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Multiple choice maths repeated multiplication exponentiation index notation and products of prime factors powers

The value of $\displaystyle\frac{2^{m+3}\times3^{2m-n}\times5^{m+n+3}6^{n+1}}{6^{m+1}\times10^{n+3}\times15^m}$ is equal to

  1. 0

  2. 1
  3. $2^m$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Numerator $=2^{m+3}\cdot3^{2m-n}\cdot5^{m+n+3}\cdot2^{n+1}\cdot3^{n+1}$
$=2^{m+n+4}\cdot3^{2m+1}\cdot5^{m+n+3}$ (i)
Denominator $=2^{m+1}\cdot3^{m+1}\cdot2^{n+3}\cdot5^{n+3}\cdot3^m\cdot5^m$
$=2^{m+n+4}\cdot3^{2m+1}\cdot5^{m+n+3}$ (ii)
Given expression $=1$

Multiple choice maths repeated multiplication exponentiation index notation and products of prime factors powers

What is the unit digit in ${({6374}^{1793}\times {625}^{317}\times{341}^{491})}$?

  1. $0$
  2. $2$
  3. $3$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Unit digit in ${6374}^{1993}=$ Unit digit in ${(4)}^{1793}$
=Unit digit in $[{({4}^{2})}^{896}\times 4]$
=Unit digit in $(6\times 4)=4$
Unit digit in ${(625)}^{317}=$ Unit digit in ${(5)}^{317}=5$
Unit digit in ${(341)}^{491}=$ Unit digit in ${(1)}^{491}=1$
Required digit$=$ Unit digit in $(4\times 5\times1)=0$

Multiple choice maths repeated multiplication exponentiation index notation and products of prime factors powers

The number of values of $x\ \epsilon \ [0,5]$ at which $f(x)=|x-\dfrac{1}{4}|+|x-2|+\tan{x}$ is not differentiable are

  1. $0$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The function contains absolute value terms and a trigonometric term. The absolute value expressions are not differentiable at their critical points where the inside equals zero, and tan(x) has discontinuities and points of non-differentiability within the interval [0, 5], culminating in 4 points total.

Multiple choice maths repeated multiplication exponentiation index notation and products of prime factors powers

The greatest of the number 
$1,2^{1/2},3^{1/3},4^{1/4},5^{1/5}, 6^{1/6}, and \ 7^{1/7}$ is

  1. $2^1/2$
  2. $3^1/3$
  3. $7^1/7$
  4. $4^1/4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To find the maximum of x^(1/x), consider the function f(x) = x^(1/x). Its derivative is f'(x) = x^(1/x) * (1 - ln(x)) / x^2. The maximum occurs at x = e (approx 2.718). Comparing values near e, 3^(1/3) is larger than 2^(1/2) and 4^(1/4).

Multiple choice maths repeated multiplication exponentiation index notation and products of prime factors powers

Find the value of $\displaystyle\frac{5^0+5^{-1}}{5^0-5^{-1}}-\left(\frac{8}{27}\right)^{\displaystyle\frac{1}{3}}-\left(\frac{36}{25}\right)^{-\displaystyle\frac{1}{3}}$

  1. 0
  2. $\displaystyle\frac{1}{2}$

  3. 1

  4. 2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle\frac{\displaystyle1+\frac{1}{5}}{\displaystyle1-\frac{1}{5}}-\left[\left(\frac{2}{3}\right)^3\right]^{\displaystyle\frac{1}{3}}-\left[\left(\frac{6}{5}\right)^2\right]^{\displaystyle-\frac{1}{2}}=\frac{\displaystyle\frac{6}{5}}{\displaystyle\frac{4}{5}}-\left(\frac{2}{3}\right)^1-\left(\frac{6}{5}\right)^{-1}=\frac{6}{4}-\frac{2}{3}-\frac{5}{6}=\frac{18-8-10}{12}=0$

Multiple choice physics dynamics - explaining motion exponentiation index notation and products of prime factors powers

Arrange the following lengths in the increasing order of their magnitudes :
1 meter, 1 centimeter, 1 kilometer, 1 millimeter .

  1. 1 meter < 1 centimeter < 1 kilometer < 1 millimeter

  2. 1 millimeter < 1 meter < 1 centimeter < 1 kilometer

  3. 1 millimeter < 1 centimeter < 1 meter < 1 kilometer

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

1 kilometer = 10$^3$ meter

1 centimeter = 10$^{-2}$ meter
1 millimeter = 10$^{-3}$ meter

Ascending order of length :
1 millimeter < 1 centimeter < 1meter < 1 kilometer

Multiple choice physics dynamics - explaining motion exponentiation index notation and products of prime factors powers

The wrong unit conversion among the following is

  1. 1 angstrom = $10^{-10} m$
  2. 1 fermi = $10^{-15}$ m
  3. 1 light year = $9.46 \times 10^{15} m$
  4. 1 astronomical unit = $1.496 \times 10^{-11} m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$A)1 $Angstrom: $1$ Ångström is equal to$ 0.1 $nanometers. Expressed in the SI base unit meters $1$ Ångström is equal to $1 \times 10^{-10}$ meters. $1 Å = 0.0000000001 m.$

$B)$Definition of fermi unit: $≡ 1\times 10^{−15} m.$

$C)$The light-year is a unit of length used to express astronomical distances and measures about $9.46$ trillion kilometres $(9.46 \times 10^{12} km)$ or $5.88$ trillion miles $(5.88 \times 10^{12} mi).$

$D)$The total radiation power received from the sun on a unit area perpendicular to the sun rays at the mean earth-sun distance, termed an astronomical unit, is called the solar constant (SC), where $1$ astronomical unit $= 1 AU = 1.496 \times 10^{11} m.$

So $D$ is wrong unit conversion 

Multiple choice physics dynamics - explaining motion exponentiation index notation and products of prime factors powers

A calorie is a unit of heat energy and its value is 4.18 J where $1 J = 1 kg m^2 s^{-2}$. Suppose we use a new system of units in which unit of mass equals $\alpha$ kg, the unit of length equals $\beta$ m and the unit of the time is $\gamma$ sec. Then the value of a calorie in the new system of units is then

  1. 4.18 $\displaystyle \frac{\gamma^2}{\alpha \beta^2}$
  2. 4.18 $\displaystyle \frac{\alpha \beta^2}{\gamma^2}$
  3. 4.18 $\displaystyle \frac{\gamma^2}{\alpha}$
  4. 4.18 $\displaystyle \frac{\beta^2}{\alpha \gamma^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$1 J = (1 kg) ( 1 m)^2 (1 sec)^{-2}$
$1 x = (\alpha kg) (\beta m)^2 (\gamma sec)^{-2}$
$\displaystyle \therefore \frac{1 J}{1x} = \left( \frac{1}{\alpha} \right) \left( \frac{1}{\beta} \right)^2 (\gamma)^2 = \frac{\gamma^2}{\alpha \beta^2} $
$\displaystyle \therefore 1 J = \frac{\gamma^2}{\alpha \beta^2}$ or $\displaystyle 1 cal = 4.18 \displaystyle \frac{\gamma^2}{\alpha \beta^2}$