Q.If $x$ is real, the minimum value of $x^2 - 8x + 17$ is:
(A) $-1$
(B) $0$
(C) $1$
(D) $2$

Showing the 12 most recent of 18 on this concept.

1. 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 …

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2. 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 $= 1\times\dfrac23=\dfrac23 = \dfrac{c}{3a} \Rightarrow c = 2a$. …

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3. 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 $b$: $12c - 5b = 12 \cdot \frac{147 + 25a}{36} - 5 \cdot \frac{5}{3}a.$ …

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4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $a, b, c \in R$ and $-a^2x^2 + bx + c > 0\ \forall x \in \left(\dfrac{3 - \sqrt{14}}{2}, \dfrac{3 + \sqrt{14}}{2}\right)$, then $c^2 - \left(\dfrac{b}{4}\right)^2 =$ (A) $4a^2$ (B) $a^3$ (C) $4a$ (D) $2a^2$

   ›Reveal solutionSolution

   The down-opening quadratic is positive exactly between its two roots, so the interval endpoints are the roots. This fixes $b = 3a^2$ and $c = \tfrac{5}{4}a^2$, and $c^2 - \left(\tfrac{b}{4}\right)^2 = a^4$. The official key marks option (A).

   Set up the roots

   The coefficient of $x^2$ is $-a^2 < 0$ (with $a \neq 0$), so $y = -a^2x^2 + bx + c$ is a downward-opening parabola and is positive only between its two real roots. Since it is positive on $\left(\dfrac{3-\sqrt{14}}{2},\ \dfrac{3+\sqrt{14}}{2}\right)$, those endpoints are exactly the roots $r_1, r_2$:

   $r_1 + r_2 = 3, \qquad r_1 r_2 = \frac{9 - 14}{4} = -\frac{5}{4}.$

   Read off $b$ and $c$

   Writing the quadratic as $-a^2(x - r_1)(x - r_2) = -a^2x^2 + a^2(r_1+r_2)x - a^2 r_1 r_2$ and comparing:

   $b = a^2(r_1 + r_2) = 3a^2, \qquad c = -a^2 r_1 r_2 = \frac{5}{4}a^2.$

   Evaluate the expression …

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5. 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 $b<0$. …

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6. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.If $(2k-1)x^2 - 2(3k-2)x + 4k > 0$ for every $x \in R$, then the sum of all possible integral values of $k$ is (A) 21 (B) 27 (C) 36 (D) 28

   ›Reveal solutionSolution

   A quadratic in $x$ is positive for all real $x$ iff its leading coefficient is positive and its discriminant is negative; solving both conditions on $k$ and summing the integers in range gives $28$.

   Concept and Intuition

   For $f(x)=ax^2+bx+c$ with $a\ne0$: $f(x)>0\ \forall x\in\mathbb R$ iff $a>0$ and $b^2-4ac<0$ (the parabola opens upward and never dips to or below the axis).

   Step-by-Step Solution

   1. Here $a=2k-1,\ b=-2(3k-2),\ c=4k$.
   2. Leading coefficient positive: $2k-1>0 \Rightarrow k>\dfrac12$.
   3. Discriminant negative:

   $b^2-4ac = 4(3k-2)^2 - 4(2k-1)(4k) < 0.$

   Divide by 4: $(3k-2)^2 - 4k(2k-1) < 0$.

   4. Expand: $(9k^2-12k+4) - (8k^2-4k) < 0 \Rightarrow k^2-8k+4<0$.

   5. Solve $k^2-8k+4=0$: $k=\dfrac{8\pm\sqrt{64-16}}{2}=\dfrac{8\pm\sqrt{48}}{2}=4\pm2\sqrt3$.

   6. So the inequality holds for $4-2\sqrt3 < k < 4+2\sqrt3$, numerically $0.536\ldots < k < 7.464\ldots$

   7. Intersecting with $k>\tfrac12$ from step 2 (which is already implied, since $0.536>0.5$), the valid range is $(0.536\ldots,\ 7.464\ldots)$. …

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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: $2c^2-b^2-c^2-b^2>0 \Rightarrow c^2-2b^2>0 \Rightarrow c^2>2b^2$. …

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8. 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. $f(4)=2(64)-15(16)+36(4)-30=128-240+144-30=2$. …

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9. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.If a tangent of slope 2 to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ touches the circle $x^2+y^2=4$, then maximum value of $ab$ is (A) $4$ (B) $12$ (C) $5$ (D) $7$

   ›Reveal solutionSolution

   The tangency conditions (tangent to the ellipse with slope 2, and this line touching the given circle) force $4a^2+b^2=20$; maximizing $ab$ under this constraint via AM–GM gives $\max(ab)=5$.

   Concept and Intuition

   A tangent of slope $m$ to $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ is $y=mx\pm\sqrt{a^2m^2+b^2}$. Requiring this line to also be tangent to a circle centred at the origin fixes its perpendicular distance from the origin to equal the circle's radius, giving one algebraic constraint linking $a,b$. Maximizing the product $ab$ under a constraint like $4a^2+b^2=\text{const}$ is a textbook AM–GM optimization.

   Step-by-Step Solution

   1. Tangent of slope 2: $y=2x+c$ where $c=\pm\sqrt{4a^2+b^2}$ (the ellipse tangency condition with $m=2$).
   2. This line, $2x-y+c=0$, touches $x^2+y^2=4$ when its distance from the origin equals the radius 2: $\dfrac{|c|}{\sqrt{4+1}}=2\Rightarrow|c|=2\sqrt5\Rightarrow c^2=20$.
   3. So $4a^2+b^2=20$.
   4. By AM–GM: $4a^2+b^2\ge2\sqrt{(4a^2)(b^2)}=2\cdot2ab=4ab$.
   5. Hence $20\ge4ab\Rightarrow ab\le5$. …

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10. 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 $\alpha=9,\ \beta=-3$. …

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11. 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. $f(2)=\dfrac43(8)-8=\dfrac{32}3-8=\dfrac83$. …

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12. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.For a particle moving on a straight line it is observed that the distance 'S' at a time 't' is given by $S = 6t - \dfrac{t^3}{2}$. The maximum velocity during the motion is (A) 3 (B) 6 (C) 9 (D) 12

    ›Reveal solutionSolution

    Differentiating the position function gives a velocity that's a downward-opening parabola in time, peaking right at $t=0$ with value $6$.

    Concept and Intuition

    Velocity is the derivative of position with respect to time. Here the velocity function turns out to be a simple downward parabola in $t$, so its maximum (over $t\ge0$, the physically meaningful domain) occurs either at its vertex (if that vertex is at $t\ge0$) or at the boundary $t=0$ — in this case both coincide.

    Step-by-Step Solution

    1. $S(t)=6t-\dfrac{t^3}{2}$.
    2. Velocity: $v(t)=\dfrac{dS}{dt}=6-\dfrac{3}{2}t^2$.
    3. To find extrema of $v(t)$: $\dfrac{dv}{dt}=-3t$. Setting this to zero: $t=0$.
    4. Second derivative: $\dfrac{d^2v}{dt^2}=-3<0$, confirming $t=0$ is a maximum (of velocity).
    5. Maximum velocity: $v(0)=6-0=6$. …

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