Question

please ans limit class 12th​

Answer 1

Solution:-As it is given in the question that f is continuous,,so

As it is given in the question that f is continuous,,soLHL=RHL

let us put sum value of n and find f(x)

for n=1

a1+sinπx. for x€[2,3]

f(x)=

b1+cosπx. for x€(1,2)

for n=0

a0+sinπx. forx€[0,1]

f(x)=

b0+ cosπx. forx€(-1,0)

for n=3

a3+sinπx. for x€[6,7]

f(x)=

b3+cosπx. forx€(5,6)

these are the values of f(x) uptto n=3

now,let us find the limit at x=2

LHL

$x - > {2}^{ - }$

seeing above ,we can say this value of x is satisfying for f(x) at n=1

so,

LHL=b1+ cosπx

putting the value of x,we get

LHL=b1+cos2π=b1+1

now,let us find RHL limit at x=2

RHL

$x - > {2}^{ + }$

seeing above we can say this value of x is satisfying under f(x) at n=1

RHL=a1+sinπx

putting x=2

RHL=a1+sin2π=a1

now,since the limit is continuous ,so

LHL=RHL

b1+1=a1

a1-b1=1

from this we can say

an-bn=1

hence option (B) is correct

now,let us find the limit at x=1

LHL at f(x) of x=1

$x - > {1}^{ - }$

this x is under the f(x) of n=0

LHL=a0+sinπ

=a0

RHL

$x - > {1}^{ + }$

this x is satisfying under f(x) of n=1

RHL=b1+cosπ=b1-1

for continuous limit

LHL=RHL

a0=b1+1

a0-b1=-1

this is satisfying,the condition

an-1 -bn=-1

Hence option D) is correct

so,both B and D) are correct