For every integer n- let an and bn be real numbers- Let function f- - be given by fx- an-sin- x- for x- -2n-2n-1-bn-cos- x- for x- 2n-1-2n - for all integers n-If f is continuous- then which of the following holds for all n-

At x=2n x→2n^+RHL=a_n+sin2nπ =a_n x→2n^ LHL =b_n+cos2nπ =b_n+1 For continuous nbsp; nbsp; a_n=b_n+1 ∴ a_n b_n=1 At nbsp; nbsp; x=2n+1 nbsp; ...

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

Byju's Answer

Standard XII

Mathematics

Modulus of a Complex Number

For every int...

Question

For every integer $n,$ let $a_{n}$ and $b_{n}$ be real numbers. Let function $f : R \to R$ be given by
$f (x) = {\begin{matrix} a_{n} + sin π x, for x \in [2 n, 2 n + 1] b_{n} + cos π x, for x \in (2 n - 1, 2 n) \end{matrix}$ , for all integers $n .$
If $f$ is continuous, then which of the following hold(s) for all $n ?$

a_{n - 1} - b_{n - 1} = 0

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

a_{n} - b_{n} = 1

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

a_{n} - b_{n + 1} = 1

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

a_{n - 1} - b_{n} = - 1

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

Open in App

Solution

The correct option is D $a_{n - 1} - b_{n} = - 1$
At $x = 2 n$
$x \to 2 n^{+}$ $R H L = a_{n} + sin 2 n π = a_{n}$
$x \to 2 n^{-} L H L = b_{n} + cos 2 n π = b_{n} + 1$
For continuous $a_{n} = b_{n} + 1$
$∴ a_{n} - b_{n} = 1$

At $x = 2 n + 1$
$x \to 2 n + 1^{+} R H L : b_{n + 1} + cos π (2 n + 1) = b_{n + 1} - 1$
$x \to 2 n + 1^{-} L H L : a_{n} + sin π (2 n + 1) = a_{n}$
For continuous $a_{n} = b_{n + 1} - 1$
$a_{n} - b_{n + 1} = - 1$

For $n = n - 1$ $a_{n - 1} - b_{n} = - 1$

Suggest Corrections

Similar questions

Q. For every integer

n

, let

a_{n}

and

b_{n}

be real numbers. Let function

f : I R \to I R

be given by

f (x) = {\begin{matrix} a_{n} + sin π x, f o r x \in [2 n, 2 n + 1] b_{n} + cos π x, f o r x \in (2 n - 1, 2 n) \end{matrix}

, f o r a l l i n t e g e r s n .

f

is continuous, then which of the following hold(s) for all

n

Q. Let f:R

\to

R be defined by f(x)=|x-2n| for x

ϵ

[2n-1,2n+1], n

ϵ

Z. Then f is periodic with period ___

Let

f

be a function on non negative integers defined as follows

f (2 n) = f (f (n))

f (2 n + 1) = f (2 n) + 1 :

-If

f (0) = 0

, find

f (n)

for every

n

Q. Assertion :For

n \geq 4

. Let

a_{n} = n \sum j = 1 n \sum k = j (\begin{matrix} n k \end{matrix}) (\begin{matrix} k j \end{matrix})

and

b_{n} = a_{n} + 2^{n + 1}

;

b_{n + 1} = 5 b_{n} - 6 b_{n - 1}

for all

n \geq 2

Reason:

a_{n} = 5 a_{n} - 6 a_{n}

for all

n \geq 2

Q. Let

s_{n}

be the sum of all integers k such that

2^{n} < k < 2^{n + 1},

for

n \geq 1.

Then

9

divides

S_{n}

, if and only if

Join BYJU'S Learning Program

For every integer n, let an and bn be real numbers. Let function f: R→R be given by f(x)={an+sinπx, for x∈[2n,2n+1]bn+cosπx, for x∈(2n−1,2n) , for all integers n. If f is continuous, then which of the following hold(s) for all n?

The correct option is D an−1−bn=−1At x=2n x→2n+ RHL=an+sin2nπ=an x→2n− LHL=bn+cos2nπ=bn+1 For continuous an=bn+1 ∴an−bn=1 At x=2n+1 x→2n+1+ RHL:bn+1+cosπ(2n+1)=bn+1−1 x→2n+1− LHL:an+sinπ(2n+1)=an For continuous an=bn+1−1 an−bn+1=−1 For n=n−1 an−1−bn=−1