Verified Solution • 2025–26 Session moderate 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}f:R→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}f(x)=asin(2π[x])+[2−x],a∈R where [t][t][t] is the greatest integer less than or equal to ttt . If limx→−1f(x)\mathop {\lim }\limits_{x \to -1 } f(x)x→−1limf(x) exists, then the value of ∫04f(x)dx\int\limits_{0}^{4} f(x) d x0∫4f(x)dx 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 Rf(x)=asin(2π[x])+[2−x]a∈R Now, ∵\because∵ limx→−1f(x)\mathop {\lim }\limits_{x \to - 1} f(x)x→−1limf(x) exist ∴\therefore∴ limx→−1−f(x)=limx→−1+f(x)\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x)x→−1−limf(x)=x→−1+limf(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⇒asin(2−2π)+3=asin(2−π)+2 ⇒−a=1⇒a=−1\Rightarrow - a = 1 \Rightarrow a = - 1⇒−a=1⇒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} }∫04f(x)dx=∫04(−sin(2π[x])+[2−x])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} } } }=∫011dx+∫12−1dx+∫23−1dx+∫34(1−2)dx =1−1−1−1=−2= 1 - 1 - 1 - 1 = - 2=1−1−1−1=−2 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring