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Maths 3 Mark Question

Operations on Algebraic Expressions · Maharashtra Board STD 8 (MSBSHSE - English Medium). 22 questions.

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3 Mark Question

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22 questions · timed · auto-graded

Question 13 Marks
If $\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$ Find the values of:
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$
Answer
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$
Squaring on both sides:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=(4)^2$
$\Rightarrow​​\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}+\frac{1}{\text{x}}=16$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}+2=16$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=16-2=14$
$\therefore\text{x}^2+\frac{1}{\text{x}^2}=14$
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Question 23 Marks
Write the quotient and remainder when we divide:
$\left(2 x^3-5 x^2+8 x-5\right)$ by $\left(2 x^2-3 x+5\right)$
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Question 43 Marks
Find the following products:
$\Big(\text{x}^4+\frac{1}{\text{x}^4})×\Big(\text{x}+\frac{1}{\text{x}}\Big)$
Answer
$\Big(\text{x}^4+\frac{1}{\text{x}^4})×\Big(\text{x}+\frac{1}{\text{x}}\Big)$$=\text{x}^4\Big(\text{x}+\frac{1}{\text{x}}\Big)+\frac{1}{\text{x}^4}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$=\text{x}^4\times\text{x}+\frac{1}{\text{x}}+\frac{1}{\text{x}^4}\times\text{x}+\frac{1}{\text{x}^4}\times\frac{1}{\text{x}}$
$=\text{x}^5+\text{x}^3+\frac{1}{\text{x}^3}+\frac{1}{\text{x}^5}$
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Question 53 Marks
Write the quotient and remainder when we divide:
$\left(x^3-6 x^2+11 x-6\right)$ by $\left(x^2-5 x+6\right)$
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Question 63 Marks
Write the quotient and remainder when we divide:
$\left(x^4-2 x^3+2 x^2+x+4\right) \text { by }\left(x^2+x+1\right)$
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Question 73 Marks
If $x+y=12$ and $x y=14$, find the value of $\left(x^2+y^2\right)$
Answer
$x+y=12$
On squaring both the sides:
$(x+y)^2=(12)^2$
$\Rightarrow x^2+y^2+2 x y=144$
$\Rightarrow x^2+y^2+2 \times 14=144$
$\Rightarrow x^2+y^2+28=144$
$\Rightarrow x^2+y^2=144-28$
$x^2+y^2=116$
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Question 83 Marks
Find following products:
$\Big(\frac{1}{3}\text{x}^2-9\Big)\Big(\frac{1}{3}\text{x}^2-9\Big)$
Answer
$\Big(\frac{1}{3}\text{x}^2-9\Big)\Big(\frac{1}{3}\text{x}^2-9\Big)$
$=\Big(\frac{1}{3}\text{x}^2-9\Big)^2$
$=\Big(\frac{1}{3}\text{x}^2\Big)^2-2\times\frac{1}{3}\text{x}^2\times9+(9)^2 $
$=\frac{1}{9}\text{x}^4-6\text{x}^2+81$
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Question 93 Marks
If $x-y=7$ and $x y=9$, find the value of $\left(x^2+y^2\right)$.
Answer
$x+y=7$
Squaring both sides:
$(x-y)^2=(7)^2$
$\Rightarrow x^2+y^2-2 x y=49$
$\Rightarrow x^2+y^2-2 \times 9=49$
$\Rightarrow x^2+y^2-18=49$
$\Rightarrow x^2+y^2=49+18=67$
$x^2+y^2=67$
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Question 103 Marks
Write the quotient and remainder when we divide:
$\left(5 x^3-12 x^2+12 x+13\right)$ by $\left(x^2-3 x+4\right)$
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Question 113 Marks
If $\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$ Find the values of:
$\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)$
Answer
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=(14)^2$
Squaring on both sides:
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)^2+2\times\text{x}^2\times\frac{1}{\text{x}^2}=196$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}+2=196$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=196-2=194$
$\therefore\text{x}^4+\frac{1}{\text{x}^4}=194$
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Question 123 Marks
Write the quotient and remainder when we divide:
$\left(8 x^4+10 x^3-5 x^2-4 x+1\right) \text { by }\left(2 x^2+x-1\right)$
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Question 143 Marks
Write the quotient and remainder when we divide : $(x^2 - 4x + 4)$ by $(x - 2)$
Answer
On dividing polynomial by a binomial we have divide every variables of polynomial
By binomial we get
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Question 153 Marks
If $\Big(\text{x}-\frac{1}{\text{x}}\Big)=5$ Find the values of:
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$
Answer
$\Big(\text{x}-\frac{1}{\text{x}}\Big)=5$
Squaring on both sides:
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=(5)^2$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}-2(\text{x})\times\Big(\frac{1}{\text{x}}\Big)\Big)=25$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-2=25$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=25+2=27$
$\therefore\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=27$
Therefore, the value of $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)\text{ is }27$
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Question 163 Marks
Find following products:
$\Big(\frac{3}{4}\text{x}-\frac{5}{6}\text{y}\Big)\Big(\frac{3}{4}\text{x}-\frac{5}{6}\text{y}\Big)$
Answer
$\Big(\frac{3}{4}\text{x}-\frac{5}{6}\text{y}\Big)\Big(\frac{3}{4}\text{x}-\frac{5}{6}\text{y}\Big)$
$=\Big(\frac{3}{4}\text{x}-\frac{5}{6}\text{y}\Big)^2​​​​​​​$
$=\Big(\frac{3}{4}\text{x}\Big)^2-2×\frac{3}{4}\text{x}×\frac{5}{6}\text{y}+\Big(\frac{5}{6}\text{y}\Big)^2 $
$=\frac{9}{16}\text{x}^2-\frac{5}{4}\text{xy}+\frac{25}{36}\text{y}^2$
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Question 173 Marks
If $\Big(\text{x}-\frac{1}{\text{x}}\Big)=5$ Find the values of:
$\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)$
Answer
$\Big(\text{x}^2+\frac{1}{\text{x}^2}=27\Big)$Squaring on both sides:
$\Rightarrow\Big(\text{x}^4-\frac{1}{\text{x}^4}-2(\text{x}^2\Big(\frac{1}{\text{x}^2}\Big)\Big)=(27)^2$ $\Rightarrow\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)-2=729$ $\Rightarrow\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)=729+2$ $\Rightarrow\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)=731$ Therefore, the value of $\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)\text { is }731.$
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Question 193 Marks
Find following products:
$\Big(\text{x}-\frac{3}{\text{x}}\Big)\Big(\text{x}-\frac{3}{\text{x}}\Big)$
Answer
$\Big(\text{x}-\frac{3}{\text{x}}\Big)\Big(\text{x}-\frac{3}{\text{x}}\Big)$
$=\Big(\text{x}-\frac{3}{\text{x}}\Big)^2$
$=(\text{x})^2-2\times\text{x}\times\frac{3}{\text{x}}+\Big(\frac{3}{\text{x}}\Big)^2 $
$=\text{x}^2-6+\frac{9}{\text{x}^2}$
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Question 203 Marks
Find following products:
$\Big(\frac{1}{2}\text{y}^2-\frac{1}{3}\text{y}\Big)\Big(\frac{1}{2}\text{y}^2-\frac{1}{3}\text{y}\Big)$
Answer
$\Big(\frac{1}{2}\text{y}^2-\frac{1}{3}\text{y}\Big)\Big(\frac{1}{2}\text{y}^2-\frac{1}{3}\text{y}\Big)$
$=\Big(\frac{1}{2}\text{y}^2-\frac{1}{3}\text{y}\Big)^2$
$=\Big(\frac{1}{2}\text{y}^2\Big)2-2\times\frac{1}{2}\text{y}^2\times\frac{1}{3}\text{y}+\Big(\frac{1}{3}\text{y}\Big)^2$
$=\frac{1}{4}\text{y}^4-\frac{1}{3}\text{y}^3+\frac{1}{9}\text{y}^2$
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Question 223 Marks
Find the following product: $\left(9 x^2-x+15\right) \times\left(x^2-x-1\right)$
Answer
We have;
$\left(9 x^2-x+15\right) \times\left(x^2-x-1\right) $
$=9 x^2\left(x^2-x-1\right)-x\left(x^2-x-1\right)+15\left(x^2-x-1\right) $
$=9 x^4-9 x^3-9 x^2-x^3+x^2+x+15 x^2-15 x-15$
Putting equal power terms together, we get,
$=9 x^4-9 x^3-x^3-9 x^2+x^2+15 x^2-15 x+x-15 $
$=9 x^4-10 x^3+7 x^2-14 x-15$
 
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3 Mark Question - Maths STD 8 Questions - Vidyadip