Q.Using the second derivative test, find the local extreme values of $f(x) = 2x^3 - 15x^2 + 36x + 10$.

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).