Column- Ι Column- ΙΙ
on 
is 
and if α = 2 + 
, then possible value(s) of | α | is (are)
is differentiable for all x ∈ R. Then possible value(s) of a is (are)
Text Solution
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( a )→ P,Q ; → P, Q ; → P,Q,S,T ; → Q, T
Sol. 
⇒ 
⇒ 3 α + α – 2 = ± 6 ⇒ 4 α = 8, –4 ⇒ α = 2, –1
(A → P, Q)
Continuous ⇒ –3a – 2 = b + a 2
differentiable ⇒ –6a = b ⇒ 6a = a 2 + 3a + 2
⇒ a 2 – 3a + 2 = 0 ⇒ a = 1,2
(B → P, Q)
⇒ ab = 2a + 2b ......(i)
q = 10 – a and 2q = 5 + b
⇒ 20 – 2a = 5 + b ⇒ 15 = 2a + b .......(ii)
From (i) and (ii) a (15 – 2a) = 2a + 2(15 – 2a)
⇒ 15a – 2a 2 = –2a + 30 ⇒ 2a 2 – 17a + 30 = 0 ⇒ a = 6, 
⇒ q = 4, 
⇒ |q – a| = 2, 5
(D → Q, T)
Let a = 3 – 3 ω + 2 ω 2
a ω = 3 ω – 3 ω 2 + 2
a ω 2 = 3 ω 2 – 3 + 2 ω
Now a 4n + 3 (1 + ω 4x + 3 + ( ω 2 ) 4n + 3 ) = 0
⇒ n should not be a multiple of 3
Hence P, Q, S, T
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