Verified Solution • 2025–26 Session moderate

QJEEM19Mad85f: JEE Main 2019 Mathematics - Class 12 JEEM19Mad85f Solution: Let [x] denote the greatest integer less than or equal to x. Then

Let [x] denote the greatest integer less than or equal to x. Then lim⁡x→0tan⁡(πsin⁡2x)+(∣x∣−sin⁡(x[x]))2x2\mathop {\lim }\limits_{x \to 0} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}

Multiple Choice Options

  • A equals π\pi + 1
  • B equals 0
  • C does not exist
  • D equals π\pi

Step-by-Step Verified Solution

R.H.L. == lim⁡x→0+tan⁡(πsin⁡2x)+(∣x∣−sin⁡(x[x]))2x2\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}

(as x →\to 0+ ⇒\Rightarrow   [x] == 0)

== lim⁡x→0+tan⁡(πsin⁡2x)+x2x2\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {x^2}} \over {{x^2}}}

== lim⁡x→0+tan⁡(πsin⁡2x)(πsin⁡2x)+1=π+1\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\left( {\pi {{\sin }^2}x} \right)}} + 1 = \pi + 1

L.H.L. == lim⁡x→0+tan⁡(πsin⁡2x)+(−x+sin⁡x)2x2\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( { - x + \sin x} \right)}^2}} \over {{x^2}}}

(as x →\to 0 −- ⇒\Rightarrow   [x] == −- 1)

lim⁡x→0+tan⁡(πsin⁡2x)πsin⁡2x . πsin⁡2xx2+(−1+sin⁡xx)2⇒π\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}}\,.\,{{\pi {{\sin }^2}x} \over {{x^2}}} + {\left( { - 1 + {{\sin x} \over x}} \right)^2} \Rightarrow \pi

R.H.L.   ≠\ne   L.H.L.

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