JEE MATHEMATICS MCQ: INCREASING & DECREASING

1 Easy

1 of 1

Analyzing +4 | -1

The function

f(x)=\tan x-x

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Correct option: A

always increases

Option A (always increases) is correct because the derivative of the function, f'(x) = sec²(x) - 1, is always non-negative, indicating that the function is always increasing.

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If the function

f\left( x \right) = \frac{{K\sin x + 2\cos x}}{{\sin x + \cos x}}

is increasing for all values of

x

, then The set of values of a for which the function

f\left( x \right) = {x^2} + ax + 1

is an increasing function on [1,2] is The value of

a

in order that

f(x)=

\sin x-\cos x-a x+b

decreases for all real values is given by The function

f(x)=x e^{x(1-x)}, x \in R

, is If

f

and

g

are two increasing functions such that

fog

is defined, then If

f\left( x \right) = {x^3} + b{x^2} + cx + d

and

0 < {b^2} < c

, then in

\left( { - \infty ,\;\infty } \right)