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Elementary Mathematics

71). If \( \Large y+\frac{1}{z}=1\ and\ x+\frac{1}{y}=1 \), then what is the value of xyz?

A). 1

B). -1

C). 0

D). \( \Large \frac{1}{2} \)

View Answer

Correct Answer: -1

Given : \( \Large y+\frac{1}{z}=1 \)

=> \( \Large yz+1 = z \)

=> \( \Large y = \frac{z-1}{z} \)

and \( \Large x+\frac{1}{y} = 1 \)

=> \( \Large xy+1 = y \)

\( \Large \therefore xy+1 = \frac{z-1}{z} \)

=> \( \Large xyz+z = z-1 \)

=>\( \Large xyz = -1 \)

72). What are the values of A and B respectively, if \( \Large \frac{\sqrt{5}-1}{\sqrt{5}+1}+\frac{\sqrt{5}+1}{\sqrt{5}-1} = A+B\sqrt{5}? \)

A). 3, 0

B). 0, 3

C). 0. 0

D). 3, 3

View Answer

Correct Answer: 3, 0

\( \Large \frac{\sqrt{5}-1}{\sqrt{5}+1}+\frac{\sqrt{5}+1}{\sqrt{5}-1} \)

= \( \Large \frac{\sqrt{5}-1}{\sqrt{5}+1} \times \frac{\sqrt{5}-1}{\sqrt{5}-1}+\frac{\sqrt{5}+1}{\sqrt{5}-1} \times \frac{\sqrt{5}+1}{\sqrt{5}+1} \)

= \( \Large \frac{\left(\sqrt{5}-1\right)^{2}}{ \left(\sqrt{5}\right)^{2}- \left(1\right)^{2} } + \frac{ \left(\sqrt{5}+1\right)^{2} }{ \left(\sqrt{5}\right)^{2}- \left(1\right)^{2} } \)

= \( \Large \frac{5+1-2\sqrt{5}}{5-1} + \frac{5+1+2\sqrt{5}}{5-1} \)

= \( \Large \frac{6-2\sqrt{5}+6+2\sqrt{5}}{4} \)

= \( \Large 3 \)

But \( \Large \frac{\sqrt{5}-1}{\sqrt{5}+1}+\frac{\sqrt{5}+1}{\sqrt{5}-1} = A+B\sqrt{5} \)

A = 3 and B = 0.

73). If \( \Large x^{4}+y^{4}=17\ and\ x+y=1 \), then what is the value of \( \Large x^{2}y^{2}-2xy \)?

A). 8

B). 10

C). 12

D). 16

View Answer

Correct Answer: 8

Given: \( \Large x^{4}+y^{4}=17 \)

and \( \Large x+y=1 \)

=> \( \Large \left(x+y\right)^{2} = 1 \)

=> \( \Large x^{2}+y^{2}+2xy = 1 \)

=> \( \Large \left(x^{2}+y^{2}+2xy\right)^{2} = 1 \)

=> \( \Large x^{4}+y^{4}+4x^{2}y^{2}+2x^{2}y^{2}+4xy^{3}+4x^{3}y = 1 \)

=> \( \Large 17+6x^{2}y^{2}+4xy \left(x^{2}+y^{2}\right) = 1 \)

=> \( \Large 17+6x^{2}y^{2}+4xy \left(1-2xy\right) = 1 \)

=> \( \Large 17+6x^{2}y^{2}+4xy-8x^{2}y^{2} = 1 \)

=> \( \Large 16 = 2x^{2}y^{2}-4xy \)

=> \( \Large x^{2}y^{2}-2xy = 8 \)

74). If \( \Large \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \), where \( \Large a+b+c \ne 0 \), \( \Large abc \ne 0 \), then what is the value of \( \Large \left(a+b\right) \left(b+c\right) \left(c+a\right)? \)

A). 0

B). 1

C). -1

D). 2

View Answer

Correct Answer: 0

Given: \( \Large \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \)

=> \( \Large \left(a+b+c\right) \left(ab+bc+ca\right)=abc \)

=> \( \Large b^{2}c+bc^{2}+abc+ca^{2}+ab^{2}+abc+a^{2}b+a^{2}c = 0 \)

=> \( \Large \left(b^{2}+bc+ba+ca\right) \left(c+a\right) = 0 \)

=> \( \Large \left(a+b\right) \left(b+c\right) \left(c+a\right) = 0 \)

75). \( \Large \frac{ \left(x+5\right) \left(x^{2}+7x+10\right) }{ \left(x-3\right) \left(x^{2}+10x+25\right) } \) is equal to

A). \( \Large \frac{ \left(x+1\right) }{x-1} \)

B). \( \Large \frac{ \left(x+2\right) }{x-3} \)

C). \( \Large \frac{ \left(x-3\right) }{x+2} \)

D). \( \Large \frac{ \left(x+7\right) }{x-7} \)

View Answer

Correct Answer: \( \Large \frac{ \left(x+2\right) }{x-3} \)

\( \Large \frac{ \left(x+5\right) \left(x^{2}+7x+10\right) }{ \left(x-3\right) \left(x^{2}+10x+25\right) } \)

=\( \Large \frac{ \left(x+5\right) \left(x^{2}+5x+2x+10\right) }{ \left(x-3\right) \left(x^{2}+5x+5x+25\right) } \)

=\( \Large \frac{ \left(x+5\right)\left[ x \left(x+5\right)+2 \left(x+5\right) \right] }{ \left(x-3\right)\left[ x \left(x+5\right)+5 \left(x+5\right) \right] } \)

= \( \Large \frac{ \left(x+5\right)\left[ \left(x+5\right) \left(x+2\right) \right] }{ \left(x-3\right)\left[ \left(x+5\right) \left(x+5\right) \right] } \)

= \( \Large \frac{x+2}{x-3} \)

76). \( \Large 40-18x-x^{2} \) is equal to

A). \( \Large \left(5-x\right) \left(8+x\right) \)

B). \( \Large \left(5-x\right) \left(10+x\right) \)

C). \( \Large \left(2+x\right) \left(20-x\right) \)

D). \( \Large \left(2-x\right) \left(20+x\right) \)

View Answer

77). \( \Large 2x+1 \) is a factor of

A). \( \Large 2x^{2}+x-1 \)

B). \( \Large 2x^{2}-x-1 \)

C). \( \Large 2x^{2}-x+1 \)

D). \( \Large 2x^{2}+x+1 \)

View Answer

Correct Answer: \( \Large 2x^{2}-x-1 \)

\( \Large 2x^{2}-x-1 = 2x^{2}-2x+x-1 \)

= \( \Large 2x \left(x-1\right)+1 \left(x-1\right) \)

= \( \Large \left(2x+1\right) \left(x-1\right) \)

Alternately \( \Large \left(2x+1\right) \) is a factor of polynomial \( \Large p \left(x\right) \)

If \( \Large p \left(-\frac{1}{2}\right) = 0 \)

Clearly, \( \Large p \left(x\right)=2x^{2}-x-1 \)

\( \Large \therefore p \left(-\frac{1}{2}\right)=2 \times \left(\frac{1}{4}\right)- \left(-\frac{1}{2}\right)-1 = 0 \)

78). \( \Large 6x^{2}+ax+7 \) when divided by \( \Large x-2 \) gives remainder 13. The value of a is

A). 0

B). -9

C). 5

D). 7

View Answer

Correct Answer: -9

By Remainder theorem, when \( \Large f \left(x\right) \) is divided by \( \Large g \left(x\right)=x-2 \), the remainder is equal to \( \Large f \left(2\right) \).

Now, \( \Large f \left(x\right)=6x^{2}+ax+7 \)

=> \( \Large f \left(2\right)=6 \times 2^{2}+a \times 2+7=2a+31 = 13 \)

=> \( \Large 2a = -18 \)

=> \( \Large a = -9 \)

79). When is the expression \( \Large x^{2}+3x-10 \) positive only?

A). \( \Large x \le - 5 \)

B). \( \Large x\ge 2 \)

C). \( \Large -5 < x < 2 \)

D). \( \Large x < -5 or x > 2 \)

View Answer

80). If \( \Large a+b+c=0 \), then what is the value of \( \Large \left(a+b-c\right)^{3}+ \left(c+a-b\right)^{3}+ \left(b+c-a\right)^{3} \)?

A). \( \Large 8 \left(a^{3}+b^{3}+c^{3}\right) \)

B). \( \Large a^{3}+b^{3}+c^{3} \)

C). 24 abc

D). -24 abc

View Answer

Correct Answer: -24 abc

\( \Large a+b-c+c+a-b+b+c-a = a+b+c = 0 \)

...\( \Large \because a+b+c=0 \)

\( \Large \therefore \left(a+b-c\right)^{3}+ \left(c+a-b\right)^{3}+ \left(b+c-a\right)^{3} \)

= \( \Large 3 \left(a+b-c\right) \left(c+a-b\right) \left(b+c-a\right) \)

=\( \Large 3 \left(-c-c\right) \left(-b-b\right) \left(-a-a\right) \)

...\( \Large \because a+b+c=0 \)

= \( \Large 3 \left(-2c\right) \left(-2b\right) \left(-2a\right) \)

= -24abc

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