Question : If $a+b=1$, then $a^4+b^4-a^3-b^3-2a^2b^2+ab$ is equal to:
Option 1: 1
Option 2: 2
Option 3: 4
Option 4: 0
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Correct Answer: 0
Solution : Given: $a+b=1$ (equation 1) We know that the algebraic identities are $(a-b)^2=a^2+b^2-2ab$, $(a^3+b^3)=(a+b)(a^2-ab+b^2)$ and $(a^2-b^2)=(a+b)(a-b)$. So, $a^4+b^4-2a^2b^2-a^3-b^3+ab$ $=(a^2-b^2)^2-(a^3+b^3)+ab$ $=(a+b)^2(a-b)^2-(a+b)(a^2-ab+b^2)+ab$ Substituting the value from equation 1 in the above expression we get, $a^4+b^4-2a^2b^2-a^3-b^3+ab=(a-b)^2-a^2+ab-b^2+ab$ $=(a-b)^2-(a^2-2ab+b^2)$ $=(a-b)^2-(a-b)^2=0$ Hence, the correct answer is 0.
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Question : If $\frac{x}{a+b}+1=\frac{x}{a-b}+\frac{a-b}{a+b}$, then $x$ is equal to:
Option 1: $2a-b$
Option 2: $a+b$
Option 3: $a-b$
Option 4: $2a+b$
Question : If $\frac{a}{1-2a}+\frac{b}{1-2b}+\frac{c}{1-2c}=\frac{1}{2}$, then the value of $\frac{1}{1-2a}+\frac{1}{1-2b}+\frac{1}{1-2c}$ is:
Option 3: 3
Option 4: 4
Question : If $5 \sin^2 A+3 \cos^2 A=4$, $0<A<90°$, then what is the value of $\tan A$?
Option 1: 0
Option 2: 3
Option 3: 1
Option 4: 2
Question : If $ax+by=1$ and $bx+ay=\frac{2ab}{a^{2}+b^{2}}$, then $(x^{2}+y^{2})(a^{2}+b^{2})$ is equal to:
Option 3: 0.5
Question : If $a+b=1$, find the value of $a^{3}+b^{3} - ab-(a^{2}-b^{2})^{2}$.
Option 1: –1
Option 2: 1
Option 3: 0
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