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If$$f(x) = \left\{ {\begin{matrix} {{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \\ q & {,x = 0} \\ {{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{{3/2}}}}}} & {,x > 0} \\ \end{matrix} } \right.$$ is continuous at x = 0, then the ordered pair (p, q) is equal to

JEE · Math · previous-year question

  1. A.$$\left( { - {3 \over 2}, - {1 \over 2}} \right)$$
  2. B.$$\left( { - {1 \over 2},{3 \over 2}} \right)$$
  3. C.$$\left( { - {3 \over 2}, {1 \over 2}} \right)$$correct
  4. D.$$\left( { {5 \over 2}, {1 \over 2}} \right)$$

Answer

C. $$\left( { - {3 \over 2}, {1 \over 2}} \right)$$

Explanation

$$f(x) = \left\{ {\begin{matrix} {{{\sin (p + 1)x + \sin x} \over x}} & {x < 0} \\ q & {x = 0} \\ {{{\sqrt {{x^2} + x} - \sqrt x } \over {{x^{{3 \over 2}}}}}} & {x > 0} \\ \end{matrix} } \right.$$ is continuous at x = 0 So f(0–) = f(0) = f (0+) ... (1) $$f({0^ - }) = \mathop {lt}\limits_{h \to 0} f(0 - h)$$ $$ \Rightarrow \mathop {lt}\limits_{h \to 0} {{\sin (p + 1)( - h) + \sin ( - h)} \over { - h}}$$ $$ \Rightarrow \mathop {lt}\limits_{h \to 0} \left[ {{{ - \sin (p + 1)h} \over { - h}} + {{\sinh } \over h}} \right]$$ $$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{\sin (p + 1)h} \over {h(p + 1)}} \times (p + 1) + \mathop {\lim }\limits_{h \to 0} {{\sinh } \over h}$$ = (p + 1) + 1 = p + 2 ...... (2) Now $$f({0^ + }) = \mathop {\lim }\limits_{h \to 0} (0 + h) = \mathop {\lim }\limits_{h \to 0} {{\sqrt {{h^2} + h} - \sqrt h } \over {{h^{3/2}}}}$$ $$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{{(h)}^{{1 \over 2}}}\left[ {\sqrt {h + 1} - 1} \right]} \over {h\left( {{h^{{1 \over 2}}}} \right)}}$$ $$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{\sqrt {h + 1} - 1} \over h} \times {{\sqrt {h + 1} + 1} \over {\sqrt {h + 1} + 1}}$$ $$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{h + 1 - 1} \over {h\left( {\sqrt {h + 1} + 1} \right)}}$$ $$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {1 \over {\sqrt {h + 1} + 1}} = {1 \over {1 + 1}} = {1 \over 2}$$ ..... (3) Now, from equation (1) f(0–) = f(0) = f(0+) p + 2 = q = 1/2 So, $$q = {1 \over 2}$$ and $$p = {1 \over 2} - 2 = {{ - 3} \over 2}$$ $$(p,q) \equiv \left( { - {3 \over 2},{1 \over 2}} \right)$$

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