1

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?

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?

Open in App

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

(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

0

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program