Q.1. Let [] denote the greatest integer function. If the domain of the function is , then is equal

Q: What is the correct answer for JEE Main Mathematics Q1?

The correct answer is A. Sol. $-1 \le \frac{4x+2[x]}{3} < 1 \Rightarrow -3 \le 4x+2[x] < 3$. We know that $x-1 < [x] \le x$. So, $2(x-1) < 2[x] \le 2x$. Substituting this into the inequality: $-3 \le 4x+2[x] < 3$. Case 1: $2[x] = 2x$. Then $-3 \le 6x < 3 \Rightarrow -1/2 \le x < 1/2$. Case 2: $2[x] = 2x-2$. Then $-3 \le 4x+2x-2 < 3 \Rightarrow -1 \le 6x < 5 \Rightarrow -1/6 \le x < 5/6$. Combining these, we get $-1/2 \le

Q: Q.1. Let [] denote the greatest integer function. If the domain of the function is , then is equal to :

The correct answer is A. Sol. $-1 \le \frac{4x+2[x]}{3} < 1 \Rightarrow -3 \le 4x+2[x] < 3$. We know that $x-1 < [x] \le x$. So, $2(x-1) < 2[x] \le 2x$. Substituting this into the inequality: $-3 \le 4x+2[x] < 3$. Case 1: $2[x] = 2x$. Then $-3 \le 6x < 3 \Rightarrow -1/2 \le x < 1/2$. Case 2: $2[x] = 2x-2$. Then $-3 \le 4x+2x-2 < 3 \Rightarrow -1 \le 6x < 5 \Rightarrow -1/6 \le x < 5/6$. Combining these, we get $-1/2 \le

Question

Q.1. Let [] denote the greatest integer function. If the domain of the function  is , then  is equal to :

FirstInTest · Accepted Answer

Sol. $-1 \le \frac{4x+2[x]}{3} < 1 \Rightarrow -3 \le 4x+2[x] < 3$. We know that $x-1 < [x] \le x$. So, $2(x-1) < 2[x] \le 2x$. Substituting this into the inequality: $-3 \le 4x+2[x] < 3$. Case 1: $2[x] = 2x$. Then $-3 \le 6x < 3 \Rightarrow -1/2 \le x < 1/2$. Case 2: $2[x] = 2x-2$. Then $-3 \le 4x+2x-2 < 3 \Rightarrow -1 \le 6x < 5 \Rightarrow -1/6 \le x < 5/6$. Combining these, we get $-1/2 \le x < 1/2$. The solution involves analyzing the inequality $-1 \le \frac{4x+2[x]}{3} < 1$. This simplifies to $-3 \le 4x+2[x] < 3$. We consider cases for $[x]$. If $x \in [n, n+1)$, then $[x]=n$. For $x \in [0,1)$, $[x]=0$, so $-3 \le 4x < 3 \Rightarrow -3/4 \le x < 3/4$. For $x \in [-1,0)$, $[x]=-1$, so $-3 \le 4x-2 < 3 \Rightarrow -1 \le 4x < 5 \Rightarrow -1/4 \le x < 5/4$. The intersection of these intervals gives the domain. The solution provided shows $-3-4x \le 2[x] \le 3-4x$. And then a graph is used to find the intersection points. The final interval is $x \in [-1/4, 3/4]$. So $\alpha = -1/4$ and $\beta = 3/4$. Then $12(\alpha + \beta) = 12(-1/4 + 3/4) = 12(2/4) = 12(1/2) = 6$.

Answer

6

Answer

8

Answer

9

Answer

4

Related Questionsin Mathematics

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions