Question

# For a fixed value of θ, if n1 denotes the number of points on the line 3x+4y=5 which are at a distance of 1+sin2θ units from (2,3) and n2 denotes the number of points on 3x+4y=5 which are at a distance of sec2θ+2 cosec2θ units from (1,3), then the value of n1+n2 is

A
2
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B
2.00
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C
2.0
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Solution

## Perpendicular distance from (2,3) to the line 3x+4y−5=0 is |3×2+4×3−5|√32+42=135 1+sin2θ< Perpendicular distance from (2,3) to the line 3x+4y−5=0 ∴n1=0 Perpendicular distance from (1,3) to the line 3x+4y−5=0 is |3×1+4×3−5|√32+42=2 Now, sec2θ+2 cosec2θ =3+tan2θ+2cot2θ ≥3+2√2 (Using AM > GM) sec2θ+2 cosec2θ> Perpendicular distance from (1,3) to the line 3x+4y−5=0 ∴n2=2

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