It's a Question of your Career!

Correct Answer:

Solution : When the given paper is unfolded, it will look like this –

Hence, the third option is correct.

Correct Answer: April and May

Solution : As per the given table,
The company produced the same number of scooters in April and May.
In both these months, the total production was 110 scooters.
Hence, the correct answer is 'April and May'.

Correct Answer: Mahesh Bhupathi

Solution : The correct answer is Mahesh Bhupathi.

In the 1990s and 2000s, Mahesh Bhupathi was regarded as one of the best doubles players. Wimbledon and the French Open were among the three doubles championships that Bhupathi and Leander Paes won in 1999. He won

Correct Answer: 2880

Solution : Given:
5, 20, 60, 240, 720, ?

To get the required missing number in the series, multiply the number by 4 and 3 alternatively –
5 × 4 = 20; 20 × 3 = 60; 60 × 4 = 240; 240 × 3 = 720;

Correct Answer: $12 \sqrt{3} \mathrm{~cm}$

Solution :
Given, $\triangle$ABC is right angled at B, where AB = 12 cm and $\angle$CAB = 60°
By using the trigonometric ratio involving AB and BC.
To $\angle$CAB, AB is the adjacent side and BC is the opposite side.
$\tan\angle CAB$ = $\tan 60°$

Correct Answer: 14

Solution : Given: $x^{4}+\frac{1}{x^{4}}=34$
Adding 2 to both sides, we get,
$⇒x^{4}+\frac{1}{x^{4}}+2=34+2$
$⇒(x^{2}+\frac{1}{x^{2}})^{2}=(6)^{2}$
$⇒x^{2}+\frac{1}{x^{2}}=6$
Subtracting 2 from both sides, we get,
$⇒x^{2}+\frac{1}{x^{2}}-2=6-2$
$⇒(x-\frac{1}{x})^{2}=(2)^{2}$
$⇒x-\frac{1}{x}=2$
Now, $(x-\frac{1}{x})^3=x^3-\frac{1}{x^3}-3×x×\frac{1}{x}(x-\frac{1}{x})$
$⇒2^3=x^3-\frac{1}{x^3}-3×2$
$\therefore x^{3}-\frac{1}{x^{3}}=14$
Hence, the correct answer is 14.

Correct Answer: 1887

Solution : The correct option is 1887.

In the year 1887, German physicist Heinrich Hertz made a pivotal discovery related to the photoelectric effect while researching radio waves. In the course of his experiments, Hertz utilised a spark gap consisting of two closely spaced, sharp electrodes capable

Correct Answer: $1$

Solution : Given:
$\operatorname{cosec}\theta-\sin\theta=l$ and $\sec\theta-\cos\theta=m$, $l^2m^2(l^2+m^2+3)$
$=(\operatorname{cosec}\theta-\sin\theta)^2(\sec\theta-\cos\theta)^2[(\operatorname{cosec}\theta-\sin\theta)^2+(\sec\theta-\cos\theta)^2+3]$
$= (\frac{1}{\sin\theta}-\sin\theta)^2(\frac{1}{\cos\theta}-\cos\theta)^2[(\frac{1}{\sin\theta}-\sin\theta)^2+(\frac{1}{\cos\theta}-\cos\theta)^2+3]$
$=(\frac{1-\sin^2\theta}{\sin\theta})^2(\frac{1-\cos^2\theta}{\cos\theta})^2[(\frac{1-\sin^2\theta}{\sin\theta})^2+(\frac{1-\cos^2\theta}{\cos\theta})^2+3]$
$=(\frac{cos^2\theta}{\sin\theta})^2(\frac{\sin^2\theta}{\cos\theta})^2[(\frac{\cos^2\theta}{\sin\theta})^2+(\frac{\sin^2\theta}{\cos\theta})^2+3]$
$=(\frac{\cos^4\theta}{\sin^2\theta})(\frac{\sin^4\theta}{\cos^2\theta})[(\frac{\cos^4\theta}{\sin^2\theta})+(\frac{\sin^4\theta}{\cos^2\theta})+3]$
$= \sin^2\theta \cos^2\theta[\frac{\cos^6\theta+\sin^6\theta+3\sin^2\theta \cos^2\theta}{\sin^2\theta \cos^2\theta}]$
$= \cos^6\theta+\sin^6\theta+3\sin^2\theta \cos^2\theta$
$=(\cos^2\theta)^3+\sin^2\theta)^3+3\sin^2\theta \cos^2\theta$
$=(\cos^2\theta+\sin^2\theta)^3-3\cos^2\theta \sin^2\theta(\cos^2\theta+\sin^2\theta)+3\sin^2\theta \cos^2\theta$
$=1^3-3\cos^2\theta \sin^2\theta(1)+3\sin^2\theta \cos^2\theta$
$=1$
Hence, the correct answer is $1$.

Correct Answer: The President in Consultation with the Election Commission

Solution : The correct answer is The President in Consultation with the Election Commission.

Depending on whether it is the Lok Sabha or the Rajya Sabha, the Speaker or the House Chairman decides on disqualification in defection-related cases. Even if

Correct Answer: Surya

Solution : The correct answer is Surya.

The sun god Surya is the object of Chhath puja. All life on Earth originates from the sun, which is visible to all beings. Chhathi Maiya is worshipped on this day together with the Sun God.