Verified Solution • 2025–26 Session moderate

QJEEM23Ma7a6f: JEE Main 2023 Mathematics - Class 12 JEEM23Ma7a6f Solution: Let be a root of the equation and f ( x ) = { 1 − cos ⁡ ( x 2 − …

Let $x=2$ be a root of the equation $x^2+px+q=0$ and $f (x) = {\begin{matrix} \frac{1 - \cos (x^{2} - 4 p x + q^{2} + 8 q + 16)}{{(x - 2 p)}^{4}}, & x \neq 2 p \\ 0, & x = 2 p \end{matrix}$

Then $\mathop {\lim }\limits_{x \to 2{p^ + }} [f(x)]$ , where $\left[ . \right]$ denotes greatest integer function, is

Multiple Choice Options

A 2
B 1
C 0
D $-1$

Step-by-Step Verified Solution

\lim \limits_{x \rightarrow 2^{+}}\left(\frac{1-\cos \left(x^{2}-4 p x+q^{2}+8 q+16\right)}{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}\right)\left(\frac{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}{(x-2 p)^{2}}\right)

\lim \limits_{h \rightarrow 0} \frac{1}{2}\left(\frac{(2 p+h)^{2}-4 p(2 p+h)+q^{2}+82+16}{h^{2}}\right)^{2}=\frac{1}{2}

Using L'Hospital's

\lim _{x \rightarrow 2 p^{+}}[f(x)]=0

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