The function f (x )=x^2+ 54x

Correct answer: B. (b) f (x )=x^2+ 54x f^ ' (x )=2x- 54x^2<0 for x ∈ (- ,0 ) ∴f (x ) is decrasing in (- ,0 ) and for critical points: f^ ' (x )=0 ⇒2x- 54x^2=0 ⇒x^3=27 ⇒x=3 So there is no critical point in (- ,0 ) ie f (x ) has neither maximum nor minimum in (- ,0 ).

JEE Previous Year Question

$The function f (x) = x^{2} + \frac{54}{x}$

Topic: JEE PYQ

Answer: B. Full step-by-step solution below.

A $is increasing and has minimum value 27 in the interval (0, \infty)$
B $is decreasing and has neither maximum nor minimum in the interval (- \infty, 0)$
C $has maximum value 27 in the interval (- \infty, \infty)$
D $is increasing and has neither maximum nor minimum values in the interval (- \infty, \infty)$

Correct answer: B

Solution

(b) f (x) = x^{2} + \frac{54}{x} f^{’} (x) = 2 x - \frac{54}{x ^{2}} < 0 for x \in (- \infty, 0) ∴ f (x) is decrasing in (- \infty, 0) and for critical points: f^{’} (x) = 0 \Rightarrow 2 x - \frac{54}{x ^{2}} = 0 \Rightarrow x^{3} = 27 \Rightarrow x = 3 So there is no critical point in (- \infty, 0) ie f (x) has neither maximum nor minimum in (- \infty, 0) .

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