Q.Find local maximum and local minimum values of the function $f$ given by $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.In the interval $(-\infty, 0)$ the function $f(x) = x^2 + \dfrac{128}{x}$ (A) has only one local minimum value at $x = 4$ (B) has one maximum and one minimum at $x = 4$ and $x = -4$ respectively (C) is increasing (D) is decreasing

   ›Reveal solutionSolution

   Checking the sign of $f'(x)=2x-128/x^2$ on $(-\infty,0)$ shows it is always negative there, so the function is monotonically decreasing throughout the interval — no local max/min occurs in this domain.

   Concept and Intuition

   A function's monotonic behaviour on an interval is read off the sign of its derivative there. Critical points (where $f'=0$) only matter if they actually lie inside the interval in question; a critical point outside the interval is irrelevant to that interval's monotonicity.

   Step-by-Step Solution

   1. $f(x) = x^2 + \dfrac{128}{x}$. Differentiate: $f'(x) = 2x - \dfrac{128}{x^2}$.
   2. Find critical points: $f'(x)=0 \Rightarrow 2x = \dfrac{128}{x^2} \Rightarrow 2x^3 = 128 \Rightarrow x^3=64 \Rightarrow x=4$.
   3. The only critical point is $x=4$, which is not in $(-\infty,0)$.
   4. Check the sign of $f'(x)$ for any $x<0$: $2x$ is negative (since $x<0$); $x^2>0$ always, so $\dfrac{128}{x^2}>0$, making $-\dfrac{128}{x^2}$ negative.
   5. So $f'(x)$ is the sum of two negative quantities for every $x<0$: $f'(x)<0$ throughout $(-\infty,0)$. …

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2. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If the function $y=g(x)$ representing the slopes of the tangents drawn to the curve $y=3x^4-5x^3-12x^2+18x+3$ is strictly increasing then the domain of $g(x)$ is (A) $\left[-\dfrac{1}{2},\dfrac{4}{3}\right]$ (B) $\left(\dfrac{-1}{2},\dfrac{4}{3}\right)$ (C) $\mathbb{R}-\left(\dfrac{-1}{2},\dfrac{3}{4}\right)$ (D) $\mathbb{R}-\left[\dfrac{-1}{2},\dfrac{4}{3}\right]$

   ›Reveal solutionSolution

   The "slope function" $g(x)$ is the derivative of the given quartic; its own strict increase is governed by $g'(x)>0$, i.e. a second derivative test producing a quadratic inequality. Answer: $\mathbb R-\left[-\frac12,\frac43\right]$.

   Concept and Intuition

   $g(x)$, "the slope of the tangent" to $y=3x^4-5x^3-12x^2+18x+3$, is exactly $y'(x)$ — a new function in its own right. Asking where this function is strictly increasing is asking where $g'(x)=y''(x)>0$, i.e. a standard increasing/decreasing analysis one derivative order up.

   Step-by-Step Solution

   1. $y=3x^4-5x^3-12x^2+18x+3$, so $g(x)=y'=12x^3-15x^2-24x+18$.
   2. $g$ is strictly increasing where $g'(x)>0$: $g'(x)=36x^2-30x-24$.
   3. Factor out 6: $g'(x)=6(6x^2-5x-4)$.
   4. Solve $6x^2-5x-4=0$: discriminant $=25+96=121=11^2$, so $x=\dfrac{5\pm11}{12}$, giving $x=\dfrac{16}{12}=\dfrac43$ and $x=\dfrac{-6}{12}=-\dfrac12$.
   5. Since the coefficient of $x^2$ is positive, $6x^2-5x-4>0$ outside the roots and $<0$ between them: positive for $x<-\dfrac12$ or $x>\dfrac43$. …

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3. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.The maximum value of 'a' such that the second derivative of $x^4 + ax^3 + \dfrac{3x^2}{2} + 1$ is positive for all real $x$ is (A) $3$ (B) $-3$ (C) $2$ (D) $-2$

   ›Reveal solutionSolution

   The second derivative is a quadratic in $x$; positivity for all $x$ requires a non-positive discriminant, which bounds $a$.

   Concept and Intuition

   An upward-opening quadratic $Ax^2+Bx+C$ (here in $x$, with $A=12>0$) is $\ge0$ for all real $x$ exactly when its discriminant $B^2-4AC\le0$.

   Step-by-Step Solution

   1. $f(x)=x^4+ax^3+\frac{3x^2}{2}+1$.
   2. $f'(x)=4x^3+3ax^2+3x$.
   3. $f''(x)=12x^2+6ax+3$.
   4. Require $f''(x)\ge0$ for all real $x$ (boundary case of the required positivity): discriminant of $12x^2+6ax+3$ is $(6a)^2-4(12)(3)=36a^2-144$.
   5. Need $36a^2-144\le0\Rightarrow a^2\le4\Rightarrow -2\le a\le2$. …

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4. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.At $x = 0$, $f(x) = \cos x - 1 + \dfrac{x^2}{2} - \dfrac{x^3}{3}$ (A) has a minimum value (B) has a maximum value (C) has no extremum value (D) is not defined

   ›Reveal solutionSolution

   The Taylor expansion of $f$ about $x=0$ has leading term $-x^3/3$ (odd power), so $f$ passes through $0$ changing sign — an inflection-type behavior, not an extremum.

   Concept and Intuition

   At a candidate critical point, if the first nonzero derivative is of odd order, the function does not have a local extremum there (it's increasing or decreasing straight through); only an even-order first-nonzero derivative gives a genuine min/max.

   Step-by-Step Solution

   1. $\cos x=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\dots$
   2. $f(x)=\cos x-1+\dfrac{x^2}{2}-\dfrac{x^3}{3}=\left(1-\dfrac{x^2}{2}+\dfrac{x^4}{24}\right)-1+\dfrac{x^2}{2}-\dfrac{x^3}{3}+\dots=-\dfrac{x^3}{3}+\dfrac{x^4}{24}+\dots$
   3. Near $x=0$, $f(x)\approx-\dfrac{x^3}{3}$: for small $x>0$, $f<0$; for small $x<0$, $f>0$ (since $-(-|x|)^3/3=|x|^3/3>0$).
   4. So $f$ changes sign through $x=0$ rather than staying one-signed on both sides — confirming this is not a local extremum. …

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5. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.Let $P(x)=x^4+ax^3+bx^2+cx+d$ be such that $x=0$ is the only real root of $P^1(x)=0$. If $P(-1)<P(1)$, then in the interval $[-1,1]$ (A) P(-1) is not minimum of P(x), but P(1) is the maximum of P(x) (B) P(-1) is minimum of P(x), but P(1) is not the maximum of P(x) (C) Neither P(-1) is the minimum nor P(1) is the maximum of P(x) (D) P(-1) is the minimum and P(1) is the maximum of P(x)

   ›Reveal solutionSolution

   This tests reading the sign of $P'$ from the structure of a cubic with a single real root; $P$ turns out to be strictly decreasing then increasing with its minimum at the interior point $x=0$, so $P(-1)$ is never the minimum, while the given inequality forces $P(1)$ to be the maximum.

   Concept and Intuition

   A quartic's monotonicity on an interval is governed by the sign of its derivative, a cubic here. If that cubic has only one real root, the other two roots are a complex-conjugate pair, so the cubic (as a real function) doesn't change sign there — it only changes sign at the single real root. That tells us $P$ has exactly one turning point on all of $\mathbb R$, at $x=0$, and it must be a minimum (since $P\to+\infty$ both ways, being a quartic with positive leading coefficient).

   Step-by-Step Solution

   1. $P'(x)=4x^3+3ax^2+2bx+c$. Since $x=0$ is a root, $P'(0)=c=0$.
   2. So $P'(x)=4x^3+3ax^2+2bx=x(4x^2+3ax+2b)$.
   3. For $x=0$ to be the only real root of $P'$, the quadratic factor $4x^2+3ax+2b$ must have no real zero, i.e. discriminant $9a^2-32b<0$. Since its leading coefficient $4>0$ and it has no real root, $4x^2+3ax+2b>0$ for all real $x$.
   4. Therefore $P'(x)=x\cdot(\text{always positive})$, so $P'(x)<0$ for $x<0$ and $P'(x)>0$ for $x>0$: $P$ is strictly decreasing on $[-1,0]$ and strictly increasing on $[0,1]$. …

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6. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.$A(1,15)$, $B(3,-12)$, $C(6,12)$ are three consecutive turning points of a continuous curve $y = f(x)$. If $f(x) = 0$ only for $x = \alpha$ and $x = \beta$, then $|\beta - \alpha| <$ (A) $27$ (B) $2$ (C) $5$ (D) $24$

   ›Reveal solutionSolution

   The two zeros lie strictly between consecutive turning points; bounding $\alpha\in(1,3)$ and $\beta\in(3,6)$ gives $|\beta-\alpha|<5$.

   Concept and Intuition

   Between a positive local extreme value and a negative one, a continuous curve must cross zero at least once (Intermediate Value Theorem). The turning points bracket where each zero can occur.

   Step-by-Step Solution

   1. At $x=1$, $f=15>0$ (local extreme); at $x=3$, $f=-12<0$ (local extreme); so by IVT, $f$ has a zero $\alpha$ strictly between $1$ and $3$: $1<\alpha<3$.
   2. At $x=3$, $f=-12<0$; at $x=6$, $f=12>0$; so $f$ has a zero $\beta$ strictly between $3$ and $6$: $3<\beta<6$. …

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7. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If $f(x) = x^x$, then the interval in which $f(x)$ decreases is (A) $\left[0, \dfrac{1}{e}\right]$ (B) $[0, e]$ (C) $\left[\dfrac{1}{e}, \infty\right]$ (D) $[0, e^e]$

   ›Reveal solutionSolution

   Differentiate $x^x$ using logarithmic differentiation and find where the derivative is negative.

   Concept and Intuition

   A function decreases where its derivative is negative. Since $x^x$ itself is always positive on its domain $x>0$, the sign of $f'(x)$ is controlled entirely by the factor $(\log x+1)$.

   Step-by-Step Solution

   1. $f(x)=x^x$. Take logs: $\log f=x\log x$.
   2. Differentiate: $\dfrac{f'}{f}=\log x+1$, so $f'(x)=x^x(\log x+1)$.
   3. Since $x^x>0$ for all $x>0$, the sign of $f'(x)$ matches the sign of $(\log x+1)$.
   4. $f'(x)<0 \iff \log x+1<0 \iff \log x<-1 \iff x<e^{-1}=\dfrac1e$. …

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8. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.For which value(s) of '$a$', $f(x) = -x^3 + 4ax^2 + 2x - 5$ is decreasing for every '$x$'? (A) $(1, 2)$ (B) $(3, 4)$ (C) $R$ (D) No value of '$a$'

   ›Reveal solutionSolution

   $f'(x)=-3x^2+8ax+2$ always attains a positive maximum for every real $a$, so $f$ can never be decreasing for all $x$ — the answer is "no value of $a$."

   Concept and Intuition

   For $f$ to be decreasing on all of $R$, its derivative must be $\le 0$ everywhere. $f'(x)$ here is a downward-opening parabola in $x$, so as $x\to\pm\infty$ it's automatically negative — the only risk is its peak (vertex) value going positive. If that peak is always positive regardless of $a$, no $a$ can work.

   Step-by-Step Solution

   1. Differentiate: $f'(x)=-3x^2+8ax+2$.
   2. This is a downward parabola in $x$ (leading coefficient $-3<0$), with vertex at $x_0=\dfrac{-8a}{2(-3)}=\dfrac{4a}{3}$.
   3. Maximum value of $f'$ at the vertex:

   $f'(x_0) = -3\left(\frac{4a}{3}\right)^2 + 8a\left(\frac{4a}{3}\right) + 2 = -\frac{16a^2}{3} + \frac{32a^2}{3} + 2 = \frac{16a^2}{3} + 2.$

   4. Since $\dfrac{16a^2}{3}\ge 0$ for every real $a$, the maximum of $f'$ is always $\ge 2 > 0$. …

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9. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If the tangent drawn to the curve $y=x^3-ax^2+x+1$ at each point $x\in\mathbb{R}$, is inclined at an acute angle with the positive direction of X-axis, then the set of all possible values of 'a' is (A) $\mathbb{R}-(-\sqrt3,\sqrt3)$ (B) $[-3,3]$ (C) $\mathbb{R}$ (D) $(-\sqrt3,\sqrt3)$

   ›Reveal solutionSolution

   The tangent slope $3x^2-2ax+1$ must stay strictly positive for all real $x$; requiring a negative discriminant gives $a\in(-\sqrt3,\sqrt3)$.

   Concept and Intuition

   "Tangent inclined at an acute angle with the positive x-axis" means the tangent's slope is strictly positive (an acute angle has $\tan\theta>0$). Since this must hold for every real $x$ (the curve's domain), the derivative — a quadratic in $x$ — must never touch or cross zero; it must be strictly positive throughout.

   Step-by-Step Solution

   1. $y = x^3-ax^2+x+1 \Rightarrow y' = 3x^2-2ax+1$.
   2. Require $y'>0$ for all $x\in\mathbb R$.
   3. A quadratic $Ax^2+Bx+C$ with $A>0$ is positive for all $x$ iff its discriminant $B^2-4AC<0$.
   4. Here $A=3,\ B=-2a,\ C=1$: discriminant $=4a^2-12$.
   5. Require $4a^2-12<0 \Rightarrow a^2<3 \Rightarrow -\sqrt3<a<\sqrt3$. …

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10. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.The cubic equation $2x^3 - 3x^2 + 6x + 2 = 0$ (A) has 3 distinct real roots (B) has only one real root in the interval $(-1, 0)$ (C) has two distinct real roots (D) has only one real root in the interval $(0, 1)$

    ›Reveal solutionSolution

    Showing $f'(x)>0$ always (via a negative-discriminant quadratic) proves the cubic is strictly increasing and has exactly one real root; sign checking at $-1$ and $0$ locates it in $(-1,0)$.

    Concept and Intuition

    A cubic with a strictly positive (or always negative) derivative is monotonic, and a monotonic continuous function can cross zero at most once — so it has exactly one real root. To locate that root, use the Intermediate Value Theorem: find two points where the function has opposite signs, and the root must lie between them.

    Step-by-Step Solution

    1. Let $f(x)=2x^3-3x^2+6x+2$. Compute $f'(x) = 6x^2-6x+6 = 6(x^2-x+1)$.
    2. Check the discriminant of $x^2-x+1$: $D = (-1)^2 - 4(1)(1) = 1-4=-3 <0$. Since the discriminant is negative and the leading coefficient is positive, $x^2-x+1>0$ for all real $x$.
    3. So $f'(x) = 6(x^2-x+1) > 0$ for all $x$ — $f$ is strictly increasing on $\mathbb{R}$, hence it can have at most (and, being an odd-degree polynomial, exactly) one real root.
    4. This immediately rules out "3 distinct real roots" and "two distinct real roots".
    5. To locate the root, evaluate $f$ at convenient points: $f(-1) = 2(-1)-3(1)+6(-1)+2 = -2-3-6+2=-9$ (negative). $f(0) = 0-0+0+2=2$ (positive). $f(1)=2-3+6+2=7$ (positive). …

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11. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.The curve represented by $x=t^5+5t^3+20t+7$ and $y=4t^3-3t^2-18t+3$ is decreasing in the interval (A) $(-2,-1)$ (B) $(3/2,2)$ (C) $(-1,3/2)$ (D) $(-2,2)$

    ›Reveal solutionSolution

    For a parametric curve with $x'(t)$ always positive, the curve decreases in $y$ exactly where $y'(t)<0$. Here that interval is $(-1,3/2)$.

    Concept and Intuition

    For a curve given parametrically, $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$. The curve is 'decreasing' (as a function $y$ of $x$) precisely where this ratio is negative. If $dx/dt$ never changes sign (stays positive throughout), then the sign of $dy/dx$ is simply the sign of $dy/dt$ — so we only need to analyze $dy/dt$.

    Step-by-Step Solution

    1. Differentiate $x=t^5+5t^3+20t+7$: $\dfrac{dx}{dt}=5t^4+15t^2+20=5(t^4+3t^2+4)$.
    2. Check the sign of $t^4+3t^2+4$: substituting $u=t^2\ge0$, this is $u^2+3u+4$, whose discriminant is $9-16=-7<0$, so it's always positive. Hence $\dfrac{dx}{dt}>0$ for every real $t$ — $x$ is strictly increasing in $t$.
    3. Differentiate $y=4t^3-3t^2-18t+3$: $\dfrac{dy}{dt}=12t^2-6t-18=6(2t^2-t-3)=6(2t-3)(t+1)$. …

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12. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.If the displacement S of a particle travelling along a straight line in t seconds is given by $S = 2t^3 + 2t^2 - 2t - 3$, then the time taken (in seconds) by the particle to change its direction is (A) $\dfrac{1}{3}$ (B) $2$ (C) $3$ (D) $\dfrac{1}{2}$

    ›Reveal solutionSolution

    A particle changes direction exactly when its velocity is zero and changes sign there. Answer: $t=\dfrac13$ s.

    Concept and Intuition

    Direction of motion is given by the sign of velocity $v=dS/dt$. The particle reverses direction only at a time where $v=0$ and $v$ actually changes sign across that instant (not merely touches zero, e.g. at an inflection with no sign change).

    Step-by-Step Solution

    1. $S=2t^3+2t^2-2t-3 \Rightarrow v=\dfrac{dS}{dt}=6t^2+4t-2$.
    2. Factor: $v=2(3t^2+2t-1)=2(3t-1)(t+1)$.
    3. $v=0 \Rightarrow t=\dfrac13$ or $t=-1$. Since time $t\ge0$, only $t=\dfrac13$ is physically valid. …

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