Let x = 2 be a root of the equation x2+px+q=0 and f(x)=1−cos⁡x2−4px+q2+8q+16(x−2p)4,x≠2p0,x=2pThen limx→2p+ fxWhere [. ] denotes greatest integer function, is

x = 2 is a root of x2+px+q=0⇒q+4=−2p=Lx→2P+1−cos⁡x2−4px+q2+8q+16x2−4px+q2+8q+162×x2−4px+q2+8q+162(x−2p)4=12Ltx→2P+x2−4px+4p22(x−2p)4=12×1=12∴Ltx→2p+[f(x)]=0

Let x = 2 be a root of the equation x2+px+q=0 and f(x)=1−cos⁡x2−4px+q2+8q+16(x−2p)4,x≠2p0,x=2pThen limx→2p+ fxWhere [. ] denotes greatest integer function, is

x = 2 is a root of x2+px+q=0⇒q+4=−2p=Lx→2P+1−cos⁡x2−4px+q2+8q+16x2−4px+q2+8q+162×x2−4px+q2+8q+162(x−2p)4=12Ltx→2P+x2−4px+4p22(x−2p)4=12×1=12∴Ltx→2p+[f(x)]=0

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Let x = 2 be a root of the equation $x^{2} + px + q = 0$ and $f (x) = \{\begin{array}{cc} \frac{1 - \cos (x^{2} - 4 px + q^{2} + 8 q + 16)}{(x - 2 p)^{4}} & , x \neq 2 p \\ 0 & , x = 2 p \end{array}$
Then $lim_{x \to 2 p^{+}} [f (x)]$
Where [. ] denotes greatest integer function, is

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Detailed Solution

x = 2 is a root of $x^{2} + px + q = 0$
$\begin{array}{l} \Rightarrow q + 4 = - 2 p \\ = \underset{x \to 2 P_{+}}{L} \{\frac{1 - \cos (x^{2} - 4 px + q^{2} + 8 q + 16)}{{(x^{2} - 4 px + q^{2} + 8 q + 16)}^{2}} \times \frac{{(x^{2} - 4 px + q^{2} + 8 q + 16)}^{2}}{(x - 2 p)^{4}}\} \end{array}$
$\begin{array}{l} = \frac{1}{2} \underset{x \to 2 P_{+}}{Lt} \frac{{(x^{2} - 4 px + 4 p^{2})}^{2}}{(x - 2 p)^{4}} \\ = \frac{1}{2} \times 1 = \frac{1}{2} \\ ∴ \underset{x \to 2 p_{+}}{Lt} [f (x)] = 0 \end{array}$