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

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

-1

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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