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

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Limits

Let

x = 2

be a root of the equation

x^{2} + p x + q = 0

and

f (x) = {\frac{1 - c o s ( x ^{2} - 4 p x + q ^{2} + 8 q + 16 )}{( x - 2 p ) ^{4}}, 0, x \neq = 2 p x = 2 p

Then

lim_{x \to 2 p^{+}} [f (x)]

where [ . ] denotes greatest integer function, is

Answer: (C)

Since

x = 2

is a root,

4 + 2 p + q = 0

, so

q + 4 = - 2 p

and the argument becomes

x^{2} - 4 p x + (q + 4)^{2} = x^{2} - 4 p x + 4 p^{2} = (x - 2 p)^{2}

. Thus

f (x) = \frac{1 - cos ( ( x - 2 p ) ^{2} )}{( x - 2 p ) ^{4}}

, and as

x \to 2 p

the standard limit

1 - cos u \sim \frac{1}{2} u^{2}

with

u = (x - 2 p)^{2}

gives

f (x) \to \frac{1}{2}

. The greatest integer not exceeding a value approaching

\frac{1}{2}

0

, so the limit of

[f (x)]

0

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option