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

If x is real, the minimum value of x² - 8x + 17 is – 2nd PUC Mathematics

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Concept understanding — Quadratic Extrema

Quadratic Extrema: From Intuition to Precision

Toss a ball straight up: it rises, slows, stops for an instant at the top, then falls. Plot its height against time and you get a parabola with exactly one turning point — a peak (maximum) or a valley (minimum). That single highest or lowest point is what quadratic extrema are about.

The Intuition First

A quadratic is $f(x) = ax^2 + bx + c$ , with $a \neq 0$ ; its graph is a parabola.

If $a > 0$ , it opens upward (a U) and has a minimum at the bottom.
If $a < 0$ , it opens downward and has a maximum at the top.

The turning point is the vertex. Every quadratic has exactly one vertex — that's the extremum.

Note

Unlike cubic or higher-degree polynomials, a quadratic never has both a maximum and a minimum. It has one or the other.

The Precise Statement

For $f(x) = ax^2 + bx + c$ with $a \neq 0$ :

Vertex (extremum) at

x = -\frac{b}{2a}

Extremum value

f\left(-\frac{b}{2a}\right) = c - \frac{b^2}{4a}

Nature: $a > 0$ → minimum; $a < 0$ → maximum.

\text{Vertex} = \left( -\frac{b}{2a},\; c - \frac{b^2}{4a} \right)

Why That $x$ ? A Quick Derivation

Complete the square:

f(x) = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)

The squared term is always $\geq 0$ . When $a > 0$ , $f(x)$ is smallest when the square is zero — at $x = -\frac{b}{2a}$ . When $a < 0$ , the largest value occurs at the same $x$ .

Tip

The vertex's $x$ -coordinate is also the average of the two roots (if they exist): $x = \frac{\text{root}_1 + \text{root}_2}{2}$ .

Common Mistake to Avoid

Watch out

Don't confuse the sign of $a$ with the sign of the extremum value. With $a > 0$ you always have a minimum, but that minimum could be positive, negative, or zero. The shape tells you max vs min, not the number itself.

Example

Find the extremum of $f(x) = 2x^2 - 8x + 5$ . …

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This is a quadratic expression that opens upward, so its minimum occurs at the vertex. Completing the square gives $(x-4)^2 + 1$ , so the minimum value is $1$ .

The expression $x^2 - 8x + 17$ is a quadratic in $x$ with a positive coefficient on $x^2$ . That means its graph is a parabola opening upward — it has a single lowest point, and no highest point. The question asks for that lowest value.

Why not just differentiate? You could, but completing the square is faster and gives the minimum directly without calculus. It also reveals the expression as a perfect square plus a constant, which is the cleanest way to see the minimum.

Here’s the step-by-step:

Complete the square. Take $x^2 - 8x$ and add and subtract the square of half the coefficient of $x$ : Half of $-8$ is $-4$ , and $(-4)^2 = 16$ . So:

x^2 - 8x + 17 = (x^2 - 8x + 16) + (17 - 16) = (x - 4)^2 + 1.

Interpret the result.

The term $(x-4)^2$ is always $\ge 0$ for real $x$ , and it equals $0$ exactly when $x = 4$ .

Therefore the whole expression is at least $0 + 1 = 1$ , and this value is actually attained at $x = 4$ .
Check the options. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2025Set 2025-E1 markMCQ
$x$
›Reveal solutionSolution
A quadratic with a maximum of 20 at $x=2$ and value 19 at $x=1$ is $f(x) = -x^2+4x+16$ , matching option (D).

A maximum at $x=2$ means the vertex is $(2,20)$ and the parabola opens downward ( $a<0$ ).

Step-by-step reasoning

Vertex form.

$f(x) = a(x-2)^2 + 20, \qquad a<0$

Use $f(1)=19$ .

$19 = a(1-2)^2 + 20 = a + 20 \quad\Rightarrow\quad a=-1$

Expand to standard form.

$f(x) = -(x-2)^2+20 = -(x^2-4x+4)+20 = -x^2+4x+16$

Match to the options. This is exactly option (D): $f(x) = -x^2+4x+16$ . …
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KCET 2024Set A-11 markMCQ
$x^2 - 10x - 16 = 0$
›Reveal solutionSolution
For any two numbers, the A.M. and G.M. are related to the sum and product of the roots. Given A.M. = 5 and G.M. = 4, the sum is 10 and the product is 16, leading to the quadratic $x^2 - 10x + 16 = 0$ .

The key idea here is that the arithmetic mean (A.M.) and geometric mean (G.M.) of the roots of a quadratic equation directly give us the sum and product of those roots. Once we have the sum and product, we can immediately write the quadratic.

For a quadratic equation $x^2 - Sx + P = 0$ , where $S$ is the sum of the roots and $P$ is the product of the roots, the roots themselves are the two numbers we are averaging. So if the roots are $\alpha$ and $\beta$ , then:

A.M. of $\alpha$ and $\beta$ = $\frac{\alpha + \beta}{2}$

G.M. of $\alpha$ and $\beta$ = $\sqrt{\alpha \beta}$

The problem gives us these means directly. We don't need to find the individual roots — just the sum and product.

Find the sum of the roots.

The A.M. is 5, so $\frac{\alpha + \beta}{2} = 5$ .

Multiplying both sides by 2 gives $\alpha + \beta = 10$ .

So the sum $S = 10$ .

Find the product of the roots.

The G.M. is 4, so $\sqrt{\alpha \beta} = 4$ .

Squaring both sides gives $\alpha \beta = 16$ .

So the product $P = 16$ .

Write the quadratic equation. …
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KCET 2024Set A-11 markMCQ
$(6, 0)$
›Reveal solutionSolution

Standard form $y^2 = 4ax$ with focus $(a,0)$ and directrix $x = -a$ , so $a = 6$ .

Step 1 — Identify the standard form. The focus $(6, 0)$ lies on the positive $x$ -axis and the directrix $x = -6$ is a vertical line on the other side of the origin, equidistant from it. The vertex is the midpoint of the perpendicular from the focus to the directrix, i.e. $\left(\frac{6 + (-6)}{2}, 0\right) = (0,0)$ . So this is the standard right-opening parabola
$y^{2} = 4ax, \qquad \text{focus } (a, 0), \quad \text{directrix } x = -a.$
Step 2 — Find $a$ . Matching: $a = 6$ . Hence
$y^{2} = 4(6)x = 24x.$
Step 3 — Derive it from the definition (check). A parabola is the locus of points equidistant from the focus and the directrix. For $P(x, y)$ : …
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KCET 2024Set A-11 markMCQ
$6$
›Reveal solutionSolution

Use the constraint $r^2 + h^2 = l^2$ to write the volume as a function of $h$ alone, then maximise it with the first derivative.

Step 1 — Set up the constraint

For a right circular cone with radius $r$ , height $h$ and slant height $l$ , the axial cross-section is a right triangle, so
$r^2 + h^2 = l^2 = 6^2 = 36 \;\Longrightarrow\; r^2 = 36 - h^2$
Step 2 — Express the volume in one variable
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (36 - h^2) h = \frac{\pi}{3}\left(36h - h^3\right), \qquad 0 < h < 6$
(Reducing to a single variable is the whole point of the constraint — only then can we use $dV/dh = 0$ .)

Step 3 — Find the critical point

$\frac{dV}{dh} = \frac{\pi}{3}\left(36 - 3h^2\right) = 0 \;\Longrightarrow\; h^2 = 12 \;\Longrightarrow\; h = 2\sqrt3$ …
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KCET 2021Set A-11 markMCQ
$y = -x^3 + 3x^2 + 2x - 27$
›Reveal solutionSolution

The slope function is $y'$ ; to find its maximum we differentiate a second time and set $y'' = 0$ — a classic "maximise the derivative" problem.

Step 1 — Write the slope as a function

For the curve $y = -x^3 + 3x^2 + 2x - 27$ , the slope of the tangent at any point is
$m(x) = \frac{dy}{dx} = -3x^2 + 6x + 2$
The question asks for the maximum value of $m(x)$ — so $m$ is now the function being optimised, not $y$ .

Step 2 — Critical point of $m$
$m'(x) = \frac{d^2y}{dx^2} = -6x + 6 = 0 \;\Longrightarrow\; x = 1$
Step 3 — Confirm it is a maximum

$m''(x) = \frac{d^3y}{dx^3} = -6 < 0$ …
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KCET 2018Set A-11 markMCQ
$\left(\dfrac{1}{x}\right)^x$
›Reveal solutionSolution
The function $f(x) = (1/x)^x$ is maximised by taking logs, differentiating, and setting the derivative to zero, which gives $x = 1/e$ and the maximum value $e^{1/e}$ .

The key idea is that when a variable appears both in the base and the exponent, the natural logarithm is our best friend. It turns the messy expression into a product we can differentiate easily. We want the maximum of $f(x) = \left(\frac{1}{x}\right)^x$ , which is defined for $x > 0$ (since raising a positive number to any real power is fine).

Let $f(x) = x^{-x}$ . Taking natural logs:
$\log f(x) = -x \log x.$
Now we maximise $\log f(x)$ instead of $f(x)$ itself — because $\log$ is a strictly increasing function, the $x$ that maximises $\log f(x)$ also maximises $f(x)$ . This is a standard trick in optimisation problems with exponentials.

Differentiate $\log f(x)$ with respect to $x$ :

$\frac{d}{dx} \bigl(-x \log x\bigr) = -\log x - x \cdot \frac{1}{x} = -\log x - 1.$

Set the derivative to zero to find critical points:

$-\log x - 1 = 0 \quad\Rightarrow\quad \log x = -1 \quad\Rightarrow\quad x = e^{-1} = \frac{1}{e}.$

Check that this is a maximum. The second derivative of $\log f(x)$ is $-\frac{1}{x}$ , which is negative for all $x > 0$ . So the function is concave down everywhere, and the critical point is indeed a global maximum.

Compute the maximum value of $f(x)$ at $x = 1/e$ : …
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KCET 2018Set A-11 markMCQ
$(x+1)^2 + (y-3)^2 = 64$
›Reveal solutionSolution

The inscribed rectangle's diagonal is the circle's diameter ( $16$ ); maximising area under $a^2+b^2 = 16^2$ gives the square, of area $d^2/2 = 128$ .

Step 1 — Read the circle.
$(x+1)^{2} + (y-3)^{2} = 64$
Centre $(-1, 3)$ , radius $r = \sqrt{64} = 8$ . (The centre is irrelevant — area is translation-invariant.)

Step 2 — The key geometric fact.

A rectangle inscribed in a circle has all four vertices on the circle, so its diagonal is a diameter:
$d = 2r = 16$
If the sides are $a$ and $b$ , then by Pythagoras
$a^{2} + b^{2} = d^{2} = 256$
Step 3 — Maximise the area $A = ab$ subject to $a^2+b^2 = 256$ .

Parametrise $a = 16\cos\theta$ , $b = 16\sin\theta$ :

$A = 256\sin\theta\cos\theta = 128\sin 2\theta$ …
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Q.If $x$ is real, the minimum value of $x^2 - 8x + 17$ is:
(A) $-1$
(B) $0$
(C) $1$
(D) $2$

Concept understanding — Quadratic Extrema

Quadratic Extrema: From Intuition to Precision

The Intuition First

The Precise Statement

Why That $x$ ? A Quick Derivation

Common Mistake to Avoid

Example

Method: Finding the Extreme Value of a Quadratic Expression

Steps

Common Mistakes

Step 1 — Set up the constraint

Step 2 — Express the volume in one variable

Step 3 — Find the critical point

Step 1 — Write the slope as a function

Step 2 — Critical point of $m$

Step 3 — Confirm it is a maximum

Q.If xxx is real, the minimum value of x2−8x+17x^2 - 8x + 17x2−8x+17 is: (A) −1-1−1 (B) 000 (C) 111 (D) 222

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

Unlock everything free for 14 days