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Question

# List I has four entries and List II has five entries. Each entry of List I is to be matched with one entry of List II. List IList II (A)If x=√6+√6+√6+⋯up to ∞, then x is equal to(P)4(B)If a and x are positive integers suchthat x<a and √a−x,√x,√a+x(Q)5are in A.P., then least possible value of a is(C)If 3a+2b+4c=0,a,b,c∈R and the line ax+by+c=0 always passesthrough a fixed point (p,q), then thevalue of 2p+q is(R)2(D)If k(sin18∘+cos36∘)2=5, then thevalue of k is(S)3(T)6 Which of the following is the only CORRECT combination?

A
CR;DS
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B
CS;DT
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C
CT;DR
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D
CR;DP
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Solution

## The correct option is D C→R;D→P(A) x=√6+√6+√6+⋯up to ∞ ⇒x=√6+x⇒x2−x−6=0⇒x=−2,3 As x>0, so x=3 A→S (B) √a−x,√x,√a+x are in A.P., then 2√x=√a−x+√a+x Squaring both the sides, we get 4x2=2a+2√(a2−x2)⇒4x=2a+2√a2−x2⇒2x−a=√a2−x2 Again squaring both the sides, we get 4x2+a2−4xa=a2−x2⇒5x2=4xa⇒x(5x−4a)=0 As x and a are positive integers, so x=4a5 Hence, the minimum value of a=5 B→S (C) 3a+2b+4c=0 ⇒b=−(3a+4c)2 Now, ax+by+c=0 ⇒ax−(3a+4c)y2+c=0⇒a(x−3y2)+c(1−2y)=0⇒(x−3y2)+ca(1−2y)=0 The fixed point is (p,q)=(34,12) Therefore, 2p+q=2 C→R (D) k(sin18∘+cos36∘)2=5⇒k(√5−14+√5+14)2=5∴k=4 D→P

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