Verified Solution • 2025–26 Session moderate

QJEEM11Ma4376: JEE Main 2011 Mathematics - Class 12 JEEM11Ma4376 Solution: The value of and for which the function f ( x ) = { sin ⁡ ( p + …

The value of pp and qq for which the function

f(x)={sin⁡(p+1)x+sin⁡xx,x<0q,x=0x+x2−xx3/2,x>0

is continuous for all xx in R, are

Multiple Choice Options

  • A p=p = 52{5 \over 2} , q=q = 12{1 \over 2}
  • B p=p = −32-{3 \over 2} , q=q = 12{1 \over 2}
  • C p=p = 12{1 \over 2} , q=q = 32{3 \over 2}
  • D p=p = 12{1 \over 2} , q=q = −32-{3 \over 2}

Step-by-Step Verified Solution

L.H.L.=lim⁡x→0f(x)=lim⁡h→0sin⁡{(p+1)(−h)}−sin⁡(−h)−hL.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}}

=lim⁡h→0−sin⁡(p+1)h−h+sin⁡(−h)−h=p+1+1=p+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.R.H.L. =lim⁡x→σ+f(x)=lim⁡h→01+h−1h= \mathop {\lim }\limits_{x \to {\sigma ^ + }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{\sqrt {1 + h} - 1} \over h}

=lim⁡h→01(1+h+1)=12= \mathop {\lim }\limits_{h \to 0} {1 \over {\left( {\sqrt {1 + h} + 1} \right)}} = {1 \over 2}

and f(0)=q⇒p=−32,q=12f\left( 0 \right) = q \Rightarrow p = - {3 \over 2},q = {1 \over 2}

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