Match the following:
Column - I | Column-II |
(i) If has integral roots and , then a can be equal to | [A]2 |
(ii) If the equation has no real roots and + 4 then integral value of c can be equal to | [B]6 |
(iii) If the equation has only negative roots then integral values of can be | [C]12 |
(iv) The largest positive term of the H.P. whose first two terms are and is | [D] 20 |
Text Solution
Verified by Experts(i) [A]; (ii) [B]; (iii) [A]; (iv) [B]
Ans.
(i) [A], [B], [C], [D]
(ii) [B], [C], [D]
(iii) [A], [C], [D]
(iv) [B]
Sol. (i) Discriminant, D = 
⇒ 1 + 4a should be a perfect square.
As 1 + 4a is always odd
⇒ 1 + 4a = (2 λ + 1) 2 , λ ∈ I + ⇒ a = λ ( λ + 1)
(ii) Let f(x) = ax
2 + 2bx + 4c – 16
Clearly f(–2) = 4a – 4b + 4c – 16
= 4 (a – b + c – 4) > 0
= f(x) > 0, ∀ ∀ x ∈ R
⇒ f(0) > 0 ⇒ 4c – 16 > 0 ⇒ c > 4
(iii) Both the roots of the equation
f(x) = x 2 + 2bx + 9b – 14 = 0 will be negative if
D ≥ 0 ⇒ 4b 2 – 4 (9b – 14) ≥ ≥ 0 …(i)
9b – 14 > 0 …(ii)
From (i) b 2 – 9b + 14 ≥ 0
⇒ (b – 2) (b – 7) ≥ 0
i.e. b ≥ 7 or b ≤ 2
From equation (ii) b > 
All these conditions on b are simultaneously satisfied if
b ≥ 7 or b ∈ 
(iv) Let the H.P. be 
, 
, 
, 
, ….
Then 
and 
from which we get a = 5/2
and d = – 7/12. Now the nth term of the H.P. is
= 
= 
So, the nth term is largest when 37 – 7n has the least positive value, which occurs for n = 5.
Hence the largest term is 
= 6.
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