Q.Let $f$ be a function defined on $[a, b]$ such that $f'(x) > 0$, for all $x \in (a, b)$. Then prove that $f$ is an increasing function on $(a, b)$.

Showing the 12 most recent of 24 on this concept.

1. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If f(x) is a differentiable function, $f'(x)\ge 5\ \forall x\in[2,6]$, $f(2)=4$ and $f(3)=15$, then a possible value of $f(6)$ (A) $=24$ (B) lies between 4 and 15 (C) $\le 15$ (D) $=5$

   ›Reveal solutionSolution

   Classic Mean Value Theorem bound: $f'(x)\ge5$ on $[2,6]$ with $f(2)=4$ forces $f(6)\ge f(2)+5(6-2)=24$; monotonicity (since $f'>0$) rules out the smaller options, leaving $f(6)=24$ as the consistent choice.

   Concept and Intuition

   When you know a lower bound on the derivative of a function over an interval, the Mean Value Theorem converts that into a lower bound on how much the function must have increased over that interval — a derivative that's 'at least 5' everywhere means the function gains at least $5\times(\text{length of interval})$ over any sub-interval. Combined with the fact that a positive derivative makes $f$ strictly increasing, this is often enough to eliminate implausible answer choices in an MCQ even without pinning down $f$ exactly.

   Step-by-Step Solution

   1. $f$ is differentiable on $[2,6]$ with $f'(x)\ge5$ throughout, and $f$ is continuous there (differentiability implies continuity), so the Mean Value Theorem applies on $[2,6]$: there is some $c\in(2,6)$ with

   $f(6)-f(2) = f'(c)\,(6-2).$

   2. Since $f'(c)\ge5$, we get $f(6)-f(2)\ge 5\times 4=20$, i.e. $f(6)\ge f(2)+20 = 4+20=24$.
   3. Also, because $f'(x)\ge5>0$ on the whole interval, $f$ is strictly increasing on $[2,6]$. In particular, since $3<6$, we must have $f(6)>f(3)=15$.
   4. Check each option against these two facts ($f(6)\ge24$ and $f(6)>15$):

   • (B) 'lies between 4 and 15' — violates $f(6)>15$. Rejected.
   • (C) '$\le 15$' — violates $f(6)>15$. Rejected.
   • (D) '$=5$' — far below 15. Rejected. …

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2. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.If the function $f(x) = ax^3 + bx^2 + 26x - 24$ satisfies the conditions of Rolle's theorem in $[2,4]$ and $f'\left(3 + \dfrac{1}{\sqrt{3}}\right) = 0$, then the value of $ab =$ (A) $-9$ (B) $9$ (C) $-3$ (D) $3$

   ›Reveal solutionSolution

   Rolle's condition $f(2)=f(4)$ plus the given stationary point, split into rational and irrational parts, pin down $a=1,\,b=-9$, giving $ab=-9$.

   Concept and Intuition

   Rolle's theorem requires $f(2)=f(4)$ for $f$ to have a critical point in $(2,4)$. The extra given condition $f'\left(3+\tfrac{1}{\sqrt3}\right)=0$ involves an irrational number; since $a,b$ must be rational (they're the polynomial's coefficients), the equation obtained by plugging this irrational value into $f'$ must have its rational part and its $\left(\tfrac1{\sqrt3}\right)$-coefficient part vanish separately.

   Step-by-Step Solution

   1. $f(x)=ax^3+bx^2+26x-24$.
   2. $f(2)=8a+4b+52-24=8a+4b+28$; $f(4)=64a+16b+104-24=64a+16b+80$.
   3. Rolle's: $f(2)=f(4) \Rightarrow 8a+4b+28=64a+16b+80 \Rightarrow 56a+12b+52=0 \Rightarrow 14a+3b+13=0$ … (Eq. 1)
   4. $f'(x)=3ax^2+2bx+26$. Let $c=3+\dfrac{1}{\sqrt3}$, so $c^2=9+\dfrac{6}{\sqrt3}+\dfrac13=\dfrac{28}{3}+2\sqrt3$ (using $\tfrac{6}{\sqrt3}=2\sqrt3$), and write $s=\tfrac1{\sqrt3}$ so $c=3+s$, $c^2=\tfrac{28}{3}+6s$.
   5. $f'(c)=3a\left(\tfrac{28}{3}+6s\right)+2b(3+s)+26=0 \Rightarrow (28a+6b+26)+s(18a+2b)=0$. …

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3. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.For all $x \in [0,2024]$ assume that $f(x)$ is differentiable, $f(0)=-2$ and $f'(x) \ge 5$. Then the least possible value of $f(2024)$ is (A) $10{,}120$ (B) $10{,}118$ (C) $10{,}122$ (D) $2024$

   ›Reveal solutionSolution

   A lower bound on the derivative gives a lower bound on the total increase via the Mean Value Theorem; the least possible value of $f(2024)$ is $10{,}118$.

   Concept and Intuition

   If a function's rate of change is bounded below by some constant $m$ everywhere on an interval, then the function itself can only increase by at least $m$ times the interval's length — it can grow faster, but never slower. This is a direct consequence of the Mean Value Theorem (or, equivalently, integrating the inequality $f'(x)\ge5$).

   Step-by-Step Solution

   1. By the Mean Value Theorem, there exists $c\in(0,2024)$ such that $f'(c)=\dfrac{f(2024)-f(0)}{2024-0}$.
   2. Since $f'(x)\ge5$ for all $x$ in $[0,2024]$, in particular $f'(c)\ge5$.
   3. So $\dfrac{f(2024)-f(0)}{2024}\ge5 \Rightarrow f(2024)\ge f(0)+5(2024)$.
   4. Substitute $f(0)=-2$: $f(2024)\ge -2+10120 = 10118$. …

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4. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If Lagrange's mean value theorem is applied on the function $f(x) = 3x + 5$ in the interval $[-2, 2]$, then the entire set of values of 'c' of the Lagrange's mean value theorem is (A) $\varnothing$ (Null Set) (B) $\{-1, 0, 1\}$ only (C) $(-2, 2) - \{-1, 0, 1\}$ (D) $(-2, 2)$

   ›Reveal solutionSolution

   Because $f$ is linear, its derivative is the constant $3$ everywhere, which already equals the required LMVT slope — so the "set of values of $c$" is the entire open interval, not a special finite set.

   Concept and Intuition

   Lagrange's Mean Value Theorem guarantees at least one $c\in(a,b)$ with $f'(c)=\dfrac{f(b)-f(a)}{b-a}$. For a general (curved) function this $c$ is typically one or a few isolated points. But for a linear function, the derivative is constant and automatically equal to the average rate of change everywhere — so the condition holds for the whole interval, not just discrete points.

   Step-by-Step Solution

   1. $f(x)=3x+5$ on $[-2,2]$; $f$ is a polynomial, hence continuous on $[-2,2]$ and differentiable on $(-2,2)$ — LMVT applies.
   2. Compute the average rate of change: $f(2)=3(2)+5=11$, $f(-2)=3(-2)+5=-1$. $\dfrac{f(2)-f(-2)}{2-(-2)} = \dfrac{11-(-1)}{4} = \dfrac{12}{4}=3$.
   3. Compute $f'(x) = 3$ — a constant, true for every real $x$, in particular every $x\in(-2,2)$. …

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5. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Let $f: \mathbb{R} \to \mathbb{R}$ be such that $f(2+x) = f(2-x) \; \forall x \in \mathbb{R}$. If $f(x)$ is twice differentiable such that $f'(1) = 0$, then which one of the following is true? (A) there exist at least one $c$ in $(0,1)$ such that $f'(c) = 0$ (B) there exist at least one $c$ in $(1,2)$ such that $f''(c) = 0$ (C) there exist at least one $c$ in $(0,1)$ such that $f''(c) = 0$ (D) there exist at least one $c$ in $(1,2)$ such that $f'(c) = 0$

   ›Reveal solutionSolution

   The symmetry condition forces $f'(2)=0$; with the given $f'(1)=0$, Rolle's theorem

   applied to $f'$ on $[1,2]$ guarantees $f''(c)=0$ for some $c\in(1,2)$.

   Concept and Intuition

   Symmetry of a function about a vertical line $x=a$ (i.e. $f(a+x)=f(a-x)$) always forces

   a horizontal tangent at $x=a$ — the graph "folds" perfectly there. Once we have two

   zeros of $f'$ (at $x=1$ and $x=2$), Rolle's theorem — applied not to $f$ but to $f'$

   itself — guarantees a zero of $f''$ strictly between them.

   Step-by-Step Solution

   1. Differentiate $f(2+x)=f(2-x)$ with respect to $x$: $f'(2+x)=-f'(2-x)$.
   2. Put $x=0$: $f'(2)=-f'(2) \Rightarrow 2f'(2)=0 \Rightarrow f'(2)=0$.
   3. We're given $f'(1)=0$ as well.
   4. $f$ is twice differentiable, so $f'$ is differentiable (hence continuous) on $[1,2]$.
   5. $f'(1)=f'(2)=0$ satisfies Rolle's theorem's hypotheses for the function $f'$ on $[1,2]$.
   6. Rolle's theorem then gives some $c\in(1,2)$ with $(f')'(c)=f''(c)=0$.

   Common Mistakes …

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6. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.Let f(x) be a differentiable function, $A(0,\alpha)$ and $B(8,\beta)$ be two points on the curve $y=f(x)$. Given $f(0)=2$ and $f'(4)=\dfrac{-3}{4}$. If the chord AB of the curve is parallel to the tangent drawn at the point $(4,f(4))$, then $\beta=$ (A) $-4$ (B) $-6$ (C) $2$ (D) $8$

   ›Reveal solutionSolution

   Equating the slope of chord $AB$ to the given tangent slope $f'(4)=-3/4$ (both being parallel lines) gives $\beta=-4$.

   Concept and Intuition

   Two line segments are parallel exactly when they have equal slopes. Here we're told the chord joining two points on the curve is parallel to the tangent line at a third point on the curve — so we simply equate the chord's slope formula to the given derivative value and solve for the unknown ordinate.

   Step-by-Step Solution

   1. Since $A(0,\alpha)$ lies on $y=f(x)$, $\alpha = f(0) = 2$.

   2. Since $B(8,\beta)$ lies on $y=f(x)$, $\beta = f(8)$ (unknown, to be found).

   3. Slope of chord $AB$:

   $m_{AB} = \frac{\beta - \alpha}{8-0} = \frac{\beta-2}{8}$

   4. Slope of the tangent at $x=4$ is given: $f'(4) = -\dfrac{3}{4}$.

   5. Since chord $AB \parallel$ tangent at $(4,f(4))$: …

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7. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If Lagrange's mean value theorem is applied on the function $f(x) = \sin^{-1} x$ in the interval $\left[\dfrac{-1}{2}, \dfrac{1}{2}\right]$, then the value of c given in Lagrange's theorem is (A) $\dfrac{\sqrt{4\pi^2 - 9}}{2\pi}$ (B) $\dfrac{\sqrt{\pi^2 - 9}}{\pi}$ (C) $0$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   LMVT on $\sin^{-1}x$ over $[-1/2,1/2]$ requires solving $f'(c)=\pi/3$, giving $c=\dfrac{\sqrt{\pi^2-9}}{\pi}$.

   Concept and Intuition

   Lagrange's Mean Value Theorem says there exists $c$ in $(a,b)$ where the instantaneous slope $f'(c)$ equals the average slope over $[a,b]$. Here the average slope is a fixed number computed from the boundary values of $\sin^{-1}x$, and we then invert $f'(x)=\dfrac{1}{\sqrt{1-x^2}}$ to find which $x=c$ achieves that slope.

   Step-by-Step Solution

   1. Compute the average slope: $\dfrac{f(1/2)-f(-1/2)}{1/2-(-1/2)} = \dfrac{\pi/6-(-\pi/6)}{1} = \dfrac{\pi}{3}$.
   2. $f'(x) = \dfrac{1}{\sqrt{1-x^2}}$, so LMVT requires $\dfrac{1}{\sqrt{1-c^2}} = \dfrac{\pi}{3}$.
   3. Invert: $\sqrt{1-c^2} = \dfrac{3}{\pi}$.
   4. Square: $1-c^2 = \dfrac{9}{\pi^2} \Rightarrow c^2 = 1-\dfrac{9}{\pi^2} = \dfrac{\pi^2-9}{\pi^2}$. …

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8. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.Consider the following functions. I) $f(x)=\begin{cases}\dfrac{1}{2}-x, & x<\dfrac{1}{2}\\ \left(\dfrac{1}{2}-x\right)^2, & x\ge \dfrac{1}{2}\end{cases}$ II) $f(x)=|3x-1|$ III) $f(x)=x|x|$ IV) $f(x)=|x|$ Then on [0, 1] Lagrange's mean value theorem is applicable to the functions (A) III, IV (B) II, III (C) I, III (D) II, IV

   ›Reveal solutionSolution

   Lagrange's Mean Value Theorem needs continuity on the closed interval and differentiability on the open interval — check each function restricted to $[0,1]$ for hidden kinks. Answer: III, IV.

   Concept and Intuition

   LMVT is often mis-applied to piecewise/absolute-value functions by overlooking a kink that happens to sit inside the interval of interest. The trick each time is to restrict the function's definition to exactly $[0,1]$ — since $x\ge0$ throughout, several of the "interesting" branches (which only differ for $x<0$) never actually come into play.

   Step-by-Step Solution

   1. I) $f(x)=\tfrac12-x$ for $x<\tfrac12$, $(\tfrac12-x)^2$ for $x\ge\tfrac12$. Both pieces agree in value at $x=\tfrac12$ (both give $0$), so $f$ is continuous on $[0,1]$. But check differentiability at $x=\tfrac12$: left-derivative of $\tfrac12-x$ is $-1$; right-derivative of $(\tfrac12-x)^2$ is $-2(\tfrac12-x)\big|_{x=1/2}=0$. Since $-1\ne0$, $f$ is not differentiable at the interior point $x=\tfrac12$ — LMVT does not apply.
   2. II) $f(x)=|3x-1|$ has a corner where $3x-1=0$, i.e. $x=\tfrac13\in(0,1)$ — slopes $-3$ and $+3$ on either side, unequal. Not differentiable there — LMVT fails.
   3. III) $f(x)=x|x|$. On $[0,1]$, since $x\ge0$, $|x|=x$, so $f(x)=x^2$ throughout — a polynomial, continuous and differentiable everywhere on $[0,1]$ (including at $x=0$, trivially). LMVT applies. …

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9. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.$f: \mathbb{R} \to \mathbb{R}$ is a function such that $|f(x) - f(y)| \le \dfrac{1}{2}|x - y|\ \forall x, y \in \mathbb{R}$ and $f'(x) \ge 1/2\ \forall x \in \mathbb{R}$, $f(1) = 1/2$. Then the number of points of intersection of curve $y = f(x)$ and the curve $y = x^2 - 2x - 5$ is (A) $1$ (B) $0$ (C) $2$ (D) infinite

   ›Reveal solutionSolution

   The Lipschitz bound forces $f$ to be exactly linear with slope $1/2$; then just count real intersections with the given parabola.

   Concept and Intuition

   The condition $|f(x)-f(y)|\le \tfrac12|x-y|$ for all $x,y$ says $f$ is Lipschitz with constant $1/2$; taking $y\to x$ this forces $|f'(x)|\le 1/2$ everywhere it's differentiable. Combined with the given $f'(x)\ge 1/2$, the only way both can hold is $f'(x)=1/2$ for all $x$. A function with constant derivative is linear.

   Step-by-Step Solution

   1. From $|f(x)-f(y)|\le\frac12|x-y|$, dividing by $|x-y|$ and taking the limit as $y\to x$ gives $|f'(x)|\le \frac12$.
   2. Given also $f'(x)\ge \frac12$, both together force $f'(x)=\frac12$ for every $x$.
   3. Integrating, $f(x)=\frac{x}{2}+c$.
   4. Using $f(1)=\frac12$: $\frac12+c=\frac12\Rightarrow c=0$. So $f(x)=\frac{x}{2}$. …

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10. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that $|f(x)-f(4)|\le 2|x-y|^{\frac{3}{2}}\ \forall x,y\in\mathbb{R}$. If $f(0)=1$, then $\displaystyle\int_0^1 f^2(x)\,dx=$ (A) $-2$ (B) $\dfrac{1}{2}$ (C) $0$ (D) $1$

    ›Reveal solutionSolution

    A Hölder-type bound with exponent $3/2>1$ forces the derivative to vanish everywhere, so $f$ must be the constant function $f\equiv1$, making the integral trivially equal to $1$.

    Concept and Intuition

    Whenever $|f(x)-f(y)|\le K|x-y|^\alpha$ with $\alpha>1$, the function is forced to be constant. This is because the difference quotient $\dfrac{f(x)-f(y)}{x-y}$ is bounded by $K|x-y|^{\alpha-1}$, which goes to $0$ as $y\to x$ (since $\alpha-1>0$). So the derivative exists and equals $0$ at every point — a stronger conclusion than Lipschitz continuity ($\alpha=1$) would give.

    Step-by-Step Solution

    1. Given: $|f(x)-f(y)|\le 2|x-y|^{3/2}$ for all real $x,y$ (reading the stray '$f(4)$' as a typo for $f(y)$, the natural reading for this classical inequality).
    2. For $x\ne y$, divide by $|x-y|$: $\left|\dfrac{f(x)-f(y)}{x-y}\right|\le 2|x-y|^{1/2}$.
    3. Take $y\to x$: the right side $\to 0$ since $|x-y|^{1/2}\to0$. By the squeeze theorem, $f'(x)=\displaystyle\lim_{y\to x}\dfrac{f(x)-f(y)}{x-y}=0$ for every $x$.
    4. A function with $f'(x)=0$ everywhere on $\mathbb R$ is constant. …

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11. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
    Q.If $f(x) = x^\alpha \log x$ and $f(0) = 0$, then the value of '$\alpha$' for which Rolle's theorem can be applied in $[0,1]$ is ____ (A) -2 (B) -1 (C) 0 (D) $\dfrac{1}{2}$

    ›Reveal solutionSolution

    Rolle's theorem needs continuity at the endpoints; only a positive exponent $\alpha$ makes $x^\alpha\log x\to0$ as $x\to0^+$, matching the given $f(0)=0$.

    Concept and Intuition

    Rolle's theorem requires: (i) $f$ continuous on $[0,1]$,

    (ii) $f$ differentiable on $(0,1)$,

    (iii) $f(0)=f(1)$. Differentiability on the open interval is automatic here for any $\alpha$ (no issue at interior points). The delicate condition is continuity at $x=0$, since $\log x\to-\infty$ there.

    Step-by-Step Solution

    1. $f(1)=1^\alpha\log 1=0$, and we are given $f(0)=0$, so $f(0)=f(1)$ holds for any $\alpha$ — this doesn't discriminate between options.
    2. For $f$ to be continuous at $x=0$ (matching the given value $f(0)=0$), we need $\displaystyle\lim_{x\to0^+}x^\alpha\log x=0$. …

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12. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.If the Rolle's theorem is applicable for the function $f(x)$ defined by $f(x) = x^3 + Px - 12$ on $[0,1]$, then the value of C of the Rolle's theorem is (A) $\pm \dfrac{1}{\sqrt{3}}$ (B) $-\dfrac{1}{\sqrt{3}}$ (C) $\dfrac{1}{\sqrt{3}}$ (D) $\dfrac{1}{3}$

    ›Reveal solutionSolution

    Rolle's theorem applying on $[0,1]$ forces $f(0)=f(1)$, which fixes $P$; then solve $f'(c)=0$ and keep only the root inside $(0,1)$.

    Concept and Intuition

    Rolle's theorem requires equal function values at the endpoints. Once $P$ is determined from that condition, the theorem guarantees at least one point $c\in(0,1)$ where the tangent is horizontal — we just need to pick the root of $f'(c)=0$ that actually lies in the open interval.

    Step-by-Step Solution

    1. $f(x)=x^3+Px-12$. Rolle's applicability requires $f(0)=f(1)$.
    2. $f(0)=-12$. $f(1)=1+P-12=P-11$.
    3. Set equal: $P-11=-12 \Rightarrow P=-1$. So $f(x)=x^3-x-12$.
    4. $f'(x)=3x^2-1$. Set $f'(c)=0$: $3c^2=1 \Rightarrow c^2=\dfrac13 \Rightarrow c=\pm\dfrac{1}{\sqrt3}$. …

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