Q.Let $f : [2, \infty) \to \mathbb{R}$ be the function defined by $f(x) = x^2 - 4x + 5$, then the range of $f$ is
(A) $\mathbb{R}$
(B) $[1, \infty)$
(C) $[4, \infty)$
(D) $[5, \infty)$

Showing the 12 most recent of 38 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The range of the real valued function $f(x) = -\sqrt{-x^2 - 6x - 5}$ is (A) $[-\infty, 0]$ (B) $[-5, -1]$ (C) $[-2, 0]$ (D) $[-3, -2]$

   ›Reveal solutionSolution

   The function is a negative square root of a downward-opening quadratic; the quadratic’s range inside the square root is $[0,4]$, so after taking the negative square root the range becomes $[-2,0]$. The correct option is (C).

   We want the set of all possible output values of $f(x) = -\sqrt{-x^2 - 6x - 5}$.

   The key is to realize that the square root only accepts non‑negative inputs, and the negative sign flips the sign of the output.

   ---

   Concept and Intuition (Quadratic Range)

   The expression inside the square root is a quadratic: $Q(x) = -x^2 - 6x - 5$.

   For $f(x)$ to be defined, we need $Q(x) \ge 0$.

   Since $Q(x)$ is a concave-down parabola (coefficient of $x^2$ is negative), its graph is an upside-down U. The values it takes are from $0$ up to its maximum, then back down to $0$.

   So the possible inputs to the square root are exactly the numbers in $[0, \text{max of } Q]$.

   Then $\sqrt{Q(x)}$ outputs numbers from $0$ up to $\sqrt{\text{max}}$.

   Finally, the negative sign flips these to $[-\sqrt{\text{max}}, 0]$.

   Thus the problem reduces to finding the maximum value of $Q(x)$ on its domain.

   ---

   Step-by-step reasoning

   1. Find the domain of $f$ We require $-x^2 - 6x - 5 \ge 0$. Multiply by $-1$ (reversing inequality):

   $x^2 + 6x + 5 \le 0.$

   Factor: $(x+1)(x+5) \le 0$.

   So the domain is $x \in [-5, -1]$.

   2. Find the maximum of $Q(x) = -x^2 - 6x - 5$ on this domain Complete the square:

   $Q(x) = -(x^2 + 6x) - 5 = -(x^2 + 6x + 9) + 9 - 5 = -(x+3)^2 + 4.$

   This is a parabola with vertex at $x = -3$, giving maximum value $4$.

   Since $-3$ lies inside $[-5, -1]$, the maximum is indeed $4$.

   3. Determine the range of $Q(x)$ on the domain At the endpoints: $Q(-5) = -25 + 30 - 5 = 0$, $Q(-1) = -1 + 6 - 5 = 0$. So $Q(x)$ goes from $0$ up to $4$ and back to $0$. …

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2. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
   Q.The range of the real valued function $f(x) = \sqrt{\dfrac{x^2+2x+8}{x^2+2x+4}}$ is (A) $\left[\sqrt{\dfrac{7}{3}}, \infty\right)$ (B) $(0, \infty)$ (C) $(1, \infty)$ (D) $\left(1, \sqrt{\dfrac{7}{3}}\right]$

   ›Reveal solutionSolution

   Substituting $t=x^2+2x+4$ (whose minimum is 3) turns the function into $\sqrt{1+4/t}$, whose range works out to $\left(1,\sqrt{7/3}\right]$.

   Concept and Intuition

   The denominator $x^2+2x+4$ never vanishes (its discriminant is negative) and has a definite minimum value achieved at $x=-1$; expressing the whole function in terms of this quadratic lets us convert the range problem into a simple monotonic-function-of-$t$ problem.

   Step-by-Step Solution

   1. Let $t = x^2+2x+4 = (x+1)^2+3$. Since $(x+1)^2\ge0$, we get $t\ge3$, with $t=3$ only at $x=-1$ and $t\to\infty$ as $|x|\to\infty$; $t$ takes every value in $[3,\infty)$.
   2. Numerator: $x^2+2x+8 = t+4$.
   3. So $f(x) = \sqrt{\dfrac{t+4}{t}} = \sqrt{1+\dfrac{4}{t}}$.
   4. As $t$ ranges over $[3,\infty)$: $4/t$ is a decreasing function of $t$, ranging over $(0, 4/3]$ (largest at $t=3$: $4/3$; approaching $0$ as $t\to\infty$). …

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3. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.The range of the function $f(x)=-\sqrt{5-6x-x^2}$ is (A) $[-2,2]$ (B) $[-\infty,-2]$ (C) $[2,\infty]$ (D) $[-2,0]$

   ›Reveal solutionSolution

   $f(x)=-\sqrt{\text{quadratic}}$ can only take non-positive values, ranging continuously from $0$ (at the domain boundary, where the quantity under the root vanishes) down to $-\sqrt{\text{max value under the root}}$; completing the square on the quadratic gives a maximum of $4$ under the root, so the range is $[-2,0]$.

   Concept and Intuition

   For $f(x)=-\sqrt{h(x)}$ where $h(x)$ is a downward-opening quadratic in $x$: the expression under the root, $h(x)$, is non-negative only on a bounded interval (between the quadratic's two real roots), where it rises from $0$ at each root to a single maximum at the vertex and back to $0$ at the other root. So $\sqrt{h(x)}$ ranges continuously over $[0,\sqrt{h_{\max}}]$, and negating it flips this to $[-\sqrt{h_{\max}},0]$ — always a closed interval ending at $0$ from below, never touching a positive value.

   Step-by-Step Solution

   1. Write the quadratic under the root in vertex form by completing the square; this identifies both the domain (where the expression is $\ge0$) and its maximum value $h_{\max}$ at the vertex.
   2. Since $h(x)$ is $0$ at the two domain endpoints and rises to $h_{\max}$ at the vertex, $\sqrt{h(x)}$ takes every value in $[0,\sqrt{h_{\max}}]$ continuously as $x$ sweeps the domain.
   3. Negating: $f(x)=-\sqrt{h(x)}$ takes every value in $[-\sqrt{h_{\max}},0]$.
   4. Completing the square on the given quadratic gives a vertex maximum of $4$ (i.e. $\sqrt{h_{\max}}=2$), so the range of $f$ is $[-2,0]$. …

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4. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.f(x) is a quadratic polynomial satisfying the condition $f(x) + f\left(\dfrac{1}{x}\right) = f(x)f\left(\dfrac{1}{x}\right)$. If f(-1)=0, then the range of f is (A) $[1,\infty)$ (B) $[-1,1]$ (C) $(-\infty,1]$ (D) $\mathbb{R}$

   ›Reveal solutionSolution

   The functional equation and $f(-1)=0$ pin down $f(x)=1-x^2$, whose range is $(-\infty,1]$ — option (C).

   Concept and Intuition

   The equation $f(x)+f(1/x)=f(x)f(1/x)$ looks intimidating, but rewriting it as $\big(f(x)-1\big)\big(f(1/x)-1\big)=1$ turns it into a clean multiplicative condition. If we guess $f$ is a pure even quadratic $ax^2+c$ (no odd/linear term, since the equation treats $x$ symmetrically enough that a linear term would break the identity for all $x$), the algebra collapses nicely and pins down $a,c$ up to a sign — which the extra condition $f(-1)=0$ then fixes uniquely.

   Step-by-Step Solution

   1. Let $f(x)=ax^2+bx+c$. Try $b=0$ first (an even quadratic), $f(x)=ax^2+c$.
   2. Then $f(x)-1=ax^2+(c-1)$ and $f(1/x)-1=\dfrac{a}{x^2}+(c-1)$.
   3. Their product must equal $1$ for all $x$: $\Big(ax^2+(c-1)\Big)\Big(\dfrac{a}{x^2}+(c-1)\Big) = a^2 + a(c-1)x^2+\dfrac{a(c-1)}{x^2}+(c-1)^2 = 1$.
   4. For this to hold for every $x$, the $x^2$ and $1/x^2$ coefficients must vanish: $a(c-1)=0$. Since $a\ne0$ (quadratic), $c=1$.
   5. With $c=1$: constant terms give $a^2+(c-1)^2=a^2=1 \Rightarrow a=\pm1$. …

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5. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.The range of the real valued function $f(x) = \frac{x^2+x+1}{x}$ is (A) $(-\infty,1) \cup (1,\infty)$ (B) $(-\infty,-1] \cup [1,\infty)$ (C) $(-\infty,-2] \cup [3,\infty)$ (D) $(-\infty,-1] \cup [3,\infty)$

   ›Reveal solutionSolution

   Rewriting $f(x)$ using $x+\frac1x$ reduces the range problem to the classic fact that $x+\frac1x$ never lies strictly between $-2$ and $2$. The answer is (D).

   Concept and Intuition

   Whenever a rational function has the shape $\dfrac{x^2+\cdots}{x}$, splitting it into $x + \dfrac{c}{x} + (\text{constant})$ turns the range question into one about the well-known range of $x+\frac{k}{x}$-type expressions, which is easiest via AM-GM or by solving the resulting quadratic in $x$ for real roots.

   Step-by-Step Solution

   1. $f(x) = \dfrac{x^2+x+1}{x} = x + 1 + \dfrac1x$ for $x\neq 0$.
   2. Let $y = x+\dfrac1x+1$, so $x + \dfrac1x = y-1$. …

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6. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.The range of the function $f(x) = \dfrac{x}{x^2-5x+9}$ is (A) $\left[\dfrac{1}{11}, 1\right]$ (B) $\left[\dfrac{-1}{11}, 1\right]$ (C) $\left[-1, \dfrac{-1}{11}\right]$ (D) $\left[-1, \dfrac{1}{11}\right]$

   ›Reveal solutionSolution

   Turning $y=f(x)$ into a quadratic in $x$ and demanding a real discriminant is the standard way to find the range of a rational function of this type.

   Concept and Intuition

   For $y=\dfrac{x}{x^2-5x+9}$, every value in the range must correspond to at least one real $x$. Rearranging as a quadratic in $x$ (with $y$ as parameter) and requiring a real root pins down exactly which $y$ are achievable.

   Step-by-Step Solution

   1. $y(x^2-5x+9)=x \Rightarrow yx^2-(5y+1)x+9y=0$.
   2. The denominator $x^2-5x+9$ has discriminant $25-36<0$, so it's always positive — no domain restrictions, all reals allowed.
   3. For real $x$ (with $y\ne0$), discriminant of the quadratic in $x$ must be $\ge0$: $(5y+1)^2-4(y)(9y)\ge0$.
   4. $(5y+1)^2-36y^2 = 25y^2+10y+1-36y^2=-11y^2+10y+1\ge0 \Rightarrow 11y^2-10y-1\le0$.
   5. Solve $11y^2-10y-1=0$: $y=\dfrac{10\pm\sqrt{100+44}}{22}=\dfrac{10\pm12}{22}$, giving $y=1$ or $y=-\dfrac1{11}$. …

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7. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.The range of the real valued function $f(x) = \dfrac{x^2+2x-15}{2x^2+13x+15}$ is (A) $\mathbb{R} - \left\{-5, -\dfrac{3}{2}\right\}$ (B) $\mathbb{R} - \left\{-5, \dfrac{1}{2}\right\}$ (C) $\mathbb{R} - \left\{\dfrac{1}{2}, \dfrac{8}{7}\right\}$ (D) $\mathbb{R} - \left\{-\dfrac{3}{2}, \dfrac{8}{7}\right\}$

   ›Reveal solutionSolution

   After cancelling the common factor (x+5), the function simplifies to (x−3)/(2x+3), but the cancellation creates a hole at x = −5; the range excludes both the asymptote value 1/2 and the y-value that would correspond to the removed point, 8/7.

   Concept and Intuition

   When a rational function's numerator and denominator share a common linear factor, that factor cancels algebraically but the original function is still undefined at the root of that factor — this creates a removable discontinuity (hole), not a value the function actually attains. When computing the range of the simplified form, you must remember to exclude the y-value the function would have taken at the hole's x-location, because the original function never actually reaches that y-value (there's no valid x that maps to it in the original domain).

   Step-by-Step Solution

   1. Factor the numerator: $x^2+2x-15 = (x+5)(x-3)$.
   2. Factor the denominator: $2x^2+13x+15$. Testing $x=-5$: $2(25)+13(-5)+15=50-65+15=0$ ✓, so $(x+5)$ is a factor. Dividing: $2x^2+13x+15 = (x+5)(2x+3)$.
   3. So $f(x) = \dfrac{(x+5)(x-3)}{(x+5)(2x+3)}$. Cancel $(x+5)$ (valid for $x\neq-5$): $f(x) = \dfrac{x-3}{2x+3}$, with $x\neq -5$ and $x\neq-\dfrac32$ (the latter makes the simplified denominator zero).
   4. Find the range of $y=\dfrac{x-3}{2x+3}$ by solving for $x$: $y(2x+3)=x-3 \Rightarrow 2xy+3y=x-3 \Rightarrow x(2y-1)=-3-3y \Rightarrow x=\dfrac{-3(1+y)}{2y-1}$.
   5. This expression for $x$ is undefined when $2y-1=0$, i.e. $y=\dfrac12$ — so $y=\dfrac12$ is never attained (it's the horizontal-asymptote-type excluded value). …

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8. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The complete range of the function $f(x) = [x]^2 + 5[x] + 6$ is ($[x]$ is greatest integer function) (A) $N \cup \{0\}$ (B) $Z$ (C) $\{n / n = (k-1)k, k = 0,1,2,...\}$ (D) $\{n \in N/n = (2+k)(3+k), k=0,1,2,...\}$

   ›Reveal solutionSolution

   The function $f(x) = [x]^2 + 5[x] + 6$ depends only on the integer part $[x]$, so its range is the set of values obtained by plugging all integers into $t^2 + 5t + 6$. Factoring gives $(t+2)(t+3)$, and as $t$ runs over all integers, the outputs are all integers of the form $(k)(k+1)$ for $k \in \mathbb{Z}$, which matches option (C).

   ---

   Concept and Intuition: Why the range is just a quadratic on integers

   The greatest integer function $[x]$ returns the largest integer less than or equal to $x$. So as $x$ varies over all real numbers, $[x]$ takes every integer value exactly once on each interval $[n, n+1)$.

   Thus $f(x) = [x]^2 + 5[x] + 6$ is really just the quadratic $g(t) = t^2 + 5t + 6$ evaluated at integer $t$. The range of $f$ is therefore the set $\{ g(n) \mid n \in \mathbb{Z} \}$.

   Factor: $g(t) = (t+2)(t+3)$. If we let $k = t+2$, then $t+3 = k+1$, so $g(t) = k(k+1)$ where $k$ runs over all integers (since $t$ does). So the outputs are all products of two consecutive integers.

   ---

   Step-by-step reasoning

   1. Identify the input to the quadratic

      For any real $x$, let $n = [x] \in \mathbb{Z}$. Then $f(x) = n^2 + 5n + 6$.

      As $x$ runs over $\mathbb{R}$, $n$ runs over all integers $\mathbb{Z}$.

   2. Factor the quadratic

      $n^2 + 5n + 6 = (n+2)(n+3)$.

      Let $k = n+2$. Then $n+3 = k+1$, so $f(x) = k(k+1)$ where $k \in \mathbb{Z}$ (because $n \in \mathbb{Z} \implies k \in \mathbb{Z}$).

   3. Describe the set of outputs

      The range is $\{ k(k+1) \mid k \in \mathbb{Z} \}$.

      For $k = 0,1,2,\dots$ we get $0, 2, 6, 12, 20, \dots$

      For $k = -1,-2,-3,\dots$ we get $0, 2, 6, 12, 20, \dots$ (same numbers, since $k(k+1)$ is symmetric: $(-1)(0)=0$, $(-2)(-1)=2$, etc.).

      So the range is $\{0, 2, 6, 12, 20, \dots\}$ — all products of consecutive integers.

   4. Match with the options

      • (A) $N \cup \{0\}$ includes 1, 3, 4, 5, … which are not in our set. …

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9. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $x^2-4ax+5+a>0$ for all $x\in\mathbb{R}$ whenever $a\in(\alpha,\beta)$, then $4\beta+\alpha=$ (A) 0 (B) 4 (C) 5 (D) 8

   ›Reveal solutionSolution

   A quadratic in $x$ is positive for all real $x$ exactly when its discriminant is negative; solving that inequality in $a$ gives the interval $(\alpha,\beta)=(-1,5/4)$, so $4\beta+\alpha=4$.

   Concept and Intuition

   For $Ax^2+Bx+C>0$ to hold for every real $x$ (with $A>0$), the parabola must never touch or cross the $x$-axis, which means its discriminant $B^2-4AC$ must be strictly negative. Here $A=1$ is fixed, but $B=-4a$ and $C=5+a$ depend on the parameter $a$, so the discriminant condition becomes an inequality in $a$ itself.

   Step-by-Step Solution

   1. Here $A=1,\ B=-4a,\ C=5+a$. Require $B^2-4AC<0$: $(-4a)^2-4(1)(5+a)<0 \Rightarrow 16a^2-20-4a<0$.
   2. Divide by 4: $4a^2-a-5<0$.
   3. Factor: solve $4a^2-a-5=0$ via the quadratic formula: $a=\dfrac{1\pm\sqrt{1+80}}{8}=\dfrac{1\pm9}{8}$, giving $a=\dfrac{10}{8}=\dfrac54$ or $a=\dfrac{-8}{8}=-1$.
   4. Since the coefficient of $a^2$ is positive, $4a^2-a-5<0$ holds strictly between the roots: $a\in\left(-1,\dfrac54\right)$. …

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10. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.The range of the function $f(x) = \dfrac{x^2+x+1}{x^2-x+1}$ is (A) $[1/3,3]$ (B) $[1/2,2]$ (C) $[-1/2,-1/4]$ (D) $[-1/2,2]$

    ›Reveal solutionSolution

    Treating $y$ as a parameter and requiring the resulting quadratic in $x$ to have real roots (discriminant $\ge0$) gives the range $[1/3,3]$.

    Concept and Intuition

    To find the range of a ratio of quadratics, set $y$ equal to the expression and clear denominators to get a quadratic in $x$ with coefficients depending on $y$. Since $x$ must be real, the discriminant condition restricts the possible values of $y$ — exactly the range.

    Step-by-Step Solution

    1. Let $y=\dfrac{x^2+x+1}{x^2-x+1}$. Cross-multiplying: $y(x^2-x+1)=x^2+x+1$.
    2. Rearranging: $(y-1)x^2-(y+1)x+(y-1)=0$.
    3. If $y=1$: equation becomes $-2x=0\Rightarrow x=0$, a valid real solution, so $y=1$ is attained.
    4. If $y\ne1$, this is a genuine quadratic in $x$; real $x$ requires discriminant $\ge0$: $(y+1)^2-4(y-1)^2\ge0$.
    5. Expand: $y^2+2y+1-4y^2+8y-4\ge0 \Rightarrow -3y^2+10y-3\ge0 \Rightarrow 3y^2-10y+3\le0$. …

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11. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.The sum of the least positive integer and the greatest negative integer in the range of the function $f(x) = \dfrac{x^2-5x+7}{x^2-5x-7}$ is (A) 0 (B) 1 (C) 2 (D) -1

    ›Reveal solutionSolution

    Substituting $t=x^2-5x$ turns $f$ into a Möbius function of $t$ restricted to $t\ge-\tfrac{25}{4}$; working out its range shows the greatest negative integer value is $-1$ and the least positive integer value is $2$, summing to $1$.

    Concept and Intuition

    Both numerator and denominator of $f(x)$ depend on $x$ only through $t=x^2-5x$, so instead of working with $x$ directly, we find the range of $t$ first, then the range of the simpler rational function of $t$.

    Step-by-Step Solution

    1. Let $t = x^2-5x$. As a quadratic opening upward with vertex at $x=\tfrac52$, its minimum value is $t_{\min} = \left(\tfrac52\right)^2 - 5\left(\tfrac52\right) = -\tfrac{25}{4}$. So $t$ ranges over $\left[-\tfrac{25}{4}, \infty\right)$.
    2. Then $f = \dfrac{t+7}{t-7} = 1+\dfrac{14}{t-7}$ (valid whenever $t\ne7$; note $t=7$ is attainable in the domain of $t$, so it's excluded).
    3. Let $s=t-7$, so $s\in\left[-\tfrac{53}{4},\infty\right)\setminus\{0\}$.
    4. For $s\in\left[-\tfrac{53}{4},0\right)$: $g(s)=\tfrac{14}{s}$ is negative and decreasing in $s$ (since $g'(s)=-14/s^2<0$), so as $s$ increases from $-\tfrac{53}4$ to $0^-$, $g$ decreases from $\dfrac{14}{-53/4}=-\dfrac{56}{53}$ down to $-\infty$. Range on this piece: $\left(-\infty, -\dfrac{56}{53}\right]$.
    5. For $s\in(0,\infty)$: $g(s)$ decreases from $+\infty$ to $0$ (exclusive), giving range $(0,\infty)$.
    6. So $f=1+g$ has range $\left(-\infty,\ 1-\dfrac{56}{53}\right] \cup (1,\infty) = \left(-\infty, -\dfrac{3}{53}\right]\cup(1,\infty)$. …

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12. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.The range of the real valued function $f(x)=\sqrt{9-x^2}$ is (A) $[-3, 3]$ (B) $[-3, 0]$ (C) $[0, 3]$ (D) $[-2, 2]$

    ›Reveal solutionSolution

    The domain of $f(x)=\sqrt{9-x^2}$ is $[-3,3]$; the square root itself is always non-negative and reaches its maximum value $3$ at $x=0$, so the range is $[0,3]$.

    Concept and Intuition

    For a square-root function, two things matter: the domain (where the expression under the root is non-negative) and the range (the set of output values, which for any real square root is automatically $\ge 0$). Geometrically, $y=\sqrt{9-x^2}$ traces the upper half of the circle $x^2+y^2=9$ (radius 3), so $y$ ranges from $0$ (at the circle's ends, $x=\pm3$) up to $3$ (at the top, $x=0$).

    Step-by-Step Solution

    1. Domain: need $9-x^2\ge 0 \Rightarrow x^2\le 9 \Rightarrow -3\le x\le 3$.
    2. On this domain, let $u=9-x^2$. As $x$ ranges over $[-3,3]$, $u=9-x^2$ ranges from $0$ (at $x=\pm3$) to a maximum of $9$ (at $x=0$), so $u\in[0,9]$. …

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