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JEE Mains · Maths · STD 11 - 12. limits

ધારોકે \(x=2\) એ સમીકરણ \(x^2+q=0\) નો બીજ છે અને \(f(x)=\left\{\begin{array}{cl}\frac{1-\cos \left(x^2-4 p x+q^2+8 q+16\right)}{(x-2 p)^4}, & x \neq 2 p \\ 0, & , x=2 p\end{array}\right.\) તો \(\lim _{x \rightarrow 2 p^{+}}[f(x)],=............\).જ્યાં [.] એ મહત્તમ પૂર્ણાક વિધેય દર્શાવે છે. 

  1. A \(2\)
  2. B \(1\)
  3. C \(0\)
  4. D \(-1\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(0\)

Step-by-step Solution

Detailed explanation

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