Given that f-x - -n - 1101 n - xn- Number of relative maximum of fx is

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Fractional Exponent

Given that ...

Question

Given that $f^{'} (x) = Π_{n = 1}^{101} (n - x)^{n}$ . Number of relative maximum of $f (x)$ is

25

No worries! We‘ve got your back. Try BYJU‘S free classes today!

26

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

50

No worries! We‘ve got your back. Try BYJU‘S free classes today!

51

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $26$

Suggest Corrections

Similar questions

If $F (x) = {(f (\frac{x}{2}))}^{2} + {(g (\frac{x}{2}))}^{2}$ where $f^{'} (x) = - f (x)$ and $g (x) = f^{'} (x)$ and given that $F (5) = 5$ , then $F (10)$ is equal to

Q. The heights of trees in a forest are given as follows. Draw a histogram to represent the data.

Heights in metre	$16 - 20$	$21 - 25$	$26 - 30$	$31 - 35$	$36 - 40$	$41 - 45$	$46 - 50$	$51 - 55$
Number of trees	$10$	$15$	$25$	$30$	$30$	$50$	$35$	$20$

Q. The total number of terms in the expansion of (x + a)⁵¹ – (x – a)⁵¹ after simplification is
(a) 102
(b) 25
(c) 26
(d) none of these

Q. If

f^{''} (x) = - f (x)

g (x) = f^{'} (x)

F (x) = {(f (\frac{x}{2}))}^{2} + {(g (\frac{x}{2}))}^{2}

and given that

F (5) = 5

, then

F (10)

Q. Assertion :Let

f (x) = x^{n}

f^{'} (x) = r!^{n} C_{r} x^{n - r}

denotes the

r^{t h}

order derivative of

f (x)

then

f (1) + \frac{f^{'} (1)}{1!} + \frac{f^{''} (1)}{2!} + . . . . . + \frac{f^{n} (1)}{n!} = 2^{n}

Reason: The sum of binomial coefficients in the expansion of

{(1 + x)}^{n}

2^{n}

Join BYJU'S Learning Program

Given that f′(x)=Π101n=1(n−x)n. Number of relative maximum of f(x) is

The correct option is B 26

Given that $f^{'} (x) = Π_{n = 1}^{101} (n - x)^{n}$ . Number of relative maximum of $f (x)$ is

The correct option is B $26$