Q102 / 160 · AP EAMCET 2019 (23 Apr Shift 1) · Mathematics

Let \([x]\) denote the greatest integer not exceeding \(x\). If \(l_1=\lim _{x \rightarrow 2^{+}}\left(x^2+[x]\right)\), \(l_2=\lim _{x \rightarrow 3^{-}}(2 x-[x])\) and \(l_3=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\cos x}{x-\frac{\pi}{2}}\right)\), then

  • A
    \(I_2 < I_3 < I_1\)
  • B
    \(I_1 < I_3 < I_2\)
  • C
    \(I_1 < I_2 < I_3\)
  • Correct answer
    D
    \(I_3 < I_2 < I_1\)

Solution

Given, \(\begin{aligned} l_1 & =\lim _{x \rightarrow 2^{+}}\left(x^2+[x]\right) \\ & =\lim _{x \rightarrow 2^{+}}\left(x^2+2\right) \quad\left[\because \text { as } x \rightarrow 2^{+},[x]=2\right] \\ & =4+2 \\ l_1 & =6 \\ l_2 & =\lim _{x \rightarrow 3^{-}}(2 x-[x]) \quad\left[\because \text { as } x \rightarrow 3^{-},[x]=2\right] \\ & =\lim _{x \rightarrow 3^{-}}(2 x-2) \quad \\ & =2(3)-2 \\ l_2 & =4 \\ l_3 & =\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\cos x}{x-\frac{\pi}{2}}\right) \end{aligned}\) Put \(x-\frac{\pi}{2}=y\) and as \(x \rightarrow \frac{\pi}{2}\), then \(y \rightarrow 0\) \(\begin{aligned} x & =\frac{\pi}{2}+y \\ & =\lim _{y \rightarrow 0} \frac{\cos \left(\frac{\pi}{2}+y\right)}{y} \\ & =\lim _{y \rightarrow 0}-\frac{\sin y}{y} \\ \quad l_3 & =-1 \\ \quad & l_3 < l_2 < l_1 \end{aligned}\) \(\therefore\) Hence, answer is (d).

Hence, the correct answer is option: 'D'.

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