Let f(θ)=1−sinθ−1cosθ1sinθ1−cosθ1=(1+sinθcosθ)−cosθ(−sinθ−cosθ)+1(−sin2θ+1)=1+sinθcosθ+sinθcosθ+=2+2sinθcosθ+cos2θ=2+sin2θ+cos2θ Now, maximum value of (1) is 2+12+12=2+2 and minimum value of (1) is 2−12+12=2−2.
If f(θ)= | … | and A and B are respectively the maximum and the minimum values of f(θ), then (A, B) is equal to:
JEE Main 2014 — Mathematics Algebra
If f(θ)=1−sinθ−1cosθ1sinθ1−cosθ1 and A and B are respectively the maximum and the minimum values of f(θ), then (A,B) is equal to:
Held on 12 Apr 2014 · Verified 6 Jul 2026.
Options
- A
(3,−1)
- B
(4,2−2)
- C
(2+2,2−2)
- D
(2+2,−1)
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