This question tests general knowledge about Indian cities and their popular nicknames. The task is to identify which city is famously known as 'The City of Palaces'.
Correct Option: A) Kolkata is widely known as 'The City of Palaces' due to its rich architectural heritage, including the Victoria Memorial, Marble Palace, and many other stately homes and mansions.
When x2 + kx - 6 can be completely divided by x + 3, the value of k is
The problem states that the polynomial \(x^2 + kx - 6\) is completely divisible by \(x + 3\). According to the Factor Theorem, if a polynomial \(P(x)\) is completely divisible by \((x - a)\), then \(P(a) = 0\). In this case, \(x + 3\) is the factor, which can be written as \(x - (-3)\). Therefore, we need to substitute \(x = -3\) into the polynomial and set the result to zero to find the value of \(k\).
Thus, the value of \(k\) is 1.
C) 1 — As calculated using the Factor Theorem, substituting \(x = -3\) into the polynomial and setting it to zero yields \(k = 1\).
The mode of a set of numbers is the value that appears most frequently in the set. To find the mode, we need to count the occurrences of each number in the given data set.
Correct Option: A) 5 is the mode because it appears 3 times, which is the highest frequency among all the numbers in the set.
The question asks to identify the government scheme launched by the Indian Prime Minister specifically to combat malnutrition and stunting among children. This requires knowledge of various government schemes and their primary objectives.
C) POSHAN Abhiyan (Prime Minister's Overarching Scheme for Holistic Nourishment) was launched by the Prime Minister of India in March 2018. Its main objective is to reduce malnutrition, stunting, under-nutrition, and low birth weight among children, as well as anemia among young children, women, and adolescent girls. It aims to achieve these targets through a multi-ministerial convergence approach and technology-enabled monitoring.
This question tests general knowledge about the Union Cabinet Ministers of India, specifically focusing on the Minister of Road Transport and Highways during a particular period (November 2018). To answer this, one needs to recall the cabinet portfolios held by prominent political figures around that time.
Correct Option: C) Nitin Gadkari was indeed the Union Minister for Road Transport and Highways, Shipping, and Water Resources, River Development & Ganga Rejuvenation in November 2018. He held the Road Transport and Highways portfolio from May 2014.
The length of the arc subtending an angle at the center of a circle of radius 14 cm is 11 cm. What is the measure of that central angle? [Use π = \( \frac{22}{7} \)]
To find the central angle subtended by an arc, we use the formula relating arc length, radius, and the central angle in radians. Then, we convert the angle from radians to degrees.
The length of an arc \(l\) subtending an angle \(\theta\) (in radians) at the center of a circle with radius \(r\) is given by:
\[ l = r \theta \]We know that \(\pi \text{ radians} = 180^\circ\). Therefore, to convert radians to degrees, we multiply by \(\frac{180}{\pi}\).
\[ \theta \text{ (in degrees)} = \frac{11}{14} \times \frac{180}{\pi} \]Substitute the value of \(\pi = \frac{22}{7}\):
\[ \theta \text{ (in degrees)} = \frac{11}{14} \times \frac{180}{\frac{22}{7}} \] \[ \theta \text{ (in degrees)} = \frac{11}{14} \times \frac{180 \times 7}{22} \]Simplify the expression:
\[ \theta \text{ (in degrees)} = \frac{11}{2 \times 7} \times \frac{180 \times 7}{2 \times 11} \]Cancel out common terms (11 and 7):
\[ \theta \text{ (in degrees)} = \frac{1}{2} \times \frac{180}{2} \] \[ \theta \text{ (in degrees)} = \frac{180}{4} \] \[ \theta \text{ (in degrees)} = 45^\circ \]Correct Option: B) 45°
Select the correct water image of the given figure.
To find the correct water image of a given figure, we need to reflect the image vertically. This means flipping the image upside down as if it is mirrored on a horizontal axis.
Correct Option: B) The circle is at the bottom left, the triangle is inverted in the center, and the star is at the top right, matching the expected water image.
The question asks for the specific category under which the Assamese film 'Dikchow Banat Palaax' received the Nargis Dutt Award at the 64th National Film Awards in 2017. This requires knowledge of the National Film Awards categories and the specific recognition received by this particular film.
Correct Option: D) Feature film on national integration
Select the correct water image of the given figure.
To find the correct water image of a given figure, we need to visualize the reflection of the figure across a horizontal axis. This involves flipping the figure vertically.
Correct Option: C) Option C is correct because it accurately represents the vertical flip of the original figure.
If tanα = \(\sqrt{15}\) + 4, then the value of tanα - cotα is ______.
4 - \(\sqrt{15}\)
\(\sqrt{15}\) - 4
2\(\sqrt{15}\)
8
The problem asks us to find the value of \(\tan\alpha - \cot\alpha\) given the value of \(\tan\alpha\). We know that \(\cot\alpha\) is the reciprocal of \(\tan\alpha\). Therefore, we can find \(\cot\alpha\) and then substitute both values into the expression.
We are given \(\tan\alpha = \sqrt{15} + 4\).
We know that \(\cot\alpha = \frac{1}{\tan\alpha}\). So, substitute the given value of \(\tan\alpha\):
\[\cot\alpha = \frac{1}{\sqrt{15} + 4}\]To simplify this expression, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{15} - 4\):
\[\cot\alpha = \frac{1}{\sqrt{15} + 4} \times \frac{\sqrt{15} - 4}{\sqrt{15} - 4}\]Using the algebraic identity \((a+b)(a-b) = a^2 - b^2\) in the denominator:
\[\cot\alpha = \frac{\sqrt{15} - 4}{(\sqrt{15})^2 - (4)^2}\]\[\cot\alpha = \frac{\sqrt{15} - 4}{15 - 16}\]\[\cot\alpha = \frac{\sqrt{15} - 4}{-1}\]\[\cot\alpha = -(\sqrt{15} - 4)\]\[\cot\alpha = 4 - \sqrt{15}\]Now substitute the values of \(\tan\alpha\) and \(\cot\alpha\) into the expression \(\tan\alpha - \cot\alpha\):
\[\tan\alpha - \cot\alpha = (\sqrt{15} + 4) - (4 - \sqrt{15})\]Distribute the negative sign:
\[\tan\alpha - \cot\alpha = \sqrt{15} + 4 - 4 + \sqrt{15}\]Combine like terms:
\[\tan\alpha - \cot\alpha = (\sqrt{15} + \sqrt{15}) + (4 - 4)\]\[\tan\alpha - \cot\alpha = 2\sqrt{15} + 0\]\[\tan\alpha - \cot\alpha = 2\sqrt{15}\]Correct Option: C) \(2\sqrt{15}\)