BITSAT 2012
Exam BITSAT
Year 2012
Subject Mathematics
Chapter Differentiation BITSAT 2012 · Mathematics · Differentiation
Let f(x+y)=f(x) · f(y) for all x, y where f(0) ≠ 0 If f(5)=2 and f^...
Let
f ( x + y ) = f ( x ) ⋅ f ( y ) f(x+y)=f(x) \cdot f(y) f ( x + y ) = f ( x ) ⋅ f ( y ) for all
x , y x, y x , y where
f ( 0 ) ≠ 0 f(0) \neq 0 f ( 0 ) = 0 If
f ( 5 ) = 2 f(5)=2 f ( 5 ) = 2 and
f ′ ( 0 ) = 3 , f^{\prime}(0)=3, f ′ ( 0 ) = 3 , then
f ′ ( 5 ) f^{\prime}(5) f ′ ( 5 ) is equal to-
Options A 6 B 0 C 1 D None of these
Step-by-step solution
f ′ ( 5 ) = lim h → 0 f ( 5 + h ) − f ( 5 ) h = lim h → 0 f ( 5 + h ) − f ( 5 + 0 ) h = lim h → 0 f ( 5 ) ⋅ f ( h ) − f ( 5 ) + f ( 0 ) h ( ∵ f ( x + y ) = f ( x ) , f ( y ) f^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h}=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5+0)}{h}=\lim _{h \rightarrow 0} \frac{f(5) \cdot f(h)-f(5)+f(0)}{h}(\because f(x+y)=f(x), f(y) f ′ ( 5 ) = lim h → 0 h f ( 5 + h ) − f ( 5 ) = lim h → 0 h f ( 5 + h ) − f ( 5 + 0 ) = lim h → 0 h f ( 5 ) ⋅ f ( h ) − f ( 5 ) + f ( 0 ) ( ∵ f ( x + y ) = f ( x ) , f ( y ) for all
x , y ) = ( lim h → 0 f ( h ) − f ( 0 ) h ) ⋅ f ( 5 ) = f ′ ( 0 ) ⋅ f ( 5 ) = 3 × 2 = 6 x, y)=\left(\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\right) \cdot f(5)=f^{\prime}(0) \cdot f(5)=3 \times 2=6 x , y ) = ( lim h → 0 h f ( h ) − f ( 0 ) ) ⋅ f ( 5 ) = f ′ ( 0 ) ⋅ f ( 5 ) = 3 × 2 = 6
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