新sat试题下载【OG数学真题】

　　新sat试题下载【OG数学真题】!新

　　新satOG数学真题精选：

　　1.Which of the following expressions is equal to 0 for some value of x?

　　A . ∣x−1∣−1∣x−1∣−1

　　B . ∣x+1∣+1∣x+1∣+1

　　C . ∣1−x∣+1∣1−x∣+1

　　D . ∣x−1∣+1

　　答案解析

　　Choice A is correct.

　　The expression ∣x−1∣−1∣x−1∣−1 will equal 0 if midx−1∣=1midx−1∣=1. This is true for x=2x=2 and for x=0x=0. For example, substituting x=2x=2 into the expression ∣x−1∣−1∣x−1∣−1 and simplifying the result yields ∣2−1∣−1=∣1∣−1=1−1=0∣2−1∣−1=∣1∣−1=1−1=0. Therefore, there is a value of x for which ∣x−1∣−1∣x−1∣−1 is equal to 0.

　　Choice В is incorrect. By definition, the absolute value of any expression is a nonnegative number. Substituting any value for xx into the expression ∣x+1∣∣x+1∣ will yield a nonnegative number as the result. Because the sum of a nonnegative number and a positive number is positive, ∣x+1∣+1∣x+1∣+1 will be a positive number for any value of xx. Therefore, ∣x+1∣+1≠0∣x+1∣+1≠0 for any value of xx. Choice С is incorrect. By definition, the absolute value of any expression is a nonnegative number. Substituting any value for xx into the expression ∣1−x∣∣1−x∣ will yield a nonnegative number as the result. Because the sum of a nonnegative number and a positive number is positive, ∣1−x∣+1∣1−x∣+1 will be a positive number for any value of xx. Therefore, ∣x−1∣+1≠0∣x−1∣+1≠0 for any value of xx. Choice D is incorrect. By definition, the absolute value of any expression is a nonnegative number. Substituting any value for x into the expression ∣x−1∣∣x−1∣ will yield a nonnegative number as the result. Because the sum of a nonnegative number and a positive number is positive, ∣x−1∣+1∣x−1∣+1 will be a positive number for any value of xx. Therefore, ∣x−1∣+1≠0∣x−1∣+1≠0 for any value of xx.

　　2.

　　3.If f(x)=−2x+5f(x)=−2x+5, what is f(−3x)f(−3x) equal to?

　　A . −6x−5−6x−5

　　B . 6x+56x+5

　　C . 6x−56x−5

　　D . 6x2−15x6x2−15x

　　答案解析

　　Choice В is correct.

　　If f(x)=−2x+5f(x)=−2x+5, then one can evaluate f(−3x)f(−3x) by substituting −3x−3x for every instance of xx. This yields f(−3х)=−2(−3x)+5f(−3х)=−2(−3x)+5, which simplifies to 6x+56x+5.

　　Choices A, C, and D are incorrect and may be the result of miscalculations in the substitution or of misunderstandings of how to evaluate f(−3x)f(−3x).

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