The function $f\left( x \right) = {x^{1/x}}$ is

Increasing in $\left( {1,\;\infty } \right)$

Decreasing in $\left( {1,\;\infty } \right)$

Decreasing in $\left( {1,\;e} \right)$ and increasing in $\left( {e,\;\infty } \right)$

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JEE • MATHEMATICS • INCREASING & DECREASING

Increasing & decreasing practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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1 Hard

1 of 7

Application +4 | -1

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

View answer and explanation

Correct option: A

\left( { - 2,\;\infty } \right)

Option A ((-2, ∞)) is correct because the function is increasing when its derivative is greater than 0 on the given interval.

2 Hard

2 of 7

Analyzing +4 | -1

If the function

f\left( x \right) = \left( {2a - 3} \right)\left( {x + 2\sin 3} \right) + \left( {a - 1} \right)\left( {{{\sin }^4}x + {{\cos }^4}x} \right) + \log 2

does not possess critical points, then

View answer and explanation

Correct option: A

a \in \left( { - \infty ,\;4/3} \right) \cup \left( {2,\;\infty } \right)

Option A (

a \in (-\infty, 4/3) \cup (2, \infty)

) is correct because for the function to have no critical points, the derivative must not equal zero, leading to the given range for 'a'.

3 Hard

3 of 7

Analyzing +4 | -1

The function

f\left( x \right) = \frac{{\left| {x - 1} \right|}}{{{x^2}}}

is monotonically decreasing on

View answer and explanation

Correct option: C

\left( { - \infty ,1} \right) \cup \left( {2,\infty } \right)

Option C (

\left( { - \infty ,1} \right) \cup \left( {2,\infty } \right)

) is correct because the function's derivative is negative in this interval, indicating monotonic decrease.

4 Hard

4 of 7

Analyzing +4 | -1

The function

f\left( x \right) = {x^{1/x}}

View answer and explanation

Correct option: C

Increasing in

\left( {1,\;e} \right)

and decreasing in

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

Option C (Increasing in (1, e) and decreasing in (e,

\infty

)) is correct because the derivative of the function is positive in (1, e) and negative in (e,

\infty

5 Hard

5 of 7

Understanding +4 | -1

Let

f\left( x \right) = {\cot ^{ - 1}}\left\{ {g\left( x \right)} \right\},

where

g\left( x \right)

is an increasing function on the interval

\left( {0,\pi } \right).

Then,

f\left( x \right)

View answer and explanation

Correct option: B

Decreasing on

\left( {0,\pi } \right)

Option B (Decreasing on (0,

\pi

)) is correct because the derivative of f(x) is negative on the interval (0,

\pi

), indicating that the function is decreasing.

6 Hard

6 of 7

Analyzing +4 | -1

In the interval [0, 1], the function

{x^2} - x + 1

View answer and explanation

Correct option: C

Neither increasing nor decreasing

Option C (Neither increasing nor decreasing) is correct because the derivative of the function changes sign within the interval [0, 1].

7 Hard

7 of 7

Analyzing +4 | -1

The function

f\left( x \right) = x\sqrt {ax - {x^2}} ,a > 0

View answer and explanation

Correct option: A

Increases on the interval

\left( {0,\;3a/4} \right)

Option A (Increases on the interval

\left( {0,\;3a/4} \right)

) is correct because the derivative of the function is positive in that interval, indicating an increasing function.

Revision notes and key takeaways

Review INCREASING & DECREASING in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 7 single-correct-answer MCQs on INCREASING & DECREASING.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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