Q.Find the value of the following: The maximum value of $[x(x-1)+1]^{\frac{1}{3}}$, $0 \le x \le 1$ is (A) $(\frac{1}{3})^{\frac{1}{3}}$ (B) $\frac{1}{2}$ (C) $1$ (D) $0$ Miscellaneous Examples

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.The function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$ $(a > 0)$ attains its maximum and minimum at p and q respectively and $p^2 = q$. Then $a =$ (A) $1$ (B) $2$ (C) $\dfrac{1}{2}$ (D) $3$

   ›Reveal solutionSolution

   The critical points of the cubic are $x=a$ (a maximum) and $x=2a$ (a minimum); the condition $p^2=q$ becomes $a^2=2a$, giving $a=2$.

   Concept and Intuition

   For a cubic with positive leading coefficient, the first critical point (smaller $x$) is a local maximum and the second (larger $x$) is a local minimum. Identifying which critical point is $p$ (max) and which is $q$ (min) correctly is essential before applying the given relation $p^2=q$.

   Step-by-Step Solution

   1. $f(x)=2x^3-9ax^2+12a^2x+1$, so $f'(x)=6x^2-18ax+12a^2=6(x^2-3ax+2a^2)=6(x-a)(x-2a)$.
   2. Critical points: $x=a$ and $x=2a$ (distinct since $a>0$).
   3. $f''(x)=12x-18a$. At $x=a$: $f''(a)=12a-18a=-6a<0$ (since $a>0$) — local maximum, so $p=a$.
   4. At $x=2a$: $f''(2a)=24a-18a=6a>0$ — local minimum, so $q=2a$.
   5. Given $p^2=q$: $a^2=2a \Rightarrow a^2-2a=0 \Rightarrow a(a-2)=0 \Rightarrow a=0$ or $a=2$. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.If the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$ attains its maximum and minimum at '$p$' and '$q$' respectively such that $p^2 = q$, then '$a$' equals ______ (A) $0$ (B) $1$ (C) $2$ (D) $-1$

   ›Reveal solutionSolution

   This uses the second-derivative test to identify which critical point is the max and which is the min, then solves the given algebraic relation between them. The answer is $a=2$.

   Concept and Intuition

   A cubic with a positive leading coefficient has a local max followed by a local min (as $x$ increases), provided the two critical points are distinct. Locating them via $f'(x)=0$ and classifying with $f''(x)$ lets us assign $p$ (max) and $q$ (min) correctly, after which the given condition $p^2=q$ becomes a simple equation in $a$.

   Step-by-Step Solution

   1. $f(x)=2x^3-9ax^2+12a^2x+1\Rightarrow f'(x)=6x^2-18ax+12a^2=6(x^2-3ax+2a^2)=6(x-a)(x-2a)$.
   2. Critical points: $x=a$ and $x=2a$ (distinct provided $a\neq0$).
   3. $f''(x)=12x-18a$. At $x=a$: $f''(a)=12a-18a=-6a$. At $x=2a$: $f''(2a)=24a-18a=6a$.
   4. Assuming $a>0$: $f''(a)=-6a<0$ (maximum at $p=a$), and $f''(2a)=6a>0$ (minimum at $q=2a$). …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.If $a^2+b^2+c^2 = 1$, $a, b, c \in \mathbb{R}$, then the set of extreme values of $ab+bc+ca$ is (A) $\{\frac{1}{2}, 2\}$ (B) $\{-1, 2\}$ (C) $\{-1, \frac{1}{2}\}$ (D) $\{\frac{-1}{2}, 1\}$

   ›Reveal solutionSolution

   Two standard inequalities — $(a+b+c)^2\ge0$ and $\frac12\sum(a-b)^2\ge0$ — bound $ab+bc+ca$ between $-\frac12$ and $1$ given $a^2+b^2+c^2=1$, and both bounds are attainable.

   Concept and Intuition

   Whenever $a^2+b^2+c^2$ is fixed, the quantity $ab+bc+ca$ is squeezed between two natural extremes: it is largest when $a,b,c$ are all equal (perfectly aligned, maximizing cross terms) and smallest when $a+b+c=0$ (as spread out/opposed as possible while respecting the fixed sum of squares).

   Step-by-Step Solution

   1. Upper bound: For any reals, $\frac12[(a-b)^2+(b-c)^2+(c-a)^2] \ge 0$. Expanding: $a^2+b^2+c^2 - ab-bc-ca \ge 0$, so $ab+bc+ca \le a^2+b^2+c^2 = 1$. Equality holds when $a=b=c=\pm\frac{1}{\sqrt3}$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.The sum of the global minimum and global maximum values of the function $f(x) = \dfrac{4}{3}x^3 - 4x$ in $[0, 2]$ is (A) $0$ (B) $8/3$ (C) $-8/3$ (D) $1$

   ›Reveal solutionSolution

   Checking $f$ at the endpoints and the critical point $x=1$ gives min $-8/3$ and max $8/3$, summing to $0$.

   Concept and Intuition

   Global extrema of a continuous function on a closed interval occur either at critical points (where $f'=0$) or at the endpoints.

   Step-by-Step Solution

   1. $f'(x)=4x^2-4=4(x-1)(x+1)$; critical point in $[0,2]$ is $x=1$.
   2. $f(0)=0$.
   3. $f(1)=\dfrac43-4=-\dfrac83$.
   4. $f(2)=\dfrac43(8)-8=\dfrac{32}3-8=\dfrac83$. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $f(x) = ax^3 + bx^2 + cx + 1$ attains an extreme value 2 at $x = 1$ and another extreme value at $x = \dfrac{2}{3}$, then $2b + 3c =$ (A) a (B) 2a (C) 3a (D) 4a

   ›Reveal solutionSolution

   This tests using the sum and product of the roots of $f'(x)=0$ (the two extreme points) to relate $b,c$ to $a$; the answer is $2b+3c=a$.

   Concept and Intuition

   Extreme values of a cubic occur where its derivative vanishes. Since we're told the extrema are at $x=1$ and $x=2/3$, these are exactly the two roots of the quadratic $f'(x)=3ax^2+2bx+c=0$, so Vieta's formulas connect $b,c$ to $a$ directly — the value $f(1)=2$ is extra information not needed to find $2b+3c$ in terms of $a$.

   Step-by-Step Solution

   1. $f'(x) = 3ax^2+2bx+c$; its roots are $1$ and $\dfrac23$.
   2. Sum of roots $= 1+\dfrac23=\dfrac53 = -\dfrac{2b}{3a} \Rightarrow b = -\dfrac{5a}{2}$.
   3. Product of roots $= 1\times\dfrac23=\dfrac23 = \dfrac{c}{3a} \Rightarrow c = 2a$. …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $ax^2+bx+c<0\ \forall x\in\mathbb{R}$ and the expressions $cx^2+ax+b$ and $ax^2+bx+c$ have their extreme values at the same point x, then for the expression $cx^2+ax+b$ (A) Minimum value $=\dfrac{4b}{3}$ (B) Maximum value $=\dfrac{4a}{3}$ (C) Minimum value $=\dfrac{3a}{4}$ (D) Maximum value $=\dfrac{3b}{4}$

   ›Reveal solutionSolution

   This tests reading sign information out of "always negative" and "same extreme point" conditions, then computing a vertex value. Answer: Maximum value $=\dfrac{3b}{4}$.

   Concept and Intuition

   A quadratic $Ax^2+Bx+C$ has its extreme (vertex) value at $x=-\frac{B}{2A}$, equal to $C-\frac{B^2}{4A}$, and that extreme is a minimum if $A>0$ and a maximum if $A<0$. The condition "$ax^2+bx+c<0$ for all $x$" is a classic sign condition forcing $a<0$ (parabola opens down, entirely below the axis) and discriminant $<0$ (no real roots). Equating the two vertex $x$-locations links $a,b,c$ together via $bc=a^2$, and that single relation is enough to simplify the second vertex value cleanly.

   Step-by-Step Solution

   1. $ax^2+bx+c<0\ \forall x$ requires $a<0$ and $b^2-4ac<0$.
   2. Vertex of $ax^2+bx+c$ is at $x=-\dfrac{b}{2a}$; vertex of $cx^2+ax+b$ is at $x=-\dfrac{a}{2c}$.
   3. Same extreme point: $-\dfrac{b}{2a}=-\dfrac{a}{2c}\Rightarrow \dfrac{b}{a}=\dfrac{a}{c}\Rightarrow bc=a^2$.
   4. From $b^2-4ac<0$ and $b^2\ge 0$, we need $4ac>0\Rightarrow ac>0$; since $a<0$, this forces $c<0$.
   5. Since $bc=a^2>0$ and $c<0$, we get $b<0$. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.Let $f(x)=x^2+2bx+2c^2$ and $g(x)=-x^2-2cx+b^2$, $x\in\mathbb{R}$. If b and c are non-zero real numbers such that $\min f(x) > \max g(x)$, then $\left|\dfrac{c}{b}\right|$ lies in the interval (A) $\left(\dfrac{1}{2},\dfrac{1}{\sqrt2}\right)$ (B) $\left(\dfrac{1}{\sqrt2},\sqrt2\right)$ (C) $(\sqrt2,\infty)$ (D) $(0,1)$

   ›Reveal solutionSolution

   Find the vertex value of each quadratic (min of the upward one, max of the downward one) and turn the given inequality into a bound on $\left|\dfrac{c}{b}\right|$; the answer is $(\sqrt2,\infty)$.

   Concept and Intuition

   For $f(x)=x^2+2bx+2c^2$ (leading coefficient $+1$, opens upward), the minimum occurs at the vertex $x=-b$. For $g(x)=-x^2-2cx+b^2$ (leading coefficient $-1$, opens downward), the maximum occurs at its vertex $x=-c$. Once both extreme values are known in terms of $b,c$, the given inequality becomes a pure algebraic condition relating $b$ and $c$.

   Step-by-Step Solution

   1. $f(x)=x^2+2bx+2c^2$: vertex at $x=-b$ (since $f'(x)=2x+2b=0$). $\min f = f(-b)=b^2-2b^2+2c^2=2c^2-b^2$.
   2. $g(x)=-x^2-2cx+b^2$: vertex at $x=-c$ (since $g'(x)=-2x-2c=0$). $\max g = g(-c)=-c^2+2c^2+b^2=c^2+b^2$.
   3. Condition: $\min f > \max g \Rightarrow 2c^2-b^2 > c^2+b^2$.
   4. Simplify: $2c^2-b^2-c^2-b^2>0 \Rightarrow c^2-2b^2>0 \Rightarrow c^2>2b^2$. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.If $\alpha, \beta$ are the roots of the equation $2x^2+6x+k=0$, then the maximum value of $\left[\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right]$ when k<0 is (A) 0 (B) 1 (C) -1 (D) -2

   ›Reveal solutionSolution

   Expressing $\alpha/\beta+\beta/\alpha$ via sum and product of roots reduces it to a function of $\alpha\beta$ alone, whose supremum over the allowed negative range of $\alpha\beta$ is $-2$.

   Concept and Intuition

   Symmetric functions of the roots of a quadratic are always expressible via the sum and product of roots (Vieta's formulas), turning a two-variable question into a one-variable optimization.

   Step-by-Step Solution

   1. For $2x^2+6x+k=0$: sum of roots $\alpha+\beta=-3$, product $\alpha\beta=\dfrac k2$.
   2. $\dfrac\alpha\beta+\dfrac\beta\alpha=\dfrac{\alpha^2+\beta^2}{\alpha\beta}=\dfrac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\dfrac{9-2\alpha\beta}{\alpha\beta}=\dfrac9{\alpha\beta}-2$.
   3. Given $k<0$, so $\alpha\beta=\dfrac k2<0$. Let $m=\alpha\beta<0$; the expression is $f(m)=\dfrac9m-2$.
   4. As $m\to0^-$, $\dfrac9m\to-\infty$, so $f(m)\to-\infty$.
   5. As $m\to-\infty$, $\dfrac9m\to0^-$, so $f(m)\to-2^-$ (approaching $-2$ from below, never reaching or exceeding it). …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If the minimum value of $f(x) = x^2 + 2bx + 2c^2$ is greater than the maximum value of $g(x) = -x^2 - 2cx + b^2$, $x$ being real, then (A) $|c| > \dfrac{|b|}{\sqrt{3}}$ (B) $-1 < c < \sqrt{2}b$ (C) $\dfrac{|c|}{\sqrt{2}} > |b|$ (D) No real values of $b$ and $c$ exist

   ›Reveal solutionSolution

   $\min f = 2c^2-b^2$ must exceed $\max g = b^2+c^2$, which reduces to $c^2>2b^2$, i.e. $\dfrac{|c|}{\sqrt{2}}>|b|$ — option (C).

   $f(x)=x^2+2bx+2c^2$ is an upward parabola, so its minimum is at $x=-b$:

   $f(-b)=b^2-2b^2+2c^2=2c^2-b^2.$

   $g(x)=-x^2-2cx+b^2$ is a downward parabola, so its maximum is at $x=-c$:

   $g(-c)=-c^2+2c^2+b^2=b^2+c^2.$

   The condition $\min f > \max g$ gives …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.The difference between the absolute maximum and absolute minimum values of the function $f(x) = 2x^3 - 15x^2 + 36x - 30$ on $[-1, 4]$ is (A) $80$ (B) $1$ (C) $85$ (D) $4$

    ›Reveal solutionSolution

    Compare $f$ at the endpoints and at all critical points inside the interval; the largest and smallest of these values are the absolute max/min. Answer: $85$.

    Concept and Intuition

    On a closed interval, a continuous function's absolute extrema occur either at the endpoints or at interior critical points where $f'=0$ (or fails to exist). Since $f$ here is a cubic, differentiable everywhere, we just need to evaluate it at the two endpoints and any critical points that fall inside $[-1,4]$, then compare all four values.

    Step-by-Step Solution

    1. $f(x)=2x^3-15x^2+36x-30$, so $f'(x)=6x^2-30x+36=6(x^2-5x+6)=6(x-2)(x-3)$.
    2. Critical points: $x=2$ and $x=3$, both lie in $[-1,4]$.
    3. Evaluate $f$ at $-1$: $f(-1)=2(-1)-15(1)+36(-1)-30=-2-15-36-30=-83$.
    4. $f(2)=2(8)-15(4)+36(2)-30=16-60+72-30=-2$.
    5. $f(3)=2(27)-15(9)+36(3)-30=54-135+108-30=-3$.
    6. $f(4)=2(64)-15(16)+36(4)-30=128-240+144-30=2$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.For a quadratic expression $ax^2 + bx + c$, if the minimum value $\dfrac{49}{12}$ exists at $x = \dfrac{-5}{6}$, then $12c - 5b =$ (A) $35$ (B) $61$ (C) $49$ (D) $37$

    ›Reveal solutionSolution

    For a quadratic with a known vertex, the minimum value gives a direct relation between coefficients; using the vertex form and expanding yields a system that determines $b$ and $c$ in terms of $a$, and the condition that the minimum is $\frac{49}{12}$ fixes $a$, leading to $12c - 5b = 61$.

    The key insight is that a quadratic $ax^2 + bx + c$ attains its extremum at $x = -\frac{b}{2a}$. Here the minimum occurs at $x = -\frac{5}{6}$, so we can match that. Then the minimum value itself is $f\!\left(-\frac{b}{2a}\right) = c - \frac{b^2}{4a}$, which we set equal to $\frac{49}{12}$. This gives two equations linking $a$, $b$, and $c$. The expression $12c - 5b$ is independent of $a$ after substitution — a neat cancellation.

    1. Vertex location gives a relation between $b$ and $a$. The vertex $x$-coordinate is $-\frac{b}{2a}$. We are told it equals $-\frac{5}{6}$.

    $-\frac{b}{2a} = -\frac{5}{6} \quad\Rightarrow\quad \frac{b}{2a} = \frac{5}{6} \quad\Rightarrow\quad b = \frac{5}{3}a.$

    2. Minimum value gives another relation. The minimum value of $ax^2 + bx + c$ (since a minimum exists, $a > 0$) is

    $f_{\min} = c - \frac{b^2}{4a}.$

    We are told this equals $\frac{49}{12}$. Substitute $b = \frac{5}{3}a$:

    $c - \frac{\left(\frac{5}{3}a\right)^2}{4a} = \frac{49}{12}.$

    Simplify the fraction:

    $\frac{\frac{25}{9}a^2}{4a} = \frac{25a}{36}.$

    So

    $c - \frac{25a}{36} = \frac{49}{12}.$

    3. Solve for $c$ in terms of $a$.

    $c = \frac{49}{12} + \frac{25a}{36} = \frac{147}{36} + \frac{25a}{36} = \frac{147 + 25a}{36}.$

    4. Form the expression $12c - 5b$. Substitute $c$ and $b$: $12c - 5b = 12 \cdot \frac{147 + 25a}{36} - 5 \cdot \frac{5}{3}a.$ …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.If $\alpha$ and $\beta$ are two double roots of $x^2+3(a+3)x-9a=0$ for different values of $a$ ($\alpha > \beta$), then the minimum value of $x^2+\alpha x-\beta=0$ is (A) $\dfrac{69}{4}$ (B) $-\dfrac{69}{4}$ (C) $-\dfrac{35}{4}$ (D) $\dfrac{35}{4}$

    ›Reveal solutionSolution

    Setting the discriminant of $x^2+3(a+3)x-9a=0$ to zero gives two values of $a$ whose double roots are $-3$ and $9$; using these as $\alpha=9,\beta=-3$, the quadratic $x^2+9x+3$ has minimum $-\dfrac{69}{4}$.

    Concept and Intuition

    A "double root" of a quadratic (in $x$, with $a$ as a parameter) is a repeated root, which happens exactly when the discriminant (in $x$) vanishes. Solving that discriminant condition for $a$ gives the specific values of $a$ for which this happens, and plugging each back gives the corresponding double-root value of $x$.

    Step-by-Step Solution

    1. The quadratic in $x$: $x^2+3(a+3)x-9a=0$. Its discriminant (with leading coefficient $1$) is:

    $D = [3(a+3)]^2 - 4(1)(-9a) = 9(a+3)^2+36a$

    2. Set $D=0$ for a double root: $9(a^2+6a+9)+36a=0 \Rightarrow 9a^2+54a+81+36a=0 \Rightarrow 9a^2+90a+81=0$.
    3. Divide by 9: $a^2+10a+9=0 \Rightarrow (a+1)(a+9)=0 \Rightarrow a=-1$ or $a=-9$.
    4. For a double root, $x = -\dfrac{3(a+3)}{2}$ (vertex of the quadratic in $x$, since $D=0$).
       • $a=-1$: $x=-\dfrac{3(2)}{2}=-3$.
       • $a=-9$: $x=-\dfrac{3(-6)}{2}=9$.
    5. These two double-root values are $\alpha$ and $\beta$ with $\alpha>\beta$, so $\alpha=9,\ \beta=-3$. …

    🔒 Start your 14-day free trial to see the full solution →