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QJEEM22Ma5a72: JEE Main 2022 Mathematics - Class 12 JEEM22Ma5a72 Solution: Let be a function defined as where is the greatest integer less …

Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be a function defined as

f(x)=asin⁡(π[x]2)+[2−x],a∈Rf(x)=a \sin \left(\frac{\pi[x]}{2}\right)+[2-x], a \in \mathbb{R} where [t][t] is the greatest integer less than or equal to tt . If lim⁡x→−1f(x)\mathop {\lim }\limits_{x \to -1 } f(x) exists, then the value of ∫04f(x)dx\int\limits_{0}^{4} f(x) d x is equal to

Multiple Choice Options

  • A −- 1
  • B −- 2
  • C 1
  • D 2

Step-by-Step Verified Solution

f(x)=asin⁡(π[x]2)+[2−x] a∈Rf(x) = a\sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]\,a \in R

Now,

∵\because lim⁡x→−1f(x)\mathop {\lim }\limits_{x \to - 1} f(x) exist

∴\therefore lim⁡x→−1−f(x)=lim⁡x→−1+f(x)\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x)

⇒asin⁡(−2π2)+3=asin⁡(−π2)+2\Rightarrow a\sin \left( {{{ - 2\pi } \over 2}} \right) + 3 = a\sin \left( {{{ - \pi } \over 2}} \right) + 2

⇒−a=1⇒a=−1\Rightarrow - a = 1 \Rightarrow a = - 1

Now, ∫04f(x)dx=∫04(−sin⁡(π[x]2)+[2−x])dx\int_0^4 {f(x)dx = \int_0^4 {\left( { - \sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]} \right)dx} }

=∫011dx+∫12−1dx+∫23−1dx+∫34(1−2)dx= \int_0^1 {1dx + \int_1^2 { - 1dx + \int_2^3 { - 1dx + \int_3^4 {(1 - 2)dx} } } }

=1−1−1−1=−2= 1 - 1 - 1 - 1 = - 2

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