In the interval [0, 1], the function ${x^2} - x + 1$ is

JEE MATHEMATICS MCQ: INCREASING & DECREASING

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Analyzing +4 | -1

In the interval [0, 1], the function

{x^2} - x + 1

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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].

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

x

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

a

f(x)=

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

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

f

g

fog

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

0 < {b^2} < c

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