Q.Find the absolute extrema of the following functions on the given closed interval.

Showing the 12 most recent of 14 on this concept.

1. CBSE 2026Set V11 markMCQ
   Q.The minimum value of $f(x) = x$, $x \in R$

   (a) 0

   (b) 1

   (c) 2

   (d) does not exist

   ›Reveal solutionSolution

   On all of $\mathbb{R}$ the identity function is unbounded below, so no minimum value exists; answer (d).

   For $f(x)=x$ with $x\in\mathbb{R}$, as $x\to-\infty$ we have $f(x)\to-\infty$. The function takes arbitrarily small (large-negative) values, so it is not bounded below and attains no least value.

   $\inf_{x\in\mathbb{R}} x=-\infty\ (\text{not attained}).$

   ✓Final answer

   (d) does not exist

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.One of the closest points on the curve $x^2-y^2=4$ to the point $(6, 0)$ is :

   (a) $(3, \sqrt5)$

   (b) $(2, 0)$

   (c) $(\sqrt{13}, -\sqrt3)$

   (d) $(\sqrt5, 1)$

   ›Reveal solutionSolution

   Substitutes the curve's constraint into the squared-distance function and minimizes over $x$ using calculus.

   1. A point on the curve $x^2-y^2=4$ satisfies $y^2=x^2-4$ (needs $x^2\ge4$).
   2. Squared distance from $(x,y)$ to $(6,0)$: $D^2=(x-6)^2+y^2=(x-6)^2+(x^2-4)$.
   3. Expand: $D^2=x^2-12x+36+x^2-4=2x^2-12x+32$.
   4. Differentiate w.r.t. $x$ and set to zero: $\dfrac{d(D^2)}{dx}=4x-12=0\Rightarrow x=3$.
   5. Second derivative $=4>0$, confirming a minimum. At $x=3$: $y^2=9-4=5\Rightarrow y=\pm\sqrt5$.
   6. So the closest points are $(3,\sqrt5)$ and $(3,-\sqrt5)$; among the options, $(3,\sqrt5)$ matches (distance$^2=9+5=14$, less than e.g. $(2,0)$'s $16$).

   ✓Final answer

   (a) $(3, \sqrt5)$

3. CBSE 2026Set ANNUAL1 mark
   Q.If $f(x) = x \cdot \log x$ then its minimum value is ______.

   ›Reveal solutionSolution

   $f'(x)=\log x+1=0$ gives $x=\tfrac{1}{e}$; $f''>0$ confirms a minimum, and $f\!\left(\tfrac{1}{e}\right)=-\tfrac{1}{e}$.

   Differentiate $f(x) = x\log x$ using the product rule:

   $f'(x) = 1\cdot\log x + x\cdot\frac{1}{x} = \log x + 1.$

   Set $f'(x) = 0$ for the critical point:

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

   Check the nature with the second derivative:

   $f''(x) = \frac{1}{x},\qquad f''\!\left(\tfrac{1}{e}\right) = e > 0,$

   so $x = \tfrac{1}{e}$ gives a minimum. Evaluate $f$ there:

   $f\!\left(\frac{1}{e}\right) = \frac{1}{e}\log\frac{1}{e} = \frac{1}{e}\cdot(-1) = -\frac{1}{e}.$

   ✓Final answer

   The minimum value is $-\dfrac{1}{e}$ (at $x = \tfrac{1}{e}$).

4. CBSE 2025Set X11 markMCQ
   Q.The absolute maximum value of the function $f$ given by $f(x) = x^3$, $x \in [-2, 2]$ is

   (a) $2$

   (b) $0$

   (c) $-2$

   (d) $8$

   ›Reveal solutionSolution

   Absolute extremum of a monotonic function on a closed interval — correct option (d).

   Since $f'(x) = 3x^2 \ge 0$, the function $f(x)=x^3$ is increasing on $[-2,2]$, so its absolute maximum occurs at the right endpoint. Evaluating, $f(2) = 2^3 = 8$.

   ✓Final answer

   (d) $8$

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.A stone is thrown up vertically. The height it reaches at time t seconds is given by $x=80t-16t^2$. The stone reaches the maximum height in time t seconds is given by :

   (a) $3$

   (b) $2$

   (c) $3.5$

   (d) $2.5$

   ›Reveal solutionSolution

   The stone reaches maximum height when its vertical velocity (the derivative of position) is zero; solving that gives $t=2.5$.

   1. Height: $x(t)=80t-16t^2$.
   2. Velocity: $v(t)=\dfrac{dx}{dt}=80-32t$.
   3. At maximum height, $v=0$ (the stone momentarily stops before falling back): $80-32t=0$.
   4. Solve: $32t=80\Rightarrow t=\dfrac{80}{32}=2.5$.
   5. Check: $\dfrac{d^2x}{dt^2}=-32<0$, confirming a maximum, not a minimum.

   ✓Final answer

   (d) $2.5$

6. CBSE 2024Set A11 markMCQ
   Q.The maximum value of the function $f(x) = x$, $x \in (1, 2)$ is

   (a) 1

   (b) do not have maximum value

   (c) 3

   (d) 2

   ›Reveal solutionSolution

   $f(x)=x$ is increasing on the open interval $(1,2)$ and its endpoint value is not attained, so there is no maximum — (b).

   The function is strictly increasing, so it approaches $2$ as $x\to 2^-$, but $x=2$ is excluded from $(1,2)$. Thus the supremum $2$ is never reached and $f$ has no maximum value on this open interval.

   ✓Final answer

   (b) do not have maximum value

7. CBSE 2023Set ANNUAL1 markMCQ
   Q.The maximum value of the function $x^2e^{-2x}$, $x>0$ is :

   (a) $\dfrac{1}{e^2}$

   (b) $\dfrac{1}{e}$

   (c) $\dfrac{4}{e^4}$

   (d) $\dfrac{1}{2e}$

   ›Reveal solutionSolution

   Setting the derivative of $x^2e^{-2x}$ to zero locates the critical point $x=1$, which gives the maximum value $1/e^2$.

   1. $f(x)=x^2e^{-2x}$. By the product rule, $f'(x)=2xe^{-2x}+x^2(-2e^{-2x})=2xe^{-2x}(1-x)$.
   2. Setting $f'(x)=0$ for $x>0$: since $2xe^{-2x}\ne0$ when $x>0$, we need $1-x=0\Rightarrow x=1$.
   3. Checking it is a maximum: $f'(x)>0$ for $0<x<1$ (function increasing) and $f'(x)<0$ for $x>1$ (function decreasing), so $x=1$ gives a maximum.
   4. Maximum value: $f(1)=1^2\cdot e^{-2(1)}=e^{-2}=\dfrac{1}{e^2}$.

   ✓Final answer

   (a) $\dfrac{1}{e^2}$

8. CBSE 2023Set ANNUAL1 markMCQ
   Q.Area of the greatest rectangle inscribed in the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ is :

   (a) $\sqrt{ab}$

   (b) $2ab$

   (c) $\dfrac{a}{b}$

   (d) $ab$

   ›Reveal solutionSolution

   Parametrising the inscribed rectangle's corner on the ellipse and maximising the resulting area $2ab\sin2\theta$ gives the greatest area $2ab$.

   1. Let one vertex of the inscribed rectangle be $(a\cos\theta,\,b\sin\theta)$ on the ellipse, $0<\theta<\pi/2$. By symmetry the rectangle has vertices $(\pm a\cos\theta,\pm b\sin\theta)$.
   2. Its sides have lengths $2a\cos\theta$ and $2b\sin\theta$, so its area is $A(\theta)=(2a\cos\theta)(2b\sin\theta)=4ab\sin\theta\cos\theta=2ab\sin2\theta$.
   3. Since $\sin2\theta\le1$ always, and equals $1$ when $2\theta=\pi/2$, i.e. $\theta=\pi/4$, the maximum area is $A_{max}=2ab(1)=2ab$.

   ✓Final answer

   (b) $2ab$

9. CBSE 2022Set ANNUAL1 markMCQ
   Q.The minimum value of the function $|3-x|+9$ is :

   (a) $6$

   (b) $0$

   (c) $9$

   (d) $3$

   ›Reveal solutionSolution

   The absolute value $|3-x|$ has minimum $0$ (at $x=3$), so $|3-x|+9$ has minimum value $9$.

   1. Let $g(x)=|3-x|+9$.
   2. For any real $x$, $|3-x|\ge 0$, with the least possible value $0$ occurring exactly when $3-x=0$, i.e. $x=3$.
   3. Since $9$ is a constant added to $|3-x|$, $g(x)$ is minimized exactly when $|3-x|$ is minimized.
   4. At $x=3$: $g(3)=|3-3|+9=0+9=9$.
   5. For any other $x$, $|3-x|>0$, so $g(x)>9$.

   ✓Final answer

   The minimum value of $|3-x|+9$ is $9$ — option (c).

10. CBSE 2020Set ANNUAL1 markMCQ
    Q.The least possible perimeter (in meter) of a rectangle of area 100 m$^2$ is :

    (a) $50$

    (b) $10$

    (c) $20$

    (d) $40$

    ›Reveal solutionSolution

    For fixed area, the perimeter of a rectangle is minimized when it is a square; with area $100$ the side is $10$ and the least perimeter is $40$.

    1. Let the sides of the rectangle be $x$ and $y$, with area $xy=100$, so $y=\dfrac{100}{x}$.
    2. The perimeter is $P(x)=2(x+y)=2\left(x+\dfrac{100}{x}\right)$, for $x>0$.
    3. To minimize, differentiate: $P'(x)=2\left(1-\dfrac{100}{x^2}\right)$.
    4. Set $P'(x)=0$: $1-\dfrac{100}{x^2}=0\Rightarrow x^2=100\Rightarrow x=10$ (taking the positive root, since $x$ is a length).
    5. Check it is a minimum: $P''(x)=\dfrac{400}{x^3}>0$ for $x>0$, confirming $x=10$ gives a minimum.
    6. At $x=10$: $y=\dfrac{100}{10}=10$, so the rectangle is actually a $10\times10$ square.
    7. The least perimeter is $P(10)=2(10+10)=40$ metres.

    ✓Final answer

    The least possible perimeter is $40$ m — option (d).

11. CBSE 2018Set ANNUAL1 markMCQ
    Q.The statement : “If $f$ has a local extremum (minimum or maximum) at $c$ and if $f'(c)$ exists then $f'(c) = 0$” is :

    (a) Law of mean

    (b) The extreme value theorem

    (c) Rolle's theorem

    (d) Fermat's theorem

    ›Reveal solutionSolution

    The stated result — a differentiable local extremum has derivative zero — is the standard statement of Fermat's theorem.

    1. Fermat's theorem (on stationary points) states: if $f$ has a local maximum or local minimum at an interior point $c$ of its domain, and if $f'(c)$ exists, then $f'(c)=0$.
    2. This is exactly the statement given in the question.
    3. Rolle's theorem is a related but different result: if $f(a)=f(b)$ and $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, then there exists some $c\in(a,b)$ with $f'(c)=0$ — it is about the existence of such a point between equal endpoint values, not a general statement about extrema.
    4. The Law of the Mean (Mean Value Theorem) relates $f'(c)$ to the average rate of change $\dfrac{f(b)-f(a)}{b-a}$, which is a different statement.
    5. The Extreme Value Theorem is about existence of a maximum/minimum on a closed interval, not about the derivative vanishing there.
    6. So the correct name for the quoted statement is Fermat's theorem.

    ✓Final answer

    The statement is Fermat's theorem — option (d).

12. CBSE 2017Set ANNUAL1 markMCQ
    Q.If $f(x) = x^2 - 4x + 5$ on $[0, 3]$ then the absolute maximum value is :

    (a) 2

    (b) 3

    (c) 4

    (d) 5

    ›Reveal solutionSolution

    On $[0,3]$, $f(x)=x^2-4x+5$ has its minimum at the interior critical point $x=2$ and its absolute maximum at the endpoint $x=0$, giving value $5$.

    1. $f'(x)=2x-4=0\Rightarrow x=2$; since $f''(x)=2>0$, $x=2$ is a local minimum, with $f(2)=4-8+5=1$.
    2. Evaluate at the endpoints of $[0,3]$: $f(0)=0-0+5=5$; $f(3)=9-12+5=2$.
    3. Compare all candidate values: $f(0)=5,\ f(2)=1,\ f(3)=2$. The largest of these is $5$, attained at $x=0$.
    4. So the absolute maximum value of $f$ on $[0,3]$ is $5$.

    ✓Final answer

    The absolute maximum value is $5$ — option (d).