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Point (a,3) inside the shaded annular sector of concentric circles of radii 4 and 2 - JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
Let C1C_1 and C2C_2 be two concentric circles centred at the origin, of radii 44 and 22 units. Shade the part of the annulus between them that lies in the first quadrant above the line y=xy=x, that is, the annular sector from 45∘45^\circ to 90∘90^\circ. If the point (a,3)(a,3) lies inside this shaded region, the interval of aa is
Aa>0a>0
B−7<a<7-\sqrt7 < a < \sqrt7
Ca<3a<3
D0<a<70 < a < \sqrt7correct
Solution
Question attachment Step 1: Inside the shaded region means inside the big circle, outside the small circle, in the first quadrant, and above the line y=xy=x. Step 2: Inside x2+y2=16x^2+y^2=16, and outside x2+y2=4x^2+y^2=4:
a2+9−16<0  ⟹  a2<7  ⟹  −7<a<7a^2+9-16 < 0 \;\Longrightarrow\; a^2 < 7 \;\Longrightarrow\; -\sqrt7 < a < \sqrt7
a2+9−4>0  ⟹  a2+5>0a^2+9-4 > 0 \;\Longrightarrow\; a^2+5 > 0
 The second holds for every real aa, so it puts no restriction. Step 3: Above the line y=xy=x, i.e. y>xy>x, and in the first quadrant:
3>a  ⟹  a<33 > a \;\Longrightarrow\; a < 3
 a>0a > 0 Step 4:
a∈(−7,7)∩R∩(−∞,3)∩(0,∞)=(0,7)a \in (-\sqrt7,\sqrt7) \cap \mathbb{R} \cap (-\infty,3) \cap (0,\infty) = \left(0,\sqrt7\right)
 since 7≈2.65<3\sqrt7 \approx 2.65 < 3, so a<3a<3 is automatically satisfied. Answer: (4) 0<a<70 < a < \sqrt7
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