We need to evaluate each conclusion based on the given statements and determine which ones logically follow. We will analyze each statement and conclusion separately to establish their validity.
A) Only conclusion 3 follows.
This is a blood relation question. We need to deduce the relationship between Sam and Jamie based on the statement made by Sam. We will break down the statement into smaller parts to understand the relationships involved.
Correct Option: B) Nephew. Sam is the son of Jamie's sister, which makes him Jamie's nephew.
If \( 2xy \cos \theta + (x^{2} - y^{2})\sin \theta = x^{2} + y^{2} \), then the value of tanθ is:
\(\frac{x^2-y^2}{x^2+y^2}\)
\(\frac{1}{\sqrt{2}}\)
\(\frac{x^2-y^2}{2xy}\)
\(\frac{y^2-x^2}{2xy}\)
The problem asks us to find the value of \( \tan \theta \) given an equation involving \( \sin \theta \) and \( \cos \theta \) along with \( x \) and \( y \). We need to rearrange the given equation to isolate \( \tan \theta \).
The given equation is:
\[ 2xy \cos \theta + (x^{2} - y^{2})\sin \theta = x^{2} + y^{2} \]To find \( \tan \theta \), we need to express the equation in terms of \( \sin \theta \) and \( \cos \theta \) such that we can divide by \( \cos \theta \). Let's rearrange the terms:
\[ (x^{2} - y^{2})\sin \theta = x^{2} + y^{2} - 2xy \cos \theta \]This approach doesn't directly lead to \( \tan \theta \). A more effective method is to divide the entire equation by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)).
\[ \frac{2xy \cos \theta}{\cos \theta} + \frac{(x^{2} - y^{2})\sin \theta}{\cos \theta} = \frac{x^{2} + y^{2}}{\cos \theta} \]\[ 2xy + (x^{2} - y^{2})\tan \theta = \frac{x^{2} + y^{2}}{\cos \theta} \]This still leaves \( \cos \theta \) on the right side. Let's try to rearrange the original equation to group terms with \( \sin \theta \) and \( \cos \theta \) on one side, and then divide by \( \cos \theta \).
Consider the structure of the equation. It resembles the form \( A \cos \theta + B \sin \theta = C \). We want to isolate \( \tan \theta \).
Let's move the \( \cos \theta \) term to the right side:
\[ (x^{2} - y^{2})\sin \theta = x^{2} + y^{2} - 2xy \cos \theta \]This is not directly helpful. Let's consider a different approach. Divide the entire equation by \( x^2+y^2 \):
\[ \frac{2xy}{x^{2} + y^{2}} \cos \theta + \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \sin \theta = 1 \]This form is \( R \cos \alpha \cos \theta + R \sin \alpha \sin \theta = R \) or similar. This is getting complicated. Let's try to express \( \sin \theta \) and \( \cos \theta \) in terms of \( \tan(\theta/2) \).
However, the options suggest a simpler algebraic manipulation. Let's revisit step 2.
The equation is:
\[ 2xy \cos \theta + (x^{2} - y^{2})\sin \theta = x^{2} + y^{2} \]Let's rearrange the equation to get \( \sin \theta \) and \( \cos \theta \) terms on one side and then divide by \( \cos \theta \) to get \( \tan \theta \). This is often done by moving one term to the other side and then factoring.
Let's move the \( \cos \theta \) term to the right side:
\[ (x^{2} - y^{2})\sin \theta = (x^{2} + y^{2}) - 2xy \cos \theta \]This doesn't seem to simplify easily to \( \tan \theta \). Let's try another rearrangement. The goal is to get something like \( A \tan \theta = B \).
Let's divide the entire equation by \( \cos \theta \) from the start:
\[ 2xy + (x^{2} - y^{2})\tan \theta = \frac{x^{2} + y^{2}}{\cos \theta} \]This is still problematic because of \( \cos \theta \) on the right side. The key is to recognize that the expression \( x^2+y^2 \) is a constant with respect to \( \theta \).
Let's rearrange the terms to group \( \sin \theta \) and \( \cos \theta \) terms:
\[ (x^{2} - y^{2})\sin \theta - (x^{2} + y^{2}) = -2xy \cos \theta \]\[ (x^{2} - y^{2})\sin \theta + 2xy \cos \theta = x^{2} + y^{2} \]This is the original equation. Let's try to isolate \( \tan \theta \) by dividing by \( \cos \theta \) after moving the \( \cos \theta \) term to the right side if it were a simpler expression. However, \( x^2+y^2 \) is not a \( \cos \theta \) term.
Let's rewrite the equation as:
\[ (x^{2} - y^{2})\sin \theta - (x^{2} + y^{2}) = -2xy \cos \theta \]This is not helpful. Let's try to divide the entire equation by \( 2xy \cos \theta \) or \( (x^2-y^2) \sin \theta \). This will also lead to complications.
Let's try to isolate \( \tan \theta \) directly. We have:
\[ (x^{2} - y^{2})\sin \theta = (x^{2} + y^{2}) - 2xy \cos \theta \]Divide both sides by \( \cos \theta \):
\[ (x^{2} - y^{2})\tan \theta = \frac{x^{2} + y^{2}}{\cos \theta} - 2xy \]This is not working. Let's re-examine the structure. The equation is of the form \( A \cos \theta + B \sin \theta = C \). To find \( \tan \theta \), we usually try to get \( \frac{\sin \theta}{\cos \theta} \).
Let's divide the entire equation by \( \cos \theta \):
\[ 2xy \frac{\cos \theta}{\cos \theta} + (x^{2} - y^{2})\frac{\sin \theta}{\cos \theta} = \frac{x^{2} + y^{2}}{\cos \theta} \]\[ 2xy + (x^{2} - y^{2})\tan \theta = \frac{x^{2} + y^{2}}{\cos \theta} \]This still has \( \cos \theta \) on the right side. This means we cannot simply divide by \( \cos \theta \) if the right side is not a multiple of \( \cos \theta \).
Let's reconsider the problem. The expression looks like a trigonometric identity or a transformation. Let's try to move all terms involving \( \theta \) to one side and constants to the other. This is not possible here as \( x^2+y^2 \) is a constant.
Let's try to rearrange the original equation to isolate \( \tan \theta \).
\[ (x^{2} - y^{2})\sin \theta = (x^{2} + y^{2}) - 2xy \cos \theta \]\[ (x^{2} - y^{2})\sin \theta + 2xy \cos \theta = x^{2} + y^{2} \]This is the original equation. Let's divide by \( \cos \theta \) to get \( \tan \theta \):
\[ (x^{2} - y^{2})\frac{\sin \theta}{\cos \theta} + 2xy \frac{\cos \theta}{\cos \theta} = \frac{x^{2} + y^{2}}{\cos \theta} \]\[ (x^{2} - y^{2})\tan \theta + 2xy = \frac{x^{2} + y^{2}}{\cos \theta} \]This is not simplifying. The problem might be simpler than this. Let's look at the options. The options are in terms of \( x \) and \( y \).
Let's try to move the \( \cos \theta \) term to the left side and \( \sin \theta \) term to the right side, or vice versa, to get a ratio.
From the given equation:
\[ 2xy \cos \theta + (x^{2} - y^{2})\sin \theta = x^{2} + y^{2} \]Let's try to isolate \( \sin \theta \) and \( \cos \theta \) terms to form \( \tan \theta \).
Consider moving \( 2xy \cos \theta \) to the right side:
\[ (x^{2} - y^{2})\sin \theta = x^{2} + y^{2} - 2xy \cos \theta \]This doesn't seem to lead to a simple \( \tan \theta \) expression. Let's try to move \( (x^2-y^2) \sin \theta \) to the right side:
\[ 2xy \cos \theta = x^{2} + y^{2} - (x^{2} - y^{2})\sin \theta \]This is also not helpful.
Let's reconsider the form \( A \cos \theta + B \sin \theta = C \). We want to find \( \tan \theta \). We can try to divide by \( \cos \theta \) or \( \sin \theta \).
Let's divide the entire equation by \( \cos \theta \):
\[ 2xy + (x^{2} - y^{2})\tan \theta = (x^{2} + y^{2})\sec \theta \]This is not directly giving \( \tan \theta \). What if we try to express \( \sin \theta \) and \( \cos \theta \) in terms of \( \tan \theta \)?
We know that \[\sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^{2} \theta}}\] and \( \cos \theta = \frac{1}{\sqrt{1 + \tan^{2} \theta}} \). Substituting these into the equation would lead to a very complex algebraic equation involving \( \tan \theta \) and \( \sqrt{1 + \tan^{2} \theta} \).
Let's look at the options again. They are simple fractions of \( x \) and \( y \).
Let's try to rearrange the original equation to isolate the \( \sin \theta \) and \( \cos \theta \) terms in a specific way.
The given equation is:
\[ 2xy \cos \theta + (x^{2} - y^{2})\sin \theta = x^{2} + y^{2} \]Let's assume \( \cos \theta \neq 0 \) and divide by \( \cos \theta \):
\[ 2xy + (x^{2} - y^{2})\tan \theta = \frac{x^{2} + y^{2}}{\cos \theta} \]This is the same issue. Let's try to rearrange the equation to get \( \sin \theta \) and \( \cos \theta \) on opposite sides, and then divide to form \( \tan \theta \).
Consider the identity \( \cos^{2} \theta + \sin^{2} \theta = 1 \). If we could express \( x^2+y^2 \) in terms of \( \cos^2 \theta + \sin^2 \theta \), it might help.
Let's try to move the \( 2xy \cos \theta \) term to the right side:
\[ (x^{2} - y^{2})\sin \theta = x^{2} + y^{2} - 2xy \cos \theta \]Now, let's try to move the \( (x^2+y^2) \) term to the left side and group it with \( \sin \theta \).
Let's try to rearrange the equation in the form \( A \sin \theta + B \cos \theta = C \). We have:
\[ (x^{2} - y^{2})\sin \theta + 2xy \cos \theta = x^{2} + y^{2} \]This is the original equation. Let's try to divide by \( x^2+y^2 \) first:
\[ \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \sin \theta + \frac{2xy}{x^{2} + y^{2}} \cos \theta = 1 \]This is in the form \( \cos \alpha \sin \theta + \sin \alpha \cos \theta = 1 \) or \( \sin(\theta + \alpha) = 1 \). This is a standard trigonometric form. Let \( \cos \alpha = \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \) and \( \sin \alpha = \frac{2xy}{x^{2} + y^{2}} \).
Then, \[\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{2xy}{x^{2} + y^{2}}}{\frac{x^{2} - y^{2}}{x^{2} + y^{2}}} = \frac{2xy}{x^{2} - y^{2}}\].
So the equation becomes \( \sin(\theta + \alpha) = 1 \).
This means \( \theta + \alpha = \frac{\pi}{2} + 2n\pi \) for some integer \( n \).
Then \( \theta = \frac{\pi}{2} - \alpha + 2n\pi \).
So \( \tan \theta = \tan(\frac{\pi}{2} - \alpha) = \cot \alpha \).
Since \( \tan \alpha = \frac{2xy}{x^{2} - y^{2}} \), then \[\cot \alpha = \frac{1}{\tan \alpha} = \frac{x^{2} - y^{2}}{2xy}\].
Therefore, \( \tan \theta = \frac{x^{2} - y^{2}}{2xy} \).
C) \( \frac{x^2-y^2}{2xy} \)
The step-by-step derivation shows that by transforming the given equation into the form \( \sin(\theta + \alpha) = 1 \), we can determine \( \tan \theta \) as \( \cot \alpha \). The value of \( \cot \alpha \) is found to be \( \frac{x^2-y^2}{2xy} \).
To find the missing number in the series, we need to identify the pattern or rule governing the sequence of numbers. We will examine the differences between consecutive terms to see if a clear mathematical progression emerges.
Thus, the number that replaces the question mark is 96.
C) 96 is the correct answer because the pattern of differences between consecutive terms is the sequence of perfect squares (\(1^2, 2^2, 3^2, 4^2, 5^2, \ldots\)). The next difference is \(6^2 = 36\), and adding this to the last term (60) gives \(60 + 36 = 96\).
Find the value of \(\frac{\sqrt{144}}{7}\) × \( \frac{14}{12} \) ×\(\frac{7}{\sqrt{196}}\).
To solve this problem, we need to simplify each term in the given expression by evaluating the square roots and then performing the multiplication of the fractions. The key is to simplify before multiplying to make the calculation easier.
First, evaluate the square roots:
\(\sqrt{144}\): We know that \(12 \times 12 = 144\), so \(\sqrt{144} = 12\).
\(\sqrt{196}\): We know that \(14 \times 14 = 196\), so \(\sqrt{196} = 14\).
Substitute these values back into the expression:
\[ \frac{12}{7} \times \frac{14}{12} \times \frac{7}{14} \]Now, multiply the fractions. We can cancel out common terms in the numerator and denominator to simplify the calculation:
\[ \frac{\cancel{12}}{7} \times \frac{14}{\cancel{12}} \times \frac{7}{14} \]The \(12\) in the numerator of the first fraction cancels with the \(12\) in the denominator of the second fraction.
The expression becomes:
\[ \frac{1}{7} \times \frac{14}{1} \times \frac{7}{14} \]Now, we can see that \(14\) in the numerator of the second fraction cancels with \(14\) in the denominator of the third fraction:
\[ \frac{1}{7} \times \frac{\cancel{14}}{1} \times \frac{7}{\cancel{14}} \]The expression simplifies to:
\[ \frac{1}{7} \times \frac{1}{1} \times \frac{7}{1} \]Finally, the \(7\) in the denominator of the first fraction cancels with the \(7\) in the numerator of the third fraction:
\[ \frac{1}{\cancel{7}} \times 1 \times \frac{\cancel{7}}{1} \]The final result is:
\[ 1 \times 1 \times 1 = 1 \]C) 1
The question asks to identify the sixth Indian cricketer inducted into the ICC Cricket Hall of Fame. This requires knowledge of the history of ICC Hall of Fame inductees from India.
Correct Option: D) Sachin Tendulkar was the sixth Indian cricketer to be inducted into the ICC Cricket Hall of Fame in 2019.
The question asks to identify the FALSE statement among the given options regarding the differences between plant and animal cells. We need to evaluate each statement based on our knowledge of cell biology.
Based on the evaluation, statement C is false.
C) Plant cells have smaller vacuoles than those of animal cells. This statement is false. Plant cells are characterized by a large, central vacuole that can occupy a significant portion of the cell's volume, playing vital roles in turgor pressure, storage, and waste disposal. Animal cells, if they have vacuoles, possess several small, temporary ones, which are not comparable in size or function to the plant cell's central vacuole.
The question asks to identify the country that launched the world's first quantum satellite for hack-proof communication. This is a factual question related to advancements in space technology and quantum communication.
Correct Option: D) China
China launched the world's first quantum satellite, "Micius," in August 2016. This satellite was designed to test quantum entanglement over long distances and establish secure, hack-proof communication channels, marking a significant milestone in quantum technology and space exploration.
The question asks about the primary components of syngas. Syngas, or synthesis gas, is a fuel gas mixture that is important in various industrial processes. Understanding its composition is key to answering this question.
Another common reaction is the partial oxidation of methane:
\[ 2CH_4 + O_2 \rightarrow 2CO + 4H_2 \]Correct Option: C) carbon monoxide and hydrogen
The question asks to identify the author of the book 'Changing India'. This requires knowledge of prominent Indian political figures and their literary works.
Correct Option: C) Dr. Manmohan Singh is the author of the book 'Changing India'. The book provides insights into his tenure as Finance Minister and Prime Minister, and his role in India's economic reforms and development.