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  • Can someone explain - Limit , continuity and differentiability - JEE Main-19

If f(x)=\left\{\begin{matrix} \frac{sin(p+1)x+sinx}{x}, x<0\\ q,\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x=0\\ \frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{\frac{3}{2}}}, \: \: \: \: \: \: \: x>0 \end{matrix}\right.

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

  • Option 1)

    (-\frac{3}{2},-\frac{1}{2})

  • Option 2)

    (-\frac{1}{2},\frac{3}{2})

  • Option 3)

    (-\frac{3}{2},\frac{1}{2})

  • Option 4)

    (\frac{5}{2},\frac{1}{2})

Answers (1)

best_answer

f(x) is continuous at x = 0 then

\lim_{x\rightarrow 0^{+}}\frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{\frac{3}{2}}}=\lim_{x\rightarrow 0^{+}}\frac{\sqrt{x+1}-1}{x}=\frac{1}{2}\: \: \: (\frac{0}{0}form)

LHL,

\lim_{x\rightarrow 0^{-}}\frac{sin(p+1)x+sinx}{x}=(p+1)+1=p+2

for continuity LHL = RHL

p+2=\frac{1}{2}

p=\frac{-3}{2}

q=\frac{1}{2}

{p,q}={{\frac{-3}{2},\frac{1}{2}}}

correct option is (3)    

 


Option 1)

(-\frac{3}{2},-\frac{1}{2})

Option 2)

(-\frac{1}{2},\frac{3}{2})

Option 3)

(-\frac{3}{2},\frac{1}{2})

Option 4)

(\frac{5}{2},\frac{1}{2})

Posted by

Plabita

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