If f -0- k - k -0- then the value of-isA- kB- 2 kC- 3 kD- 4 k

The correct option is C 3klim_x→ 02f (x) 3f (2x) +f (4x)x^2 nbsp; As its is 00 form so,we can use L'Hospital Rule twice lim_x→ 02f” (x) 12f ”(2x) +16f” (4x ...

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Standard XII

Mathematics

Continuity in an Interval

If f ''0= k ,...

Question

If $f^{''} (0) = k, k \neq 0,$ then the value of
$lim x \to 0 \frac{2 f (x) - 3 f (2 x) + f (4 x)}{x^{2}}$ is

k

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2 k

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3 k

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4 k

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Solution

The correct option is C $3 k$
$lim x \to 0 \frac{2 f (x) - 3 f (2 x) + f (4 x)}{x^{2}}$
As its is $\frac{0}{0}$ form so,we can use L'Hospital Rule twice
$lim x \to 0 \frac{2 f^{''} (x) - 12 f^{''} (2 x) + 16 f^{''} (4 x)}{2} = \frac{6 k}{2} = 3 k$

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If f′′(0)=k,k≠0, then the value of limx→02f(x)−3f(2x)+f(4x)x2is

The correct option is C 3klimx→02f(x)−3f(2x)+f(4x)x2 As its is 00 form so,we can use L'Hospital Rule twice limx→02f′′(x)−12f′′(2x)+16f′′(4x)2=6k2=3k

If $f^{''} (0) = k, k \neq 0,$ then the value of
$lim x \to 0 \frac{2 f (x) - 3 f (2 x) + f (4 x)}{x^{2}}$ is

The correct option is C $3 k$
$lim x \to 0 \frac{2 f (x) - 3 f (2 x) + f (4 x)}{x^{2}}$
As its is $\frac{0}{0}$ form so,we can use L'Hospital Rule twice
$lim x \to 0 \frac{2 f^{''} (x) - 12 f^{''} (2 x) + 16 f^{''} (4 x)}{2} = \frac{6 k}{2} = 3 k$