Q.Find a point on the curve $y = (x - 3)^2$, where the tangent is parallel to the chord joining the points $(3, 0)$ and $(4, 1)$.

Showing the 12 most recent of 24 on this concept.

1. 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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2. 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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3. 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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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If the function $f(x) = x^3 + bx^2 + ax$ satisfies the conditions of Rolle's theorem in $[1, 3]$ with $c = 2 + \dfrac{1}{\sqrt{3}}$ then $(a, b) =$ (A) $(11, 6)$ (B) $(11, -6)$ (C) $(6, 11)$ (D) $(6, -11)$

   ›Reveal solutionSolution

   Use both Rolle conditions — $f(1)=f(3)$ and $f'(c)=0$ — to pin down $a$ and $b$; the pair $(11,-6)$ satisfies both exactly.

   Concept and Intuition

   Rolle's theorem for $f$ on $[1,3]$ requires $f(1)=f(3)$, and then guarantees some $c\in(1,3)$ with $f'(c)=0$. Here we're told the specific $c$, so we get two independent equations in $a,b$: the endpoint-equality condition and the stationary-point condition at the given $c$.

   Step-by-Step Solution

   1. $f(x)=x^3+bx^2+ax$, so

   $f(1)=1+b+a,\qquad f(3)=27+9b+3a.$

   Setting $f(1)=f(3)$:

   $1+b+a=27+9b+3a\ \Rightarrow\ 8b+2a=-26\ \Rightarrow\ 4b+a=-13.$

   2. $f'(x)=3x^2+2bx+a$, and $f'(c)=0$ at $c=2+\dfrac1{\sqrt3}$. Rather than substitute the messy $c$ directly, test the given options against $f'(x)=3x^2+2bx+a=0$ and see which produces exactly this root.
   3. Try $(a,b)=(11,-6)$: $f'(x)=3x^2-12x+11$. Solving,

   $x=\frac{12\pm\sqrt{144-132}}{6}=\frac{12\pm\sqrt{12}}{6}=2\pm\frac{1}{\sqrt3}.$

   The root $x=2+\dfrac1{\sqrt3}$ appears exactly as required. …

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5. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The value c of the Rolle's theorem for the function $f(x)=2\sin x+\sin 2x$ in the interval $[0,\pi]$ is (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{6}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   Rolle's theorem needs $f(0)=f(\pi)$; then $c$ is found by solving $f'(c)=0$ inside $(0,\pi)$, using the double-angle identity to reduce to a quadratic in $\cos x$. Answer: $c=\dfrac{\pi}{3}$.

   Concept and Intuition

   Rolle's theorem guarantees at least one point $c$ in $(a,b)$ where $f'(c)=0$, provided $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$. Here $f$ is a smooth trig function (always continuous/differentiable), so only the endpoint condition and the root-finding for $f'$ need checking — and any root that lands exactly on an endpoint must be discarded, since Rolle's $c$ must be strictly interior.

   Step-by-Step Solution

   1. Check hypotheses: $f(0)=2\sin0+\sin0=0$; $f(\pi)=2\sin\pi+\sin2\pi=0+0=0$. So $f(0)=f(\pi)=0$ — Rolle's theorem applies, guaranteeing some $c\in(0,\pi)$ with $f'(c)=0$.
   2. Differentiate: $f'(x)=2\cos x+2\cos2x$.
   3. Set $f'(x)=0$: $\cos x+\cos2x=0$. Use $\cos2x=2\cos^2x-1$:

   $\cos x + 2\cos^2x - 1 = 0 \implies 2\cos^2x+\cos x-1=0.$

   4. Factor the quadratic in $\cos x$: $2c^2+c-1=(2c-1)(c+1)=0$ where $c=\cos x$. So $\cos x=\dfrac12$ or $\cos x=-1$.
   5. $\cos x=\dfrac12 \Rightarrow x=\dfrac{\pi}{3}$ (lies in the open interval $(0,\pi)$ — valid). …

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6. 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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7. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If the function $f(x) = x^3 + bx^2 + cx - 6$ satisfies all the conditions of Rolle's theorem in [1, 3] and $f'\left(\dfrac{2\sqrt{3}+1}{\sqrt{3}}\right) = 0$, then $bc =$ (A) $18$ (B) $-66$ (C) $38$ (D) $-46$

   ›Reveal solutionSolution

   Rolle's theorem forces $f(1)=f(3)$; combining that with the given root of $f'$ pins down $b$ and $c$, giving $bc=-66$.

   Concept and Intuition

   Rolle's theorem guarantees a point where $f'=0$ between two points of equal function value. Here we're told which point that is, so instead of proving existence we use it as an equation to solve for the unknown constants $b,c$, alongside the equal-endpoints condition.

   Step-by-Step Solution

   1. Equal endpoints: $f(1)=1+b+c-6=b+c-5$ and $f(3)=27+9b+3c-6=21+9b+3c$. Setting $f(1)=f(3)$: $b+c-5=21+9b+3c \Rightarrow -8b-2c=26 \Rightarrow 4b+c=-13$.
   2. Use the given critical point: $x_0=\dfrac{2\sqrt3+1}{\sqrt3}=2+\dfrac1{\sqrt3}$. $f'(x)=3x^2+2bx+c$, so $f'(x_0)=3x_0^2+2bx_0+c=0$.
   3. Compute $x_0^2=\left(2+\tfrac1{\sqrt3}\right)^2=4+\tfrac{4}{\sqrt3}+\tfrac13$, so $3x_0^2=12+\tfrac{12}{\sqrt3}+1=13+4\sqrt3$. Also $2bx_0=4b+\tfrac{2b}{\sqrt3}$. …

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8. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If the Lagrange's mean value theorem is applied to the function $f(x) = e^x$ defined on the interval $[1, 2]$ and the value of $c \in (1, 2)$ is k, then $e^{k-1} =$ (A) $2$ (B) $e - 1$ (C) $e + 1$ (D) $1$

   ›Reveal solutionSolution

   Applying Lagrange's Mean Value Theorem to $f(x)=e^x$ on $[1,2]$ directly gives $e^{k-1}=e-1$ without needing to solve for $k$ explicitly.

   Concept and Intuition

   LMVT guarantees a point $c=k$ in $(1,2)$ where the instantaneous rate of change equals the average rate of change over the interval. The elegance here is that we never need to isolate $k$ itself — the quantity asked for, $e^{k-1}$, falls out directly from the LMVT equation after dividing by $e$.

   Step-by-Step Solution

   1. LMVT: there exists $k\in(1,2)$ such that $f'(k)=\dfrac{f(2)-f(1)}{2-1}$.
   2. $f(x)=e^x \Rightarrow f'(x)=e^x$, $f(2)=e^2$, $f(1)=e$.
   3. So $e^k = e^2-e = e(e-1)$. …

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9. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.Consider the quadratic equation $ax^2 + bx + c = 0$, where $2a + 3b + 6c = 0$ and let $g(x) = \dfrac{ax^3}{3} + \dfrac{bx^2}{2} + cx$. Statement-I: The given quadratic equation $ax^2 + bx + c = 0$ has atleast one root in $(0, 1)$. Statement-II: Rolle's theorem is applicable to $g(x)$ on $[0, 1]$. Then (A) Statement-I is false, Statement-II is true (B) Statement-I is true, Statement-II is false (C) Statement-I is true, Statement-II is true but Statement-II is not a correct explanation of Statement-I (D) Statement-I is true, Statement-II is true and Statement-II is a correct explanation of Statement-I

   ›Reveal solutionSolution

   The condition $2a+3b+6c=0$ is exactly what makes $g(0)=g(1)=0$, so Rolle's theorem on $g$ directly forces a root of the quadratic in $(0,1)$. Answer: (D).

   Concept and Intuition

   This is a classic use of Rolle's theorem to prove existence of a root without solving the quadratic: construct an antiderivative $g$ of the quadratic such that $g$ vanishes at both endpoints of an interval, then Rolle's theorem guarantees $g'$ (i.e. the original quadratic) vanishes somewhere strictly inside. The condition $2a+3b+6c=0$ is engineered precisely to make $g(1)=0$.

   Step-by-Step Solution

   1. $g(x)=\dfrac{ax^3}3+\dfrac{bx^2}2+cx$, so $g(0)=0$ trivially.
   2. $g(1)=\dfrac a3+\dfrac b2+c$. Put over a common denominator $6$: $g(1)=\dfrac{2a+3b+6c}{6}$.
   3. Given $2a+3b+6c=0$, so $g(1)=0$.
   4. $g$ is a polynomial, hence continuous on $[0,1]$ and differentiable on $(0,1)$, and $g(0)=g(1)=0$ — all three hypotheses of Rolle's theorem hold. Statement-II is TRUE.
   5. By Rolle's theorem, there exists $k\in(0,1)$ with $g'(k)=0$.
   6. But $g'(x)=ax^2+bx+c$, which is exactly the quadratic in question. So $ak^2+bk+c=0$ for some $k\in(0,1)$ — the quadratic has a root in $(0,1)$. Statement-I is TRUE. …

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10. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.If Rolle's theorem is applicable for the function $f(x)=x(x+3)e^{-x/2}$ on $[-3,0]$, then the value of c is (A) $3$ (B) $3$ and $-2$ (C) $-2$ (D) $-1$

    ›Reveal solutionSolution

    Solving $f'(x)=0$ gives two roots, $x=3$ and $x=-2$, but only $x=-2$ actually lies inside the interval $(-3,0)$, so that's the Rolle's-theorem point.

    Concept and Intuition

    Rolle's theorem says that if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists $c\in(a,b)$ with $f'(c)=0$ — a horizontal tangent somewhere strictly inside the interval. Solving $f'(x)=0$ can produce extra roots outside the interval; only the ones actually inside $(a,b)$ count as valid "$c$" values.

    Step-by-Step Solution

    1. Check the hypothesis: $f(-3)=(-3)(0)e^{3/2}=0$ and $f(0)=(0)(3)e^0=0$, so $f(-3)=f(0)=0$ — Rolle's applies.
    2. Write $f(x)=(x^2+3x)e^{-x/2}$ and differentiate using the product rule: $f'(x) = (2x+3)e^{-x/2} + (x^2+3x)\left(-\tfrac12\right)e^{-x/2} = e^{-x/2}\left[(2x+3)-\tfrac12(x^2+3x)\right]$. …

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11. 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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12. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.If the function $f(x)=\sqrt{x^2-4}$ satisfies the Lagrange's mean value theorem on [2, 4], then the value of C is (A) $2\sqrt{3}$ (B) $-2\sqrt{3}$ (C) $\sqrt{6}$ (D) $-\sqrt{6}$

    ›Reveal solutionSolution

    Apply Lagrange's Mean Value Theorem: find $c\in(2,4)$ where the instantaneous slope equals the average slope of $f(x)=\sqrt{x^2-4}$ over $[2,4]$.

    Concept and Intuition

    LMVT guarantees a point $c$ in the open interval where the tangent slope equals the secant slope between the endpoints — a formal statement of "at some instant, the instantaneous rate equals the average rate."

    Step-by-Step Solution

    1. $f(2)=\sqrt{4-4}=0$, $f(4)=\sqrt{16-4}=\sqrt{12}=2\sqrt3$.
    2. Average rate of change $=\dfrac{f(4)-f(2)}{4-2}=\dfrac{2\sqrt3}{2}=\sqrt3$.
    3. $f'(x)=\dfrac{1}{2\sqrt{x^2-4}}\cdot 2x=\dfrac{x}{\sqrt{x^2-4}}$.
    4. Set $f'(c)=\sqrt3$: $\dfrac{c}{\sqrt{c^2-4}}=\sqrt3\Rightarrow c^2=3(c^2-4)\Rightarrow c^2=3c^2-12\Rightarrow 2c^2=12\Rightarrow c^2=6$. …

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