TSI Math FREE Sample Practice Questions

3- C
\(f(x) = x –\frac{5}{3}⇒ y = x – \frac{5}{3}⇒ y+ \frac{5}{3}=x\)
\(f^{-1 }= x+ \frac{5}{3}\)
\(f ^{–1}(5) = 5 +\frac{5}{3}=\frac{20}{3}\) For additional educational resources, .

4- C
\(cos\)\( 30^\circ = \frac{\sqrt 3}{2}\) For additional educational resources, .

5- C For additional educational resources, .

sin\(θ=\frac{3}{5}\)⇒ we have following triangle, then:
\(c=\sqrt {(5^2-3^2 )}=\sqrt{(25-9)}=\sqrt 16=4\)
cos\(θ=\frac{4}{5}\) For additional educational resources, .

6- B
\(-2x- y = -9\)
\(5x-2y= 18\)
⇒ Multiplication \((–2)\) in first equation
\(4x +2y =18\)
\(5x-2y= 18\)
Add two equations together ⇒\( 9x =36 ⇒ x= 4\) then: \(y = 1\) For additional educational resources, .

7- C
\(|9 – (12 ÷ | 2 – 5 |)| = |(9-(12÷|-3|))|=|9-(12÷3)|=|9-4|=|5|=5\) For additional educational resources, .

8- D
METHOD ONE
\(log_2⁡x = 5\)
Apply logarithm rule:\(a = log_b⁡(b^a)\)
\(5 = log_2⁡(2^5) = log_2⁡(32)\)
\(log_2⁡x = log_2⁡(32)\)
When the logs have the same base: \(log_b⁡(f(x)) = log_b⁡(g(x))⇒ f(x) = g(x)\)
then: \(x = 32\)
METHOD TWO
We know that:
\(log_a⁡b=c⇒b=a^c\)
\(log_2⁡x=5⇒x=2^5=32\) For additional educational resources, .

9- D
\(\frac{x^3}{16}\)
⇒ reciprocal is : \(\frac{16}{x^3}\) For additional educational resources, .

10- A
\(f(x) = ln (2x + 1)\)
\(y = ln (2x + 1)\)
Change variables \( x\) and \(y: x = ln (2y + 1)\)
solve: \(x = ln (2y + 1)\)
\(y = \frac{e^{x}-1}{2}=\frac{1}{2}(e^{x} – 1)\) For additional educational resources, .

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