What is the ratio of the <strong>GREATEST</strong> to the <strong>SMALLEST</strong> value of 2-2 \sin x-\sin^{2}x, for 0 \leq x \leq \frac{\pi}{2}?

Explanation

Let \sin x = t, where 0 \leq t \leq 1. The function is f(t) = 2 - 2t - t^2. Over , the greatest value is at t=0 (f(0)=2) and smallest is at t=1 (f(1)=-1). Ratio = -2. Since -2 is not an option, this question was dropped by UPSC.

Related questions on Trigonometry

• Two poles are situated 24 m apart and their heights differ by 10 m. What is the distance between their tips?
• If \frac{\cos \theta}{1 - \sin \theta} + \frac{\cos \theta}{1 + \sin \theta} = 4, then which one of the following is a value of $(\tan^2 \...
• For 0 < \theta < \frac{\pi}{2}, consider the following : I. $(\tan^4 \theta + \tan^6 \theta)(\cot^4 \theta + \cot^6 \theta) = \sec^2 \the...
• If 3\sin \theta + 4\cos \theta = 5, then what is a value of 4\tan \theta + 3\cot \theta ?
• At a point on level ground, the tangent of the angle of elevation of the top of a tower is found to be \frac{5}{6}. On walking 70 m toward...