1 Mark Question

Question 11 Mark

If $a=3$ and $b=-2$, find the values of: $a+b^{ b } b$

Answer

We have,
$a^b+b^a$
$=(3+(-2))^{3(-2)}$
$=(3-2)^{-6}$
$=(1)^{-6}=1$

Question 21 Mark

If a = 3 and b = -2, find the values of:$\text{a}^\text{b}+\text{b}^\text{a}$

Answer

We have,$\text{a}^\text{b}+\text{b}^\text{a}$
$=3^{-2}+(-2)^{3}$
$=\big(\frac{1}{3}\big)^2+(-2)^3$
$=\frac{1}{9}+-8$
$=-\frac{71}{9}$

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Question 31 Mark

Simplify the following:
$\left(2 x^{-2} y^3\right)^3$

Answer

$\left(2 x^{-2} y^3\right)^3$
$\left(2^3 \times x^{-2 \times 3} y^3 \times 3\right)=8 x^{-6} y^9$

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Question 41 Mark

If a = 3 and b = -2, find the values of:$\text{a}^\text{a}+\text{b}^\text{b}$

Answer

We have,$\text{a}^\text{a}+\text{b}^\text{b}$
$=3^3+(-2)^{-2}$
$=3^3+(-\frac{1}{2})^2$
$=27+\frac{1}{4}$
$=\frac{109}{4}$

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Question 51 Mark

If $493992=a^4 b^2 c^3$, find the values of $a, b$ and $c$ are and are different positive primes.

Answer

Taking out the LCM , the factors are $2^4, 3^2$ and $7^3 a ^4 b^2 c ^3=2^4, 3^2$ and $7^3 a =2, b=3$ and $c =7[$ since, a b and c are primes].

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Question 61 Mark

Simplify the following:
$3\left(a^4 b^3\right)^{10} \times 5\left(a^2 b^2\right)^3$

Answer

$3\left(a^4 b^3\right)^{10} \times 5\left(a^2 b^2\right)^3$
$=3\left(a^{40} b^{39}\right) \times 5\left(a^6 b^6\right)$
$=15\left(a^{46} b^{36}\right)$

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Question 71 Mark

Given $4725=3^\text{a}\ 5^\text{b}\ 7^\text{c}$ find,The value of $2^{-\text{a}}3^\text{b}7^\text{c}$

Answer

The value of $2^{-\text{a}}\times3^\text{b}\times7^\text{c}$ Sol:$2^{-a}\times3^\text{b}\times7^\text{c}=2^{-3}\times3^2\times7^1$
$2^{-3}\times3^2\times7^1=\frac{1}{8}\times9\times7$
$\frac{63}{8}$

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Question 81 Mark

Given $4725=3^\text{a}\ 5^\text{b}\ 7^\text{c}$ find,The integral values of a, b and c

Answer

Taking out the LCM of 4725, we get$3^3\times5^2\times7^1=3^\text{a}\times5^\text{b}\times7^\text{c}$
By comparing, we get$\text{a}=\text{3},\ \text{b}=2$ and $\text{c}=1.$

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Question 91 Mark

Prove that:$\Big(\frac{\text{x}^\text{a}}{\text{x}^\text{b}}\Big)^\text{c}\times\Big(\frac{\text{x}^\text{b}}{\text{x}^\text{c}}\Big)^\text{a}\times\Big(\frac{\text{x}^\text{c}}{\text{x}^\text{a}}\Big)^\text{b}=1$

Answer

To prove,$\Big(\frac{\text{x}^\text{a}}{\text{x}^\text{b}}\Big)^\text{c}\times\Big(\frac{\text{x}^\text{b}}{\text{x}^\text{c}}\Big)^\text{a}\times\Big(\frac{\text{x}^\text{c}}{\text{x}^\text{a}}\Big)^\text{b}=1$
Left hand side (LHS) = Right hand side (RHS) Considering LHS,$=\Big(\frac{\text{x}^{\text{ac}}}{\text{x}^{\text{bc}}}\Big)\times\Big(\frac{\text{x}^{\text{ba}}}{\text{x}^{\text{ca}}}\Big)\times\Big(\frac{\text{x}^{\text{bc}}}{\text{x}^{\text{ab}}}\Big)$
$=\text{x}^{\text{ac}-\text{bc}}\times\text{x}^{\text{ba}-\text{ca}}\times\text{x}^{\text{bc}-\text{ab}}$
$=\text{x}^{\text{ac}-\text{bc}+\text{ba}-\text{ca}+\text{bc}-\text{ab}}$
$=\text{x}^0$
$=1$
Therefore, LHS = RHS Hence proved.

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