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

જો \(f\left( x \right)\left\{ {\begin{array}{*{20}{c}}
  {\frac{{\sin \,\left( {p + 1} \right)x + \sin \,x}}{x},\,\,}&{x < 0} \\ 
  {q\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,}&{x = 0} \\ 
  {\frac{{\sqrt {x + {x^2}}  - \sqrt x }}{{x/2}},}&{x > 0} 
\end{array}} \right.\) એ \(x = 0\) આગળ સતત હોય તો ક્રમયુક્ત જોડ \((p, q)\) મેળવો.

  1. A \(\left( { - \frac{3}{2}, - \frac{1}{2}} \right)\)
  2. B \(\left( {\frac{5}{2},\frac{1}{2}} \right)\)
  3. C \(\left( { - \frac{1}{2},\frac{3}{2}} \right)\)
  4. D \(\left( { - \frac{3}{2}, \frac{1}{2}} \right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\left( { - \frac{3}{2}, \frac{1}{2}} \right)\)

Step-by-step Solution

Detailed explanation

\(RHL = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt {x + {x^2}} - \sqrt x }}{{{x^{3/2}}}}\) \( = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt {1 + x} - 1}}{x} = \frac{1}{2}\)…
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