Q.Divide the number 30 into two parts such that their product is maximum.

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 an …

   🔒 Start your 14-day free trial to see the full solution →
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: $\dfrac{d(D^2)}{dx}=4x-12=0\Rightarrow x=3$. …

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

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

   …

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

   🔒 Start your 14-day free trial to see the full solution →
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): $80-32t=0$. …

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

   🔒 Start your 14-day free trial to see the full solution →
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 $1-x=0\Rightarrow x=1$. …

   🔒 Start your 14-day free trial to see the full solution →
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 $(\pm a\cos\theta,\pm b\sin\theta)$. …

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

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

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

    🔒 Start your 14-day free trial to see the full solution →
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 $[0,3]$: $f(0)=0-0+5=5$; $f(3)=9-12+5=2$. …

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