JEE MATHEMATICS MCQ: MAXIMA & MINIMA

1 Easy

1 of 1

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\left| {{x^3} + {x^2} + 3x + \sin x} \right|\left( {3 + \sin \frac{1}{x}} \right),\;x \ne 0}\\ {0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x = 0} \end{array}} \right.

then number of points (where

f\left( x \right)

attains its minimum value) is

View answer and explanation

Correct option: A

Option A (1) is correct because the function attains its minimum value of 0 only at x=0, due to the absolute value and the term (3 + sin(1/x)).

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