Let f : R − { 0 } → R be a function such that f ( x ) − 6 f ( x 1 ) = 3 x 35 − 2 5 . If the lim x → 0 ( αx 1 + f ( x ) ) = β ; α , β ∈ R , then α + 2 β is equal to

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Let f : R − { 0 } → R be a function such that f ( x ) − 6 f ( x 1 ​ ) = 3 x 35 ​ − 2 5 ​ . If the lim x → 0 ​ ( αx 1 ​ + f ( x ) ) = β ; α , β ∈ R , then α + 2 β is equal to

TestHub · Accepted Answer

​ f ( x ) − 6 f ( x 1 ​ ) = 3 x 35 ​ − 2 5 ​ ... ( 1 ) 6 ( f ( x 1 ​ ) − 6 f ( x ) = 3 35 x ​ − 2 5 ​ ) 6 f ( x 1 ​ ) − 36 f ( x ) = 3 210 x ​ − 2 30 ​ ... ( 2 ) ( 1 ) + ( 2 ) − 35 f ( x ) = 3 35 ​ [ x 1 ​ + 6 x ] − 2 5 ​ ( 1 + 6 ) − f ( x ) = 3 1 ​ ( x 1 ​ + 6 x ) − 2 1 ​ f ( x ) = − 3 x 1 ​ − 2 x + 2 1 ​ x → 0 lim ​ [ αx 1 ​ − 3 x 1 ​ − 2 x + 2 1 ​ ] = β ⇒ α = 3 β + 2 β = 3 + 2 × 2 1 ​ = 4 ​

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