JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Hard

1 of 1

Analyzing +4 | -1

Suppose

f(x)

is twice differentiable for all real Further suppose

|f(x)| \leq 1

and

\left|f^{\prime \prime}(x)\right| \leq 1

for all rea

x

. Then

\left|f^{\prime}(x)\right|

is
where

f(x+1) =f(x)+\frac{f^{\prime}(x)}{\lfloor}+\frac{f^{\prime}(\xi)}{\lfloor 2}

\text { and } \quad f(x-1) =f(x)-\frac{f^{\prime}(x)}{\lfloor}+\frac{f^{\prime \prime}(\eta)}{\lfloor 2}

for some

\xi

and

\eta

View answer and explanation

Correct option: C

less than or equal to

3 / 2

The correct answer is less than or equal to

3/2

because by using Taylor's expansion and the given conditions, we can bound the value of the first derivative.

More Questions from COMPLEX NUMBERS

Practice All Chapter MCQs →

{\left| {1 - i} \right|^x} = {2^x}

|z|=1

z \neq \pm 1

z, \frac{z}{1-z^2}

\left|z_r-r\right| \leq r, \forall r=1,2,3, \ldots, n

\left|\sum\limits_{r=1}^n z_r\right|

(z+a b)^3=a^3

a \neq 0

\alpha, \beta

\gamma

|\alpha-\beta|

{z^2} + {\left| z \right|^2} = 0,

z \in C

2x = - 1 + \sqrt 3 i

{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6}