#16 of June 2018 Grade 12 - Natural Sciences CSAT (Korean SAT) Mock test

Calculus Level 4

As shown on the right, there is a sector O A B \rm OAB with radius 1 1 and central angle π 2 . \dfrac{\pi}{2}.

Let H \rm H be the foot of perpendicular from P , \rm P, a point on arc A B , \rm AB, to O A ‾ . \overline{\rm OA}.

Pick a point Q \rm Q on arc B P \rm BP so that ∠ P O H = ∠ P H Q = θ . \angle \rm POH = \angle PHQ = \theta.

Then, the area of triangle O H Q \rm OHQ is S ( θ ) . S(\theta).

Find the value of lim ⁡ θ → 0 + S ( θ ) θ . \displaystyle \lim_{\theta\to 0^{+}}\dfrac{S(\theta)}{\theta}.

  • It is expected that you solve this in 5 minutes.

This problem is a part of <Grade 12 CSAT Mock test> series .

1 + 2 2 \dfrac{1+\sqrt{2}}{2} 5 + 2 2 \dfrac{5+\sqrt{2}}{2} 3 + 2 2 \dfrac{3+\sqrt{2}}{2} 4 + 2 2 \dfrac{4+\sqrt{2}}{2} 2 + 2 2 \dfrac{2+\sqrt{2}}{2}

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Jun 9, 2018

Note that O H ‾ = cos ⁡ θ \overline{\rm OH} = \cos \theta and O Q ‾ = 1 , \overline{\rm OQ} = 1, along with ∠ O H Q = π 2 − θ . \angle \rm OHQ = \dfrac{\pi}{2}-\theta.

Use the cosine law to find H Q ‾ = x : \overline{\rm HQ}=x:

O Q ‾ 2 = O H ‾ 2 + H Q ‾ 2 − 2 O H ‾ ⋅ H Q ‾ ⋅ cos ⁡ ( π 2 − θ ) 1 = cos ⁡ 2 θ + x 2 − 2 x cos ⁡ θ sin ⁡ θ x 2 − 2 x cos ⁡ θ sin ⁡ θ − sin ⁡ 2 θ = 0 x = cos ⁡ θ sin ⁡ θ + cos ⁡ 2 θ sin ⁡ 2 θ + sin ⁡ 2 θ \overline{\rm OQ}^2 = \overline{\rm OH}^2 + \overline{\rm HQ}^2 - 2\overline{\rm OH}\cdot\overline{\rm HQ}\cdot \cos \left(\dfrac{\pi}{2}-\theta\right) \\ 1 = \cos^2\theta + x^2 - 2x \cos \theta \sin \theta \\ x^2 - 2x\cos\theta\sin\theta - \sin^2\theta = 0 \\ x = \cos\theta\sin\theta + \sqrt{\cos^2\theta\sin^2\theta + \sin^2\theta}

Then we proceed to find the limit.

lim ⁡ θ → 0 + S ( θ ) θ = lim ⁡ θ → 0 + 1 2 cos ⁡ θ ⋅ x ⋅ sin ⁡ ( π 2 − θ ) θ = 1 2 lim ⁡ θ → 0 + sin ⁡ ( π 2 − θ ) cos ⁡ θ ⋅ lim ⁡ θ → 0 + cos ⁡ θ sin ⁡ θ + sin ⁡ θ cos ⁡ 2 θ + 1 θ = 1 2 ( 1 + 2 ) = 1 + 2 2 . \lim_{\theta\to 0^{+}}\dfrac{S(\theta)}{\theta} = \lim_{\theta\to 0^{+}} \dfrac{\dfrac{1}{2}\cos\theta \cdot x \cdot \sin\left(\dfrac{\pi}{2}-\theta\right)}{\theta} \\ =\frac{1}{2}\lim_{\theta\to 0^{+}}\sin\left(\dfrac{\pi}{2}-\theta\right)\cos \theta\cdot\lim_{\theta\to 0^{+}} \frac{\cos\theta\sin\theta+\sin\theta\sqrt{\cos^2\theta+1}}{\theta} \\ =\frac{1}{2}(1+\sqrt{2}) \\ =\boxed{\dfrac{1+\sqrt{2}}{2}}.

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This is what happens when you solve all correctly and mark your answer wrong

Shivam Hinduja - 2 years, 11 months ago

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