JEE MAIN - Mathematics (2011 - No. 2)
The value of $$p$$ and $$q$$ for which the function
$$f\left( x \right) = \left\{ {\matrix{ {{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \cr q & {,x = 0} \cr {{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{3/2}}}}} & {,x > 0} \cr } } \right.$$
is continuous for all $$x$$ in R, are
$$f\left( x \right) = \left\{ {\matrix{ {{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \cr q & {,x = 0} \cr {{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{3/2}}}}} & {,x > 0} \cr } } \right.$$
is continuous for all $$x$$ in R, are
$$p =$$ $${5 \over 2}$$, $$q = $$ $${1 \over 2}$$
$$p =$$ $$-{3 \over 2}$$, $$q = $$ $${1 \over 2}$$
$$p =$$ $${1 \over 2}$$, $$q = $$ $${3 \over 2}$$
$$p =$$ $${1 \over 2}$$, $$q = $$ $$-{3 \over 2}$$
Explanation
$$L.H.L. = \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{\sin \left\{ {\left( {p + 1} \right)\left( { - h} \right)} \right\} - \sin \left( { - h} \right)} \over { - h}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{ - \sin \left( {p + 1} \right)h} \over { - h}} + {{\sin \left( { - h} \right)} \over { - h}} = p + 1 + 1 = p + 2$$
$$R.H.L.$$ $$ = \mathop {\lim }\limits_{x \to {\sigma ^ + }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{\sqrt {1 + h} - 1} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {1 \over {\left( {\sqrt {1 + h} + 1} \right)}} = {1 \over 2}$$
and $$f\left( 0 \right) = q \Rightarrow p = - {3 \over 2},q = {1 \over 2}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{ - \sin \left( {p + 1} \right)h} \over { - h}} + {{\sin \left( { - h} \right)} \over { - h}} = p + 1 + 1 = p + 2$$
$$R.H.L.$$ $$ = \mathop {\lim }\limits_{x \to {\sigma ^ + }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{\sqrt {1 + h} - 1} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {1 \over {\left( {\sqrt {1 + h} + 1} \right)}} = {1 \over 2}$$
and $$f\left( 0 \right) = q \Rightarrow p = - {3 \over 2},q = {1 \over 2}$$
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