QJEEM19Mae2cd: JEE Main 2019 Mathematics - Class 12 JEEM19Mae2cd Solution: If f ( x ) = { sin ⁡ ( p + 1 ) x + sin ⁡ x x , x 0 is…

Question

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If

f (x) = {\begin{matrix} \frac{\sin (p + 1) x + \sin x}{x} & , x < 0 \\ q & , x = 0 \\ \frac{\sqrt{x + x^{2}} - \sqrt{x}}{x^{3 / 2}} & , x > 0 \end{matrix}

is continuous at x = 0, then the ordered pair (p, q) is equal to

Accepted Answer

f (x) = {\begin{matrix} \frac{\sin (p + 1) x + \sin x}{x} & x < 0 \\ q & x = 0 \\ \frac{\sqrt{x^{2} + x} - \sqrt{x}}{x^{\frac{3}{2}}} & x > 0 \end{matrix}

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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