Question : If $\frac{3}{(x+2)(2x+1)}=\frac{a}{2x+1}+\frac{b}{x+2}$ is an identity, the value of $b$ is:
Option 1: 0
Option 2: –1
Option 3: 2
Option 4: 3
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Correct Answer: –1
Solution : Given: $\frac{3}{(x+2)(2x+1)}=\frac{a}{2x+1}+\frac{b}{x+2}$ $⇒\frac{3}{(x+2)(2x+1)}=\frac{a(x+2)+b(2x+1)}{(2x+1)(x+2)}$ $⇒3=a(x+2)+b(2x+1)$ To eliminate the term '$a$' we have to put $x = -2$, $⇒3=a(–2+2)+b(2×(–2)+1)$ $⇒3=–3b$ $⇒b=–1$ Hence, the correct answer is –1.
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Question : If $\frac{3–5x}{2x}+\frac{3–5y}{2y}+\frac{3–5z}{2z}=0$, the value of $\frac{2}{x}+\frac{2}{y}+\frac{2}{z}$ is:
Option 1: 20
Option 2: 5
Option 3: 10
Option 4: 15
Question : If $2x=\sqrt{a}+\frac{1}{\sqrt{a}}, a>0$, then the value of $\frac{\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}$ is:
Option 1: $a+1$
Option 2: $\frac{1}{2}(a+1)$
Option 3: $\frac{1}{2}(a–1)$
Option 4: $a–1$
Question : If $2x-\frac{2}{x}=1(x \neq 0)$, then the the value of $(x^3-\frac{1}{x^3})$ is:
Option 1: $\frac{13}{4}$
Option 2: $\frac{13}{8}$
Option 3: $\frac {17}{4}$
Option 4: $\frac{17}{8}$
Question : If $(x+\frac{1}{x})^{2}=3$, then the value of $(x^{3}+\frac{1}{x^{3}})$ is:
Option 2: 1
Option 4: –1
Question : If $x^{2}+\frac{1}{x^{2}} = 98(x>0)$, then the value of $x^{3}+\frac{1}{x^{3}}$ is:
Option 1: 970
Option 2: 1030
Option 3: –970
Option 4: –1030
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