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Question : 61 of 160
Marks:
+1 ,
-0
Solution:
Since,
x ⟶ 3 − , we have
x < 3 .
∴ | x | = x ‌ and ‌ | 3 − x | = 3 − x As,
x ⟶ 3 − , 3 x ⟶ 9 − , so
9 − 3 x ⟶ 0 + Thus,
[ 9 − 3 x ] = 0 for
x slightly less than 3 and
[ 3 x − 9 ] = − 1 for
x slightly less than 3 .
∴ ‌ ( 3 − x + s i n ‌ ( 3 − x ) ) ‌ cos ( 0 ) ( 3 − x ) ( − 1 )
‌ ‌ = ‌ ( 3 − x + s i n ‌ ( 3 − x ) ) ( 3 − x ) ( − 1 )
Let
t = 3 − x .
As,
x ⟶ 3 − , t ⟶ 0 + So,
‌ ‌ ‌ ( 3 − x + s i n ‌ ( 3 − x ) ) ( 3 − x ) ( − 1 )
‌ ⇒ ‌ ‌ ‌ ‌ ⇒ ‌ ‌ ( − 1 − ‌ ) ‌ ⇒ ‌ ‌ − 1 − 1 ‌ ‌ [ ∵ ‌ = 1 ] ‌ ⇒ ‌ ‌ − 2 ‌ ∴ ‌ ( 3 − | x | + s i n ‌ | 3 − x | ) ‌ cos [ 9 − 3 x ] | 3 − x | [ 3 x − 9 ]
‌ = − 2 © examsiri.com
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