If x > 1, and x + \( \frac{1}{x} \)= \(\sqrt{29}\), what is the value of x - \( \frac{1}{x} \)?
The question gives the value of the sum of a number and its reciprocal. We need to find the difference between the number and its reciprocal using an algebraic identity. The condition x > 1 helps us determine the positive value of the difference.
Identify the given information:
\[ x+\frac{1}{x}=\sqrt{29} \]Also:
\[ x>1 \]We need to find:
\[ x-\frac{1}{x} \]Recall the algebraic identity:
We know that:
\[ (a-b)^2=(a+b)^2-4ab \]Taking:
\[ a=x,\quad b=\frac{1}{x} \]Since:
\[ ab=x\times\frac{1}{x}=1 \]The identity becomes:
\[ \left(x-\frac{1}{x}\right)^2 =\left(x+\frac{1}{x}\right)^2-4 \]Substitute the given value:
Since:
\[ x+\frac{1}{x}=\sqrt{29} \]Substituting into the identity:
\[ \left(x-\frac{1}{x}\right)^2 =(\sqrt{29})^2-4 \] \[ =29-4 \] \[ =25 \]Find the square root:
\[ x-\frac{1}{x}=\pm\sqrt{25} \] \[ =\pm5 \]Since x > 1, we have:
\[ x>\frac{1}{x} \]Therefore, the difference must be positive.
\[ x-\frac{1}{x}=5 \]Determine the final answer:
The required value is 5.
Correct Option: C) 5 — Using the identity:
\[ \left(x-\frac{1}{x}\right)^2 =\left(x+\frac{1}{x}\right)^2-4 \]
Substituting the given value:
\[ =29-4=25 \]
Since x > 1, the difference is positive:
\[ x-\frac{1}{x}=5 \]
Hence, Option C is the correct answer.
A) 4: This is incorrect because the square of the required difference is 25, whereas the square of 4 is 16.
B) 2: This is incorrect because the required difference is 5, not 2.
D) 3: This is incorrect because the square of 3 is 9, whereas the algebraic identity gives 25.
The question asks about the language in which Al-Biruni wrote his famous book 'Kitab-ul-Hind'. To answer this, one needs to recall historical facts about Al-Biruni and his literary works.
Correct Option: A) Arabic
This question tests the ability to identify implicit assumptions in a given statement. An assumption is something taken for granted or supposed without proof. For an assumption to be implicit, it must be a necessary underlying belief for the statement to be made. We need to evaluate if the statement "Nowadays, Indian cricket team plays very well" logically requires either of the given assumptions to be true.
D) Neither assumptions I nor II are implicit.
To arrange the historical events in chronological order, we need to recall the approximate dates or periods for each event mentioned. Once the dates are established, they can be ordered from earliest to latest.
Based on these dates, the chronological order is:
Thus, the correct sequence is C, A, D, B.
Correct Option: C) C, A, D, B
The question asks to identify the country that has participated in every FIFA Football World Cup since its inception. This requires knowledge of FIFA World Cup history and the participation records of major footballing nations.
Correct Option: D) Brazil has participated in all 22 FIFA World Cup tournaments held from 1930 to 2022, making them the only country with this distinction.
The question describes a fundamental principle relating force and momentum. To answer correctly, we need to recall Newton's Laws of Motion and identify which law specifically defines the relationship between the rate of change of momentum and the applied force.
Correct Option: A) Newton's Second Law of Motion
Find the value of r such that the mean of the first r odd natural numbers is \( \frac{r^2}{16} \).
To solve this problem, we need to recall the formula for the sum of the first \(r\) odd natural numbers and the formula for the mean of a set of numbers. We will then equate the calculated mean with the given mean expression and solve for \(r\).
Identify the first \(r\) odd natural numbers:
The first \(r\) odd natural numbers are \(1, 3, 5, \ldots, (2r-1)\).
Calculate the sum of the first \(r\) odd natural numbers:
The sum of the first \(r\) odd natural numbers is given by the formula \(S_r = r^2\).
Alternatively, this is an arithmetic progression with first term \(a=1\), common difference \(d=2\), and number of terms \(r\). The sum is:
\[ S_r = \frac{r}{2} [2a + (r-1)d] = \frac{r}{2} [2(1) + (r-1)2] = \frac{r}{2} [2 + 2r - 2] = \frac{r}{2} [2r] = r^2 \]Calculate the mean of the first \(r\) odd natural numbers:
The mean is the sum divided by the number of terms. So, the mean (\(\text{Mean}\)) is:
\[ \text{Mean} = \frac{S_r}{r} = \frac{r^2}{r} = r \]Equate the calculated mean with the given mean:
The problem states that the mean of the first \(r\) odd natural numbers is \( \frac{r^2}{16} \). So, we set our calculated mean equal to the given mean:
\[ r = \frac{r^2}{16} \]Solve for \(r\):
We can rearrange the equation:
\[ 16r = r^2 \]Move all terms to one side:
\[ r^2 - 16r = 0 \]Factor out \(r\):
\[ r(r - 16) = 0 \]This gives two possible solutions for \(r\):
\[ r = 0 \quad \text{or} \quad r - 16 = 0 \implies r = 16 \]Since \(r\) represents the number of odd natural numbers, \(r\) must be a positive integer. Therefore, \(r=0\) is not a valid solution in this context.
Thus, \(r = 16\).
A) 16 is the correct answer because when \(r=16\), the mean of the first 16 odd natural numbers is \(16\), which is equal to \[\frac{16^2}{16} = \frac{256}{16} = 16\].
The question asks to identify the recipient of the Bal Shakti Puraskar-2019 for a specific innovation: a breakthrough technology for diagnosing pre-symptomatic protein energy malnutrition non-invasively using paper. This requires knowledge of the awardees and their innovations for that particular year.
Correct Option: D) Mohammed Suhail Chinya Salimpasha was awarded the Bal Shakti Puraskar in 2019 for his innovation in developing a breakthrough technology for diagnosing pre-symptomatic protein energy malnutrition non-invasively using paper. This innovation aims to provide an affordable and accessible method for early detection of malnutrition, particularly in remote areas.
The question asks to identify the mountain pass that does NOT connect Uttarakhand with Tibet. This requires knowledge of the geographical location of various Himalayan passes.
Correct Option: D) Shipki La
This problem involves compound interest. The key concept here is that for a fixed interest rate, if the principal amount multiplies by a certain factor in a given time period, it will multiply by the same factor again in the same time period. We can use the formula for compound interest or a more intuitive approach based on the multiplicative nature of compound interest.
Let the principal sum be \(P\) and the rate of interest be \(R\) per annum. Let \(n\) be the number of years.
The formula for the amount (\(A\)) after \(n\) years on compound interest is:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]According to the problem, the sum becomes 4 times in 3 years. So, \(A = 4P\) when \(n = 3\).
\[ 4P = P \left(1 + \frac{R}{100}\right)^3 \]Dividing both sides by \(P\), we get:
\[ 4 = \left(1 + \frac{R}{100}\right)^3 \quad \ldots(1) \]Now, we need to find the time (\(n'\)) when the sum becomes 64 times, i.e., \(A' = 64P\).
\[ 64P = P \left(1 + \frac{R}{100}\right)^{n'} \]Dividing both sides by \(P\), we get:
\[ 64 = \left(1 + \frac{R}{100}\right)^{n'} \quad \ldots(2) \]We know that \(64 = 4^3\). Substitute this into equation (2):
\[ 4^3 = \left(1 + \frac{R}{100}\right)^{n'} \]From equation (1), we know that \(4 = \left(1 + \frac{R}{100}\right)^3\). Substitute this into the equation from step 7:
\[ \left(\left(1 + \frac{R}{100}\right)^3\right)^3 = \left(1 + \frac{R}{100}\right)^{n'} \]Using the exponent rule \((a^b)^c = a^{bc}\):
\[ \left(1 + \frac{R}{100}\right)^{3 \times 3} = \left(1 + \frac{R}{100}\right)^{n'} \] \[ \left(1 + \frac{R}{100}\right)^9 = \left(1 + \frac{R}{100}\right)^{n'} \]Comparing the exponents, we find:
\[ n' = 9 \text{ years} \]D) 9 years