Question : If $(x+y)^3+8(x-y)^3=(3 x+A y)\left(3 x^2+B x y+C y^2\right)$, then the value of (A + B + C) is:
Option 1: 0
Option 2: 4
Option 3: 2
Option 4: 3
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Correct Answer: 0
Solution : $(x+y)^3+8(x-y)^3=(3 x+A y)\left(3 x^2+B x y+C y^2\right)$ Taking L.H.S. $(x+y)^3+8(x-y)^3$ $=(x+y)^3+[2(x-y)]^3$ Using identity, $a^3+b^3=(a+b)(a^2-ab+b^2)$ $=[x+y+2(x-y)][(x+y)^2-2(x+y)(x-y)+[2(x-y)]^2]$ $=(3x-y)[(x^2+y^2+2xy) -2(x^2-y^2)+[4(x^2+y^2-2xy]]$ $=(3x-y)[x^2+y^2+2xy -2x^2+2y^2+4x^2+4y^2-8xy]$ $=(3x-y)(3x^2 -6xy+7y^2)$ On comparing the coefficient with R.H.S., we get, ⇒ A = –1 ⇒ B = –6 ⇒ C = 7 $\therefore$ (A + B + C) = – 1 – 6 + 7 = 0 Hence, the correct answer is 0.
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Question : If $27 x^3-64 y^3=(A x+B y)\left(C x^2-D y^2+12 x y\right)$, then the value of $4 A+B+3 C+2 D$ is:
Option 1: 5
Option 2: 3
Option 3: –3
Option 4: –4
Question : The factors of $x^2+4 y^2+4 y-4 x y-2 x-8$ are:
Option 1: $(x-2 y-4)(x-2 y+2)$
Option 2: $\left(x^2-2 y-4\right)\left(x^2-2 y+2\right)$
Option 3: $(x+2 y-4)(x+2 y+2)$
Option 4: $\left(x^2-2 y-4\right)\left(x^2+2 y+2\right)$
Question : If $\sin Y=x$, then what will be the value of $\cos 2Y\left(\right.$ where $\left.0 \leq Y \leq 90^{\circ}\right)$?
Option 1: $(\sqrt{ 2} -1) x$
Option 2: $\sqrt{2} x$
Option 3: $1-2 x$
Option 4: $1-2 x^2$
Question : If $x^{2} -3x +1=0$, then the value of $\frac{\left(x^4+\frac{1}{x^2}\right)}{\left(x^2+5 x+1\right)}$ is:
Option 1: $\frac{9}{4}$
Option 2: $\frac{27}{8}$
Option 3: $\frac{5}{2}$
Option 4: $2$
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$?
Option 1: 1
Option 2: 0
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