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Q.

Let f be continuous at x = 0 f" (0) = 4. Then Ltx→02f(x)−3f(2x)+f(4x)x2=

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a

0

b

6

c

12

d

8

answer is C.

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Detailed Solution

Ltx→02f(x)−3f(2x)+f(4x)x200form by L'Hospital rule =Ltx→02f1(x)-6f1(2x)+4f1(4x)2x00form by L'Hospital rule =Ltx→02f"(x)-12f"(2x)+16f"(4x)2=(2-12+16)42=6(2)=12

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