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Limits: Let Denotes Fractional Part

JEE Maths question with a full step-by-step solution.

Question
Let f(x)=cos⁡−1(1−{x})sin⁡−1(1−{x})2{x}(1−{x})f(x)=\dfrac{\cos^{-1}(1-\{x\})\sin^{-1}(1-\{x\})}{\sqrt{2\{x\}}(1-\{x\})} where {x}\{x\} denotes the fractional part of xx, then
Alim⁡x→0−f(x)=π22\displaystyle\lim_{x\to 0^{-}}f(x)=\dfrac{\pi}{2\sqrt{2}}correct
Blim⁡x→0+f(x)=2lim⁡x→0−f(x)\displaystyle\lim_{x\to 0^{+}}f(x)=\sqrt{2}\lim_{x\to 0^{-}}f(x)correct
Clim⁡x→0−f(x)=π42\displaystyle\lim_{x\to 0^{-}}f(x)=\dfrac{\pi}{4\sqrt{2}}
Dlim⁡x→0+f(x)=π2\displaystyle\lim_{x\to 0^{+}}f(x)=\dfrac{\pi}{2}correct
Solution
Step 1: Compute the RHL (x→0+x\to 0^{+}). For xx slightly greater than 00, {x}=x\{x\}=x, so {x}→0+\{x\}\to 0^{+}. Let h={x}h=\{x\}:
lim⁡h→0+cos⁡−1(1−h)⋅sin⁡−1(1−h)2h⋅(1−h)\lim_{h\to 0^{+}}\dfrac{\cos^{-1}(1-h)\cdot\sin^{-1}(1-h)}{\sqrt{2h}\cdot(1-h)}
 Step 2: For small hh: 1−(1−h)2=2h−h2=h(2−h)\sqrt{1-(1-h)^{2}}=\sqrt{2h-h^{2}}=\sqrt{h(2-h)}. Using the right triangle with sides 2h−h2\sqrt{2h-h^{2}}, (1−h)(1-h) and hypotenuse 11: cos⁡−1(1−h)∼2h−h2\cos^{-1}(1-h)\sim\sqrt{2h-h^{2}} for small hh (since the angle equals approximately the opposite side when hypotenuse is 11). Step 3: As h→0+h\to 0^{+}: sin⁡−1(1−h)→sin⁡−1(1)=π2\sin^{-1}(1-h)\to\sin^{-1}(1)=\dfrac{\pi}{2}. Step 4: Substitute:
lim⁡h→0+2h−h2⋅sin⁡−1(1−h)2h⋅(1−h)=lim⁡h→0+2h1−h2⋅sin⁡−1(1−h)2h⋅(1−h)\lim_{h\to 0^{+}}\dfrac{\sqrt{2h-h^{2}}\cdot\sin^{-1}(1-h)}{\sqrt{2h}\cdot(1-h)}=\lim_{h\to 0^{+}}\dfrac{\sqrt{2h}\sqrt{1-\dfrac{h}{2}}\cdot\sin^{-1}(1-h)}{\sqrt{2h}\cdot(1-h)}
=lim⁡h→0+1−h/2⋅sin⁡−1(1−h)1−h=1⋅π21=π2=\lim_{h\to 0^{+}}\dfrac{\sqrt{1-h/2}\cdot\sin^{-1}(1-h)}{1-h}=\dfrac{1\cdot\dfrac{\pi}{2}}{1}=\dfrac{\pi}{2}
 So RHL =π2=\dfrac{\pi}{2}. Option (4) is correct. Step 5: Compute the LHL (x→0−x\to 0^{-}). For xx slightly less than 00, {x}=1+x\{x\}=1+x (for −1<x<0-1<x<0). Let x=−hx=-h where h→0+h\to 0^{+}, so {x}=1−h\{x\}=1-h. Then 1−{x}=h1-\{x\}=h:
lim⁡h→0+cos⁡−1(h)⋅sin⁡−1(h)2(1−h)⋅h\lim_{h\to 0^{+}}\dfrac{\cos^{-1}(h)\cdot\sin^{-1}(h)}{\sqrt{2(1-h)}\cdot h}
 Step 6: As h→0+h\to 0^{+}: cos⁡−1(h)→π2\cos^{-1}(h)\to\dfrac{\pi}{2} and sin⁡−1(h)∼h\sin^{-1}(h)\sim h:
=lim⁡h→0+π2⋅h2(1−h)⋅h=π/22=π22=\lim_{h\to 0^{+}}\dfrac{\dfrac{\pi}{2}\cdot h}{\sqrt{2(1-h)}\cdot h}=\dfrac{\pi/2}{\sqrt{2}}=\dfrac{\pi}{2\sqrt{2}}
 So LHL =π22=\dfrac{\pi}{2\sqrt{2}}. Option (1) is correct. Step 7: Check (2): π2=2⋅π22=π2\dfrac{\pi}{2}=\sqrt{2}\cdot\dfrac{\pi}{2\sqrt{2}}=\dfrac{\pi}{2}. True. Option (2) is correct.
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