx+c$. The derivative $f'(0)=5$ tells us $m=5$. The value $f(0)=7$ gives $c=7$. So

   $f(x)=5x+7.$

   2. Plug into the functional equation. The left side:

   $f(ax+by)=5(ax+by)+7 = 5ax+5by+7.$

   The right side:

   $af(x)+bf(y)=a(5x+7)+b(5y+7)=5ax+7a+5by+7b.$

   3. Equate for all $x,y$. For these to be equal for every real $x$ and $y$, the coefficients of $x$ and $y$ already match ($5a$ and $5b$ on both sides). The constant terms must also match:

   $7 = 7a + 7b \quad\Rightarrow\quad 1 = a+b.$

   So the functional equation forces $a+b=1$.

   Watch out

   A common mistake is to forget that the equation must hold for all $x,y$. If you just plug in specific numbers, you might miss the condition $a+b=1$. Always compare coefficients when dealing with identities.

   4. Compute $f(1)+f(-1)$.

   $f(1)=5(1)+7=12,\qquad f(-1)=5(-1)+7=2.$

   Hence

   $f(1)+f(-1)=12+2=14.$

   5. Divide by $a+b$. Since $a+b=1$,

   $\frac{f(1)+f(-1)}{a+b} = \frac{14}{1}=14.$

   Tip

   The condition $a+b=1$ is not given — it’s deduced from the functional equation. This is the key insight: the equation only makes sense for all $x,y$ if the coefficients of the constant term balance.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

4. 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 $f(\sin x) = u = x - \dfrac{\pi}{6}$.
   6. Let $t=\sin x$; then $x=\arcsin t$ (principal branch), so $f(t)=\arcsin t - \dfrac{\pi}{6}$ for all $t$ in the range of $\sin$.
   7. Differentiate: $f'(t)=\dfrac{1}{\sqrt{1-t^2}}$, so $f'(x)=\dfrac{1}{\sqrt{1-x^2}}$.

   Common Mistakes

   • Forgetting to actually generate the second equation via the $\pi/2-x$ substitution and instead trying to differentiate the original equation directly (which leaves two unknown derivatives, $f'(\sin x)$ and $f'(\cos x)$, unresolved).
   • Sign errors when eliminating $v$ between the two linear equations.

   ✓Final answer

   The correct option is (D) — $\dfrac{1}{\sqrt{1-x^2}}$.

   ANSWER: D

5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $f(x)=\dfrac{1}{x^2}\displaystyle\int_{3}^{x} (2t-3f'(t))\,dt$, then $f'(3)=$ (A) $-\dfrac{1}{2}$ (B) $\dfrac{1}{2}$ (C) $-\dfrac{1}{3}$ (D) $\dfrac{1}{3}$

   ›Reveal solutionSolution

   Differentiating $x^2f(x)=\int_3^x(2t-3f'(t))\,dt$ via the product rule and the Fundamental Theorem of Calculus turns the implicit definition into a solvable equation for $f'(3)$, giving $1/2$.

   Concept and Intuition

   Here $f$ appears both outside the integral (as $f(x)$) and inside it (as $f'(t)$, the integrand). This self-referential setup is handled by differentiating both sides with respect to $x$: the Fundamental Theorem of Calculus turns $\frac{d}{dx}\int_3^x g(t)\,dt$ into $g(x)$, converting the integral equation into an ordinary equation involving $f(x)$ and $f'(x)$ at the same point, which can then be evaluated at the specific point $x=3$.

   Step-by-Step Solution

   1. Multiply through by $x^2$: $x^2 f(x) = \displaystyle\int_3^x (2t-3f'(t))\,dt$.
   2. Setting $x=3$: the right side is $\int_3^3(\cdots)\,dt = 0$, so $9f(3)=0 \Rightarrow f(3)=0$.
   3. Differentiate both sides with respect to $x$ using the product rule on the left and FTC on the right:

   $\dfrac{d}{dx}\big[x^2f(x)\big] = 2xf(x)+x^2f'(x)$

   $\dfrac{d}{dx}\int_3^x(2t-3f'(t))\,dt = 2x-3f'(x)$

   4. So $2xf(x)+x^2f'(x) = 2x-3f'(x)$.
   5. Evaluate at $x=3$, using $f(3)=0$: $2(3)(0)+9f'(3) = 2(3)-3f'(3)$.
   6. $9f'(3) = 6-3f'(3) \Rightarrow 12f'(3)=6 \Rightarrow f'(3)=\dfrac12$.

   Common Mistakes

   • Forgetting to first establish $f(3)=0$ from the integral being zero at $x=3$ (needed to simplify the differentiated equation).
   • Sign error differentiating $-3f'(t)$ inside the integral (it just becomes $-3f'(x)$ by FTC, no extra chain rule needed since the integrand itself, not a composed function, is being evaluated at the limit).

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{2}$.

   ANSWER: B

6. 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

   • Trying to first solve for the explicit form of $f$ (e.g. assuming $f(x)=e^{cx}$) rather than directly differentiating the functional equation, which is unnecessary and can obscure why the answer must hold for the general (only once-differentiable) $f$.
   • Differentiating with respect to $x$ instead of $y$, which does not directly isolate $f'$ at the point where derivability is guaranteed ($x=0$).

   ✓Final answer

   The correct option is (B) — $f(x)f'(0)$.

   ANSWER: B

7. KCET 2026Set UNKNOWN1 markMCQ
   Q.$\int x f(x)\, dx + \frac{f(x)}{2} = 0$, then $f(x)$ is equal to (A) $e^{-2x}$ (B) $e^{2x}$ (C) $e^{-x^2}$ (D) $e^{x^2}$

   ›Reveal solutionSolution

   Differentiate the given integral relation to turn it into a separable first-order differential equation in $f(x)$.

   Step 1 — Differentiate both sides

   $\int x f(x)\,dx + \frac{f(x)}{2} = 0$

   Differentiating with respect to $x$:

   $x f(x) + \frac12 f'(x) = 0$

   Step 2 — Separate variables

   $f'(x) = -2x f(x) \implies \frac{f'(x)}{f(x)} = -2x$

   Step 3 — Integrate

   $\ln f(x) = -x^2 + C \implies f(x) = A e^{-x^2}$

   Taking the particular solution $A=1$:

   $f(x) = e^{-x^2}$

   Check: $\int x e^{-x^2}\,dx = -\tfrac12 e^{-x^2} + C = -\dfrac{f(x)}{2}+C$, so $\int x f(x)\,dx + \dfrac{f(x)}{2} = C$, which vanishes for this particular solution.

   ✓Final answer

   The correct option is (C) — $f(x) = e^{-x^2}$.

8. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If $\dfrac{d}{dx} \left\{ \dfrac{x-1}{x-\sqrt{x}} e^{2x+1} \right\} = \dfrac{x-1}{x-\sqrt{x}} e^{2x+1} f(x)$, then $f(4)=$ (A) $0$ (B) $1$ (C) $\dfrac{35}{24}$ (D) $\dfrac{47}{24}$

   ›Reveal solutionSolution

   The problem gives a derivative identity that essentially defines $f(x)$ as the logarithmic derivative of the given function. By simplifying the function first and then differentiating, we find $f(4) = \frac{47}{24}$, which corresponds to option (D).

   We are told that

   $\frac{d}{dx} \left\{ \frac{x-1}{x-\sqrt{x}} e^{2x+1} \right\} = \frac{x-1}{x-\sqrt{x}} e^{2x+1} f(x).$

   This means $f(x)$ is exactly the logarithmic derivative of the function $g(x) = \frac{x-1}{x-\sqrt{x}} e^{2x+1}$. That is, if $g'(x) = g(x) f(x)$, then $f(x) = \frac{g'(x)}{g(x)} = \frac{d}{dx} \log g(x)$. So instead of doing a messy product/quotient differentiation directly, we can compute $\log g(x)$, differentiate, and simplify — that’s the clean path.

   ---

   1. Simplify the function first

   Notice $x - \sqrt{x} = \sqrt{x}(\sqrt{x} - 1)$. So

   $\frac{x-1}{x-\sqrt{x}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + \frac{1}{\sqrt{x}}.$

   Thus

   $g(x) = \left(1 + \frac{1}{\sqrt{x}}\right) e^{2x+1}.$

   2. Take the natural log

   $\log g(x) = \log\left(1 + x^{-1/2}\right) + (2x+1).$

   3. Differentiate

   $\frac{d}{dx} \log g(x) = \frac{-\frac{1}{2} x^{-3/2}}{1 + x^{-1/2}} + 2.$

   Simplify the fraction: multiply numerator and denominator by $x^{3/2}$:

   $\frac{-\frac{1}{2}}{x^{3/2} + x} = -\frac{1}{2x^{3/2} + 2x}.$

   But better: write $1 + x^{-1/2} = \frac{\sqrt{x}+1}{\sqrt{x}}$, so

   $\frac{-\frac{1}{2} x^{-3/2}}{1 + x^{-1/2}} = -\frac{1}{2} \cdot \frac{x^{-3/2}}{(\sqrt{x}+1)/\sqrt{x}} = -\frac{1}{2} \cdot \frac{x^{-3/2} \cdot \sqrt{x}}{\sqrt{x}+1} = -\frac{1}{2} \cdot \frac{x^{-1}}{\sqrt{x}+1} = -\frac{1}{2x(\sqrt{x}+1)}.$

   Thus

   $f(x) = -\frac{1}{2x(\sqrt{x}+1)} + 2.$

   4. Evaluate at $x=4$

   $\sqrt{4}=2$, so

   $f(4) = -\frac{1}{2 \cdot 4 \cdot (2+1)} + 2 = -\frac{1}{24} + 2 = \frac{48}{24} - \frac{1}{24} = \frac{47}{24}.$

   Tip

   Recognizing $f(x)$ as the logarithmic derivative turns a messy differentiation into a clean algebraic simplification. Always check if the given function can be simplified before differentiating.

   Watch out

   A common mistake is to differentiate the original unsimplified form using the quotient rule directly — it’s doable but error-prone. Simplifying first avoids that trap.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

9. 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: $\dfrac{dy}{dx}=\dfrac{1}{16}(10x+4)=\dfrac{5x+2}{8}$.
   7. At $x=1$: $\dfrac{dy}{dx}=\dfrac{5(1)+2}{8}=\dfrac78$.

   Common Mistakes

   • Trying to differentiate $y=xf(x)$ using the original implicit functional equation directly (product/chain rule on an unknown $f$) instead of first solving explicitly for $f(x)$ — solving explicitly is far safer here since it's possible.
   • Sign errors during the elimination step (mixing up which equation to multiply by 5 vs 3).

   ✓Final answer

   The correct option is (B) — $\dfrac{7}{8}$.

   ANSWER: B

10. WBJEE 2025Set math-20251 markMCQ
    Q.A function $f : \mathbb{R} \to \mathbb{R}$, satisfies $f\left(\frac{x+y}{3}\right) = \frac{f(x) + f(y) + f(0)}{3}$ for all $x, y \in \mathbb{R}$. If the function $f$ is differentiable at $x = 0$, then $f$ is (A) linear (B) quadratic (C) cubic (D) biquadratic

    ›Reveal solutionSolution

    Define $g(x)=f(x)-f(0)$; the condition reduces to Cauchy-type additivity, and differentiability at $0$ makes $f$ affine (linear).

    Given $f\!\left(\dfrac{x+y}{3}\right)=\dfrac{f(x)+f(y)+f(0)}{3}$. Put $x=y=0$: $f(0)=\dfrac{3f(0)}{3}=f(0)$ (consistent). Let $g(x)=f(x)-f(0)$. Then

    $g\!\left(\tfrac{x+y}{3}\right)=\tfrac{g(x)+g(y)}{3}.$

    Setting $y=0$ gives $g(x/3)=g(x)/3$. Using this, the relation becomes $g(x+y)=g(x)+g(y)$ (additive Cauchy equation). With differentiability at $0$, the only solutions are $g(x)=mx$, so $f(x)=mx+f(0)$ — a first-degree (linear) function.

    ✓Final answer

    $f$ is linear — option (A).

    ANSWER: A

11. 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: $f'(x) = \dfrac35 - \dfrac{2}{5x^2}$.
    6. $f'(2) = \dfrac35 - \dfrac{2}{5(4)} = \dfrac35-\dfrac1{10} = \dfrac{6-1}{10}=\dfrac{5}{10}=\dfrac12$.

    Common Mistakes

    • Sign errors while eliminating $f(1/x)$ between the two equations.
    • Differentiating $\dfrac{2}{5x}$ incorrectly (forgetting the negative power rule gives $-\dfrac{2}{5x^2}$).

    ✓Final answer

    The correct option is (B) — $\dfrac{1}{2}$.

    ANSWER: B

12. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.If $f(x) = |x-1| + |x-2|$, then $f'(-2023) + f'\left(\dfrac{2024}{2023}\right) + f'(2023) =$ (A) $1$ (B) $-1$ (C) $0$ (D) $3$

    ›Reveal solutionSolution

    The function is piecewise linear with slopes that change at the "kinks" $x=1$ and $x=2$. The derivative at any point not at a kink is just the sum of the slopes of the absolute-value pieces. Evaluating at the given points gives $f'(-2023) = -2$, $f'(2024/2023) = 0$, $f'(2023) = 2$, and their sum is $0$.

    ---

    The key idea: $|x-a|$ has derivative $+1$ for $x > a$ and $-1$ for $x < a$ (it's undefined at $x=a$). So $f(x) = |x-1| + |x-2|$ is a sum of two such V-shaped functions. Its derivative is simply the sum of the derivatives of each piece, as long as we avoid the points $x=1$ and $x=2$ where the absolute values have corners.

    We just need to figure out, for each given $x$, whether it lies to the left or right of $1$ and $2$, then add the corresponding $\pm 1$ contributions.

    ---

    1. For $x = -2023$

       This is far to the left of both $1$ and $2$.

       • For $|x-1|$: since $x < 1$, derivative is $-1$.
       • For $|x-2|$: since $x < 2$, derivative is $-1$. So $f'(-2023) = (-1) + (-1) = -2$.

    2. For $x = \dfrac{2024}{2023}$

       Note that $\dfrac{2024}{2023} \approx 1.0005$, so it's just slightly greater than $1$ but still less than $2$.

       • For $|x-1|$: since $x > 1$, derivative is $+1$.
       • For $|x-2|$: since $x < 2$, derivative is $-1$. So $f'\left(\dfrac{2024}{2023}\right) = (+1) + (-1) = 0$.

    3. For $x = 2023$

       This is far to the right of both $1$ and $2$.

       • For $|x-1|$: since $x > 1$, derivative is $+1$.
       • For $|x-2|$: since $x > 2$, derivative is $+1$. So $f'(2023) = (+1) + (+1) = 2$.

    4. Sum them up

       $(-2) + 0 + 2 = 0$.

    Tip

    A quick mental picture: $f(x)$ is shaped like a "V" with a flat bottom between $1$ and $2$ (slope $0$ there), slope $-2$ to the left of $1$, and slope $+2$ to the right of $2$. The three points given are one on each slope region and one on the flat part, so their derivatives cancel symmetrically.

    Watch out

    A common mistake is to forget that the derivative of $|x-a|$ is not defined at $x=a$, but here none of the given points equal $1$ or $2$, so we're safe.

    ✓Final answer

    The correct option is (C).

    ANSWER: C