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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

વિધેય  \(f(x)=\frac{\mathrm{P}(\mathrm{x})}{\sin (\mathrm{x}-2)}, \quad \mathrm{x} \neq 2\) \(\quad \quad \quad \quad 7, \quad\quad\quad \mathrm{x}=2\)  આપેલ છે  કે જ્યાં \(P(x)\) એ બહુપદી છે કે જેથી \(P^{\prime \prime}(x)\) એ હંમેશા અચળ થાય છે અને \(P(3)=9\) છે જો વિધેય \(f(x)\) એ \(x=2\) આગળ સતત હોય તો \(P(5)\) ની કિમંત મેળવો.

  1. A \(41\)
  2. B \(40\)
  3. C \(39\)
  4. D \(71\)
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Answer & Solution

Correct Answer

(C) \(39\)

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Detailed explanation

\(f(x)=\frac{\mathrm{P}(\mathrm{x})}{\sin (\mathrm{x}-2)}, \quad \mathrm{x} \neq 2\) \(\quad \quad \quad \quad 7, \quad\quad\quad \mathrm{x}=2\) \(\mathrm{P}^{\prime \prime}(\mathrm{x})=\) const. \(\Rightarrow \mathrm{P}(\mathrm{x})\) is a 2 degree polynomial \(f(x)\) is cont.…
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