Let $f\left( x \right) = \mathop \smallint \limits_0^x \frac{{\cos t}}{t}dt$ Then, at $x = \left( {2n + 1} \right)\frac{\pi }{2},\;f\left( x \right)$ has

Maxima when $n = - 2,\; - 4,\; - 6$ … and minima when $n = - 1,\; - 3,\; - 5, \ldots $

Minima when $n = 0,\;2,\;4, \ldots $ and maxima when $n = 1,\;3,\;5, \ldots $

JEE MATHEMATICS MCQ: INCREASING & DECREASING

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Analyzing +4 | -1

Let

f\left( x \right) = \int \limits_0^x \frac{{\cos t}}{t}dt

Then, at

x = \left( {2n + 1} \right)\frac{\pi }{2},\;f\left( x \right)

has

View answer and explanation

Correct option: B

Maxima when

n = - 1,\; - 3,\; - 5

...and minima when

n = 1,\;3,\;5, \ldots

The correct answer is Maxima when

n = - 1, - 3, - 5

...and minima when

n = 1, 3, 5, ...

because the sign of the second derivative determines whether the function has a maxima or minima at the given points.