The USA is the largest importer of Indian goods as of 2023.

State Bank of India has the highest number of branches in India as of 2023.

Solar energy has the highest capacity among renewable sources in India as of 2023.

The services sector is the largest contributor to India's GDP as of 2023.

At the equinox, the sun is directly overhead at the equator, which is at 0 degrees latitude.

During the summer solstice, the sun is directly overhead at the Tropic of Cancer, which is at 23.5°N latitude.

The sun does not set during the summer solstice at latitudes above the Arctic Circle (66.5°N).

The longest day occurs at the Arctic Circle during the summer solstice, where the sun does not set.

3 hours ahead means 3 * 15° = 45°. Therefore, 0° + 45° = 135°E.

3 hours ahead means 3 * 15 = 45 degrees East, so 0° + 45° = 135° East.

Each hour corresponds to 15 degrees. Therefore, 3 hours ahead means 3 * 15 = 45 degrees east of GMT, which is 45°E.

6 hours ahead corresponds to 6 * 15° = 90°. Therefore, 90°E is the correct answer.

150°E corresponds to UTC+6, as each 15° represents one hour.

The local time is the same as that of the Prime Meridian at 0° longitude.

The area under the curve is given by ∫(from 1 to 4) (2x + 1) dx = [x^2 + x] from 1 to 4 = (16 + 4) - (1 + 1) = 20.

The area under the curve is given by ∫(from 0 to 2) x^3 dx = [x^4/4] from 0 to 2 = (16/4) - (0) = 4.

The determinant of J is calculated as 1*(1*1 - 0*3) - 2*(0*1 - 0*2) + 1*(0*3 - 1*2) = 1 - 0 - 2 = -1.

Det(I) = (3*4) - (2*1) = 12 - 2 = 10.

Using the distance formula: d = √[(4 - 1)² + (5 - 1)²] = √[9 + 16] = √25 = 5.

Using the distance formula: d = √[(1 - 1)² + (5 - 2)²] = √[0 + 9] = √9 = 3.

Using the distance formula: d = √[(2 - 6)² + (3 - 8)²] = √[16 + 25] = √41.

Using the distance formula: d = √((6 - 6)² + (2 - 8)²) = √(0 + 36) = √36 = 6.

∫(2 to 3) (x^3) dx = [x^4/4] from 2 to 3 = (81/4 - 16/4) = 65/4 = 16.25.

∫(2 to 5) (4x - 1) dx = [2x^2 - x] from 2 to 5 = (50 - 5) - (8 - 2) = 40.

Using L'Hôpital's Rule, we differentiate the numerator and denominator to find lim (x -> 0) (1/(1 + x)) = 1.

Since |sin(1/x)| ≤ 1, we have |x^2 sin(1/x)| ≤ |x^2|. Thus, lim (x -> 0) x^2 sin(1/x) = 0.

Using the fact that sin(x) ~ x as x approaches 0, we find that lim (x -> 0) (x^3)/(sin(x)) = 0.

Factoring gives lim (x -> 2) ((x - 2)(x^2 + 2x + 4))/(x - 2) = lim (x -> 2) (x^2 + 2x + 4) = 12.

Dividing numerator and denominator by x^2 gives lim (x -> ∞) (3 + 2/x^2)/(5 - 4/x^2) = 3/5.

Dividing numerator and denominator by x^2 gives lim (x -> ∞) (3 + 2/x^2)/(5 - 4/x + 1/x^2) = 3/5.