Show that the maximum value of the function x³+1/x³ is less than its minimum value.

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Q. Show that the maximum value of the function x³+1/x³ is less than its minimum value.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2025Subjective· 5mImportance★★★★★

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$f(x)=x^3+1/x^3$ has separate local extrema for $x>0$ and $x<0$ ; find them via the first/second derivative test.

Let $f(x)=x^3+x^{-3}$ (domain $x\ne0$ ). Differentiate:

$f'(x) = 3x^2-3x^{-4} = \frac{3(x^6-1)}{x^4}$

Setting $f'(x)=0$ : $x^6=1 \Rightarrow x=\pm1$ (the only real critical points).

Second derivative: $f''(x)=6x+12x^{-5}$ .

At $x=1$ : $f''(1)=6+12=18>0$ — a local minimum. Value: $f(1)=1+1=2$ .

At $x=-1$ : $f''(-1)=-6+12(-1)=-18<0$ — a local maximum. Value: $f(-1)=-1+(-1)=-2$ .

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

CBSE 2026Set SEM31 markMCQ
$y = \dfrac{\log_e x}{x}$
(a) $e$
(b) $e^2$
(c) $\dfrac{1}{e}$
(d) $\dfrac{1}{e^2}$
›Reveal solutionSolution

Differentiate $y=\dfrac{\log_e x}{x}$ , set $y'=0$ to get $x=e$ , and evaluate; the maximum value is $\dfrac1e$ .

Maximising a function using the first derivative is a standard NCERT/CBSE Class 12 application of derivatives problem.

Differentiate with the quotient rule:

$y' = \frac{\tfrac{1}{x}\cdot x - \log_e x \cdot 1}{x^2} = \frac{1-\log_e x}{x^2}.$

…

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CBSE 2026Set SEM31 markMCQ
$f(x) = 3 + |x - 3|$
(a) Statement-I is true, Statement-II is false
(b) Statement-I is false, Statement-II is true
(c) Statements-I and II both are true
(d) Statements-I and II both are false
›Reveal solutionSolution

Both statements are correct, so the answer is (c).

These are qualitative NCERT/CBSE Class 12 maxima–minima facts.

Statement-I: $|x-3| \ge 0$ for all $x$ , with equality only at $x=3$ . Hence $f(x)=3+|x-3| \ge 3$ , and the least value $3$ is achieved at $x=3$ . So it has a local (indeed global) minimum value $3$ . True.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Maximum value of 5 - (x-1)² is$
(a) 5
(b) 4
(c) 6
(d) 3
›Reveal solutionSolution

$(x-1)^2 \ge 0$ always, so $5-(x-1)^2$ is largest exactly when the squared term is $0$ .

For any real $x$ , $(x-1)^2 \ge 0$ , with equality exactly when $x=1$ . So $5-(x-1)^2 \le 5$ for all $x$ , and this maximum value $5$ is attained at $x=1$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. The value of the function f(x) = 4x - x² - 3 will be maximum when:$
(a) x = 3
(b) x = 2
(c) x = -2
(d) x = -3
›Reveal solutionSolution

Set the derivative to zero to find the critical point, then confirm it's a maximum with the second derivative.

Given $f(x)=4x-x^2-3$ .

First derivative:

$f'(x) = 4-2x$

Set $f'(x)=0$ :

$4-2x=0 \Rightarrow x=2$

Second derivative test: …

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Q.Show that the maximum value of the function x³+1/x³ is less than its minimum value.

Concept understanding — Maxima and Minima -- First Derivative Test

Q.Show that the maximum value of the function x³+1/x³ is less than its minimum value.

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