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Statement 1: Minimum number of points of...

Statement 1: Minimum number of points of discontinuity of the function `f(x)=(g(x)[2x-1]AAx in (-3,-1)` , where [.] denotes the greatest integer function and `g(x)=a x^3=x^2+1` is zero. Statement 2: `f(x)` can be continuous at a point of discontinuity, say `x=c_1of[2x-1]ifg(c_1)=0.` Statement 1 is True, Statement 2 is True, Statement 2 isa correct explanation for Statement 1. Statement 1 is True, Statement 2 is True, Statement 2 is NOT a correct explanation for statement 1. Statement 1 is True, Statement 2 is False Statement 1 is False, Statement 2 is True.

A

Statement 1 is True, Statement 2 is True, Statement 2 is a correct explaination for Statement 1.

B

Statement 1 is True, Statement 2 is True, Statement 2 is NOT a correct explanation for Statement 1/

C

Statement 1 is True, Statement 2 is False.

D

Statement 1 is False, Statement 2 is True.

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the two statements provided about the function \( f(x) = g(x)(2x - 1) \) where \( g(x) = ax^3 + x^2 + 1 \) and \( x \in (-3, -1) \). ### Step 1: Identify Points of Discontinuity The term \( 2x - 1 \) is a linear function and is continuous everywhere. However, the function \( g(x) \) is a polynomial, which is also continuous everywhere. The potential points of discontinuity in \( f(x) \) arise from the greatest integer function (denoted by [.]). The expression \( 2x - 1 \) will change its integer value at points where \( 2x - 1 \) is an integer. Setting \( 2x - 1 = n \) for integers \( n \), we can solve for \( x \): \[ 2x = n + 1 \implies x = \frac{n + 1}{2} \] ### Step 2: Determine Specific Points To find the specific points of discontinuity in the interval \( (-3, -1) \), we can calculate: - For \( n = -5 \): \( x = \frac{-5 + 1}{2} = -2 \) - For \( n = -4 \): \( x = \frac{-4 + 1}{2} = -\frac{3}{2} \) - For \( n = -3 \): \( x = \frac{-3 + 1}{2} = -1 \) Thus, the points of discontinuity in the interval \( (-3, -1) \) are: - \( x = -2 \) - \( x = -\frac{3}{2} \) ### Step 3: Analyze Statement 1 The minimum number of points of discontinuity is determined by the discontinuities found. Since we found two points of discontinuity, Statement 1, which claims that the minimum number of points of discontinuity is zero, is **False**. ### Step 4: Analyze Statement 2 Statement 2 states that \( f(x) \) can be continuous at a point of discontinuity \( x = c_1 \) if \( g(c_1) = 0 \). If \( g(c_1) = 0 \), then \( f(c_1) = 0 \cdot (2c_1 - 1) = 0 \). This means that even if \( f(x) \) has a discontinuity at \( c_1 \), it can still be continuous if \( g(c_1) = 0 \). Thus, Statement 2 is **True**. ### Conclusion - Statement 1 is False. - Statement 2 is True. ### Final Answer **Statement 1 is False, Statement 2 is True.**
To solve the problem, we need to analyze the two statements provided about the function \( f(x) = g(x)(2x - 1) \) where \( g(x) = ax^3 + x^2 + 1 \) and \( x \in (-3, -1) \). ### Step 1: Identify Points of Discontinuity The term \( 2x - 1 \) is a linear function and is continuous everywhere. However, the function \( g(x) \) is a polynomial, which is also continuous everywhere. The potential points of discontinuity in \( f(x) \) arise from the greatest integer function (denoted by [.]). The expression \( 2x - 1 \) will change its integer value at points where \( 2x - 1 \) is an integer. Setting \( 2x - 1 = n \) for integers \( n \), we can solve for \( x \): \[ 2x = n + 1 \implies x = \frac{n + 1}{2} ...
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Statement 1: Minimum number of points of discontinuity of the function f(x)=(g(x)[2x-1]AAx in (-3,-1) , where [.] denotes the greatest integer function and g(x)=a x^3+x^2+1 is zero. Statement 2: f(x) can be continuous at a point of discontinuity, say x=c_1 of [2x-1] if g(c_1)=0.

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