If n is a positive integer for the function fx-5xx-2n-x-0-2- By Rolle-s theorem value of c is 15- then n is equal to

F^'(c)=0 and nbsp; c=15 ⇒ 5 x · n(x 2)^n 1+5(x 2)^n=0 . ⇒ 5 ·15· n(15 2)^n 1+5(15 2)^n=0 ⇒ n(15 2)^n 1+5(15 2)^n=0 ⇒ n(15 2)^ 1+5=0 ⇒ n( 59)+5=0 ⇒5 n9=5 ⇒ ...

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

Byju's Answer

Standard XII

Mathematics

Integration as Antiderivative

If n is a pos...

Question

If $n$ is a positive integer for the function $f (x) = 5 x (x - 2)^{n}, x \in [0, 2] .$ By Rolle's theorem value of $c$ is $\frac{1}{5},$ then $n$ is equal to

3

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

6

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

9

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

10

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

Open in App

Solution

The correct option is C $9$
$f^{'} (c) = 0$ and $c = \frac{1}{5}$
$\Rightarrow 5 x \cdot n (x - 2)^{n - 1} + 5 (x - 2)^{n} = 0 .$

$\Rightarrow 5 \cdot \frac{1}{5} \cdot n {(\frac{1}{5} - 2)}^{n - 1} + 5 {(\frac{1}{5} - 2)}^{n} = 0$

$\Rightarrow n {(\frac{1}{5} - 2)}^{n - 1} + 5 {(\frac{1}{5} - 2)}^{n} = 0$

$\Rightarrow n {(\frac{1}{5} - 2)}^{- 1} + 5 = 0$

$\Rightarrow n (\frac{- 5}{9}) + 5 = 0$

$\Rightarrow \frac{5 n}{9} = 5 \Rightarrow n = 9 .$

Suggest Corrections

Similar questions

Q. If the value of c prescribed in Rolle's theorem for the function
f (x) = 2x (x − 3)ⁿ on the interval

[0, 2 \sqrt{3}] is \frac{3}{4},

write the value of n (a positive integer).

Q. Let

n ϵ N

. If the value of

c

prescribed in Rolle's theorem for the function

f (x) = 2 x {(x - 3)}^{n}

[0, 3]

\frac{3}{4}

, then

n

is equal to

Q. If

f (x) = (x - a)^{m} (x - b)^{n} m, n

are positive integers, satisfies the conditions of Rolle's theorem on

[a, b]

, then find '

c

Q. What are the value of

c

for which Rolle's theorem for the function

f (x) = x^{3} - 3 x^{2} + 2 x

in the interval

[0, 2]

is verified?

Q. Verify Rolle's theorem for the function f(x) = x(x −2)² on the interval [0, 2].

Join BYJU'S Learning Program

Explore more

Integration as Antiderivative

Standard XII Mathematics

Join BYJU'S Learning Program

If n is a positive integer for the function f(x)=5x(x−2)n,x∈[0,2]. By Rolle's theorem value of c is 15, then n is equal to

The correct option is C 9f′(c)=0 and c=15 ⇒5x⋅n(x−2)n−1+5(x−2)n=0. ⇒5⋅15⋅n(15−2)n−1+5(15−2)n=0 ⇒n(15−2)n−1+5(15−2)n=0 ⇒n(15−2)−1+5=0 ⇒n(−59)+5=0 ⇒5n9=5⇒n=9.

If $n$ is a positive integer for the function $f (x) = 5 x (x - 2)^{n}, x \in [0, 2] .$ By Rolle's theorem value of $c$ is $\frac{1}{5},$ then $n$ is equal to