The question asks for the year King George III established the Supreme Court in Madras. This is a historical fact related to the judicial system during the British Raj in India. We need to recall the specific year of this establishment.
Correct Option: A) 1800 — The Supreme Court in Madras was established by King George III in 1800.
The question asks about the economic impact of global warming on India, specifically the percentage by which its economy shrunk between 1961 and 2010, according to a 2019 Stanford study. This is a factual recall question based on a specific research finding.
Correct Option: B) 31% — A 2019 Stanford study found that global warming caused India's economy to shrink by 31% between 1961 and 2010.
The question asks for the approximate percentage of female literacy in India according to the 2011 Census. This requires recalling specific data from the Census of India, 2011.
Correct Option: A) 65.46% is the correct approximate percentage of female literacy in India as per the 2011 Census.
To determine the time difference between Indian Standard Time (IST) and Greenwich Mean Time (GMT), we need to understand the concept of longitudes and their relationship with time zones. The Earth rotates \(360^\circ\) in 24 hours. This rotation rate allows us to calculate the time difference for each degree of longitude.
Correct Option: D) 5 h, 30 min
The question asks to identify the airline that suspended its operations on April 17, 2019, due to financial difficulties. This requires knowledge of significant events in the Indian aviation sector around that time.
Correct Option: B) Jet Airways suspended all its operations for an indefinite period on April 17, 2019, due to a severe liquidity crisis and inability to secure further funding. This was a major event in the Indian aviation industry.
The question asks for the full form of the acronym JNNURM. This requires knowledge of common government schemes and their official names, which is a factual recall question.
Correct Option: B) Jawaharlal Nehru National Urban Renewal Mission
JNNURM was a massive city-modernization scheme launched by the Government of India under the Ministry of Urban Development. It aimed to encourage reforms and fast-track planned development of identified cities. The mission focused on urban infrastructure and governance, basic services to the urban poor, and community participation.
The question asks for the SI unit of the power of a lens. This is a fundamental concept in optics. We need to recall the definition of lens power and its associated unit.
Correct Option: C) Dioptre is the SI unit of the power of a lens. One dioptre is defined as the power of a lens whose focal length is 1 meter.
This problem requires logical deduction based on the given statements. We need to analyze each statement and then evaluate the conclusion's truthfulness against these statements. This is a classic syllogism-type question often found in logical reasoning sections of competitive exams.
C) The conclusion is definitely false.
The conclusion directly contradicts Statement 2, which states that "Gold is glittery." If gold is glittery, then gold is one of the things that glitters. Therefore, it cannot be true that *everything* that glitters is something *other than* gold.
The given series consists of squares of even numbers. We can factor out a common term and use the formula for the sum of squares of the first 'n' natural numbers.
Identify the series:
\( S = 2^{2} + 4^{2} + 6^{2} + \dots + 22^{2} \)
Rewrite each term by factoring out \(2^2\):
\( S = (2 \times 1)^{2} + (2 \times 2)^{2} + (2 \times 3)^{2} + \dots + (2 \times 11)^{2} \)
Factor out \(2^2\) from the entire series:
\( S = 2^{2} (1^{2} + 2^{2} + 3^{2} + \dots + 11^{2}) \)
\( S = 4 (1^{2} + 2^{2} + 3^{2} + \dots + 11^{2}) \)
Recall the formula for the sum of the squares of the first 'n' natural numbers:
\[ \sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6} \]
In our case, \(n = 11\). Substitute this value into the formula:
\[ 1^{2} + 2^{2} + \dots + 11^{2} = \frac{11(11+1)(2 \times 11+1)}{6} \]
\[ = \frac{11(12)(22+1)}{6} \]
\[ = \frac{11 \times 12 \times 23}{6} \]
Simplify the expression:
\[ = 11 \times 2 \times 23 \]
\[ = 22 \times 23 \]
\[ = 506 \]
Substitute this sum back into the expression for \(S\):
\( S = 4 \times 506 \)
\( S = 2024 \)
Correct Option: B) 2024
If α and β are the zeros of the polynomial \( f(x) = kx^{2} + 4x + 4 \) such that α\( ^{2} + \) β\( ^{2} = 24 \), then find the positive value of k.
\( \frac{4}{3} \)
\( \frac{1}{3} \)
\( \frac{3}{4} \)
\( \frac{2}{3} \)
The problem involves finding the value of 'k' for a quadratic polynomial given a relationship between its zeros. We will use Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots (zeros). For a quadratic polynomial \(ax^2 + bx + c\), the sum of the zeros \(\alpha + \beta = -b/a\) and the product of the zeros \(\alpha\beta = c/a\). We are also given a condition involving the squares of the zeros, \(\alpha^2 + \beta^2 = 24\). We can express \(\alpha^2 + \beta^2\) in terms of \(\alpha + \beta\) and \(\alpha\beta\).
Identify coefficients:
The given polynomial is \( f(x) = kx^{2} + 4x + 4 \). Comparing this with the standard quadratic form \( ax^2 + bx + c \), we have:
Apply Vieta's formulas:
Let \(\alpha\) and \(\beta\) be the zeros of the polynomial.
Use the given condition:
We are given that \( \alpha^{2} + \beta^{2} = 24 \).
We know the identity: \( (\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta \).
Rearranging this, we get: \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \).
Substitute and solve for k:
Substitute the expressions for \( \alpha + \beta \) and \( \alpha\beta \) from step 2 into the identity from step 3:
\[ 24 = \left(-\frac{4}{k}\right)^2 - 2\left(\frac{4}{k}\right) \]\[ 24 = \frac{16}{k^2} - \frac{8}{k} \]Simplify the equation:
Multiply the entire equation by \( k^2 \) to eliminate denominators (assuming \( k \neq 0 \)):
\[ 24k^2 = 16 - 8k \]\[ 24k^2 + 8k - 16 = 0 \]Solve the quadratic equation for k:
Divide the equation by 8 to simplify:
\[ 3k^2 + k - 2 = 0 \]Factor the quadratic equation:
We need two numbers that multiply to \( 3 \times (-2) = -6 \) and add to 1. These numbers are 3 and -2.
\[ 3k^2 + 3k - 2k - 2 = 0 \]\[ 3k(k + 1) - 2(k + 1) = 0 \]\[ (3k - 2)(k + 1) = 0 \]Find the possible values of k:
This gives two possible values for k:
Choose the positive value of k:
The question asks for the positive value of k. Therefore, \( k = \frac{2}{3} \).
D) \( \frac{2}{3} \)