If f:R→R is defined by fx=1+3x2−cos2xx2,x≠0k,x=0 is continuous at x=0, then k=

Ltx→01+3x2−cos2xx2K=Ltx→06x+2sin2x2x=3+2⇒K=5

If f:R→R is defined by fx=1+3x2−cos2xx2,x≠0k,x=0 is continuous at x=0, then k=

Ltx→01+3x2−cos2xx2K=Ltx→06x+2sin2x2x=3+2⇒K=5

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If $f : R \to R$ is defined by $f (x) = \{\begin{matrix} \frac{1 + 3 x^{2} - \cos 2 x}{x^{2}}, & x \neq 0 \\ k, & x = 0 \end{matrix}$ is continuous at $x = 0,$ then $k =$

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$\underset{x \to 0}{L t} \frac{1 + 3 x^{2} - \cos 2 x}{x^{2}}$

$K = \underset{x \to 0}{L t} \frac{6 x + 2 \sin 2 x}{2 x} = 3 + 2 \Rightarrow K = 5$

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If f:R→R is defined by fx=1+3x2−cos2xx2,x≠0k,x=0 is continuous at x=0, then k=

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If $f : R \to R$ is defined by $f (x) = \{\begin{matrix} \frac{1 + 3 x^{2} - \cos 2 x}{x^{2}}, & x \neq 0 \\ k, & x = 0 \end{matrix}$ is continuous at $x = 0,$ then $k =$