Questions

1 Marks Question

🎯

Test yourself on this topic

21 questions · timed · auto-graded

Question 11 Mark
If $a=3$ and $b=-2$, find the values of:
$a+b^{a b}$
Answer
We have, $a^b+b^a=(3+(-2))^{3(-2)}$
$=(3-2)^{-6}$
$=(1)^{-6}$
$=1$
View full question & answer
Question 21 Mark
If $a = 3$ and $b = -2,$ find the values of:
$\text{a}^\text{b}+\text{b}^\text{a}$
Answer
We have,$\text{a}^\text{b}+\text{b}^\text{a}$
$=3^{-2}+(-2)^{3}$
$=\big(\frac{1}{3}\big)^2+(-2)^3$
$=\frac{1}{9}+-8$
$=-\frac{71}{9}$
View full question & answer
Question 31 Mark
Simplify the following:
$\left(2 x^{-2} y^3\right)^3$
 
Answer
$\left(2 x^{-2} y^3\right)^3$
$\left(2^3 \times x^{-2} \times 3 y^{3 \times 3}\right)=8 x^{-6} y^9$
View full question & answer
Question 41 Mark
If $a = 3$ and $b = -2,$ find the values of:
$\text{a}^\text{a}+\text{b}^\text{b}$
Answer
We have,
$\text{a}^\text{a}+\text{b}^\text{b}$
$=3^3+(-2)^{-2}$
$=3^3+(-\frac{1}{2})^2$
$=27+\frac{1}{4}$
$=\frac{109}{4}$
View full question & answer
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$].$
View full question & answer
Question 61 Mark
Simplify the following:
$3\left(a^4 b^3\right)^{10} \times 5\left(a^2 b^2\right)^3$
 
Answer
$3\left(a^4 b^3\right)^{10} \times 5\left(a^2 b^2\right)^3$
$=3\left(a^{40} b^{30}\right) \times 5\left(a^6 b^6\right)$
$=15\left(a^{46} b^{36}\right)$
View full question & answer
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: $2^{-a}\times3^​\text{b}​\times7^\text{c}=2^{-3}\times3^2\times7^1$ $2^{-3}\times3^2\times7^1=\frac{1}{8}\times9\times7$ $\frac{63}{8}$
View full question & answer
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 $\text{c}=1.$
View full question & answer
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.
View full question & answer
1 Marks Question - Maths STD 9 Questions - Vidyadip