Match the following :
Column I | Column II |
(i) If a ≥ b > 1, then the largest possible value of the expression loga + logb is | [A] 0 |
(ii) The angle bisector of ∠A in the Δ ABC, where A (–8, 5), B (–15, –19) and C (1, –7) is 13x + by + c = 0 then the value of b + c is equal to | [B] 128 |
(iii) If the tangent to the curve xy + ax + by = 0 at (1, 1) is inclined at an angle tan–1 2 with x axis then a – b is equal to | [C] 2 |
(iv) If 3 sin θ + 4 cosθ = 5, then the value of 4 sinθ – 3 cosθ is | [D] 3 |
Text Solution
Verified by Experts(i) [A]; (ii) [B]; (iii) [D]; (iv) [A]
Ans.
(i) [A]
(ii) [B]
(iii) [D]
(iv) [A]
Sol. (i) log a 
+ log b 
= log a a –log a b + log b b – log b a
= 2 – (log a b+ log b a)
⇒ maximum value = 2 – 2 = 0
( log a b + log b a ≥ 2)
(ii) AB = 25 and AC = 15
Let M is the point of intersection of angle bisector and BC. Then M will divide BC in the ratio of 5: 3. Hence coordinates of M will be (–5, 23/2)
∴ Equation of AM is
y – 5 = 
(x + 8)
⇒ 13x – 6y + 134 = 0
b = – 6, c = 134
⇒ b + c = 128
(iii) xy + ax + by = 0
diff. w.r. t. x
x 
+ y + a + b 
= 0
∴ 
= 
= – 
= 2
⇒ –a – 1 = 2 + 2b
⇒ a + 2b = –3 …..(i)
also curve passes through (1, 1)
so a + b = –1 …….(ii)
b = –2, a = 1
∴ a – b = 1 – (–2) = 3
(iv) 3 sin θ + 4cos θ = 5
⇒ 9 sin 2 θ + 16 cos 2 θ + 24 sin θ cos θ = 25
⇒ 9 – 9 cos 2 θ + 16 – 16 sin 2 θ + 24 sin θ cos θ = 25
⇒ 9 cos 2 θ + 16 sin 2 θ + 24 sin θ cos θ = 0
⇒ 4 sin θ – 3 cos θ = 0
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