Correct Answer: Quicken the pulse

Solution : The third option is the correct choice.

The expression "quicken the pulse" means to cause excitement or increase enthusiasm. In the context of the sentence, the announcement of the new tourism policy has made some people excited or interested, making "quicken the pulse"

Correct Answer: reptiles that became extinct

Solution : The correct option is reptiles that became extinct.

Dinosaurs were a varied group of reptiles that lived during the Mesozoic Era millions of years ago. Dinosaurs became extinct at the end of the Cretaceous Period about 65 million years ago. The Cretaceous-Paleogene

Correct Answer: The disparity in access to digital technologies 

Solution : The digital divide refers to the disparity in access to digital technologies, such as computers and the internet, between different groups of people, often based on socioeconomic status, geography, or demographics.

Correct Answer: Imperial Bank of India

Solution : The correct option is Imperial Bank of India.

The Imperial Bank of India operated from 1921 to 1955. It came into existence through the merger of three presidency banks: the Bank of Calcutta, the Bank of Bombay, and the Bank of Madras.

Correct Answer: $\sqrt{2}+1$

Solution : $\sin\theta+\cos\theta=\sqrt{2}\cos\theta$
⇒ $\sin \theta = (\sqrt{2}-1)\cos\theta$
⇒ $\frac{\sin \theta}{(\sqrt{2}-1)}=\cos\theta$
⇒ $\frac{\sin \theta(\sqrt{2}+1)}{2-1}=\cos\theta$
⇒ $\frac{\sin \theta(\sqrt{2}+1)}{2-1}=\cos\theta$
⇒ $\frac{\cos \theta}{\sin \theta}=\sqrt{2}+1$
$\therefore\cot \theta=\sqrt{2}+1$
Hence, the correct answer is $\sqrt{2}+1$.

Correct Answer: $4-2 \sqrt{2}$

Solution : This is a telescoping series, where each term cancels out a part of the next term. 
$\frac{1}{4-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+\frac{1}{\sqrt{14}-\sqrt{13}}-\frac{1}{\sqrt{13}-\sqrt{12}}+\frac{1}{\sqrt{12}-\sqrt{11}}-\frac{1}{\sqrt{11}-\sqrt{10}}+\frac{1}{\sqrt{10}-3}-\frac{1}{3-\sqrt{8}}$
Rationalising the denominator and use the formula $a^2 - b^2 = (a+b)(a-b)$, we get,
$=\frac{\sqrt{15}+4}{1}-\frac{\sqrt{15}+\sqrt{14}}{1}+\frac{\sqrt{14}+\sqrt{13}}{1}-\frac{\sqrt{13}+\sqrt{12}}{1}+\frac{\sqrt{12}+\sqrt{11}}{1}-\frac{\sqrt{11}+\sqrt{10}}{1}+\frac{\sqrt{10}+3}{1}-\frac{3+\sqrt{8}}{1}$
$=\sqrt{15}+4-\sqrt{15}-\sqrt{14}+\sqrt{14}+\sqrt{13}-\sqrt{13}-\sqrt{12}+\sqrt{12}+\sqrt{11}-\sqrt{11}-\sqrt{10}+\sqrt{10}+3-3-\sqrt{8}$
$=4-\sqrt{8}$
$=4-2\sqrt{2}$
Hence, the correct answer is $4-2\sqrt{2}$.

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