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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)The minimum value of ab if roots ofx3−ax2+bx−2=0 are positive, is(P)36(B)The number of quadrilateral formed in an octagon having two sides common(Q)24with the octagon, is(C)If 2nC4, 2nC5 and 2nC6 are in A.P.,then the value of 2n is (R)18(D)The value of 72sinπ18sin5π18sin7π18 is(S)14(T)9

Which of the following is the only CORRECT combination?

List IList II (A)The minimum value of ab if roots ofx3−ax2+bx−2=0 are positive, is(P)36(B)The number of quadrilateral formed in an octagon having two sides common(Q)24with the octagon, is(C)If 2nC4, 2nC5 and 2nC6 are in A.P.,then the value of 2n is (R)18(D)The value of 72sinπ18sin5π18sin7π18 is(S)14(T)9

Which of the following is the only CORRECT combination?

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Solution

The correct option is **B** A→R;B→P

(A) Let x1,x2,x3 be the roots of the equation x3−ax2+bx−2=0, then

A.M.≥H.M.⇒x1+x2+x33≥3x1x2x3x1x2+x2x3+x3x1⇒a3≥3×2b∴ab≥18

Hence, the minimum value of ab is 18.

A→R

(B) Number of quadrilateral having two adjacent sides common

= 8C1× 3C1=8×3=24

Number of quadrilateral not having two adjacent sides common

=12( 8C1× 3C1)=12

Hence, total number of quadrilateral =24+12=36

(C) 2nC4, 2nC5 and 2nC6 are in A.P., so

2× 2nC5= 2nC4+ 2nC6⇒2= 2nC4 2nC5+ 2nC6 2nC5⇒2=52n−4+2n−56⇒2n2−21n+49=0⇒(n−7)(2n−7)=0∴n=7 (∵n∈N)

Hence, 2n=14

C→S

(D) 72sinπ18sin5π18sin7π18 =72sin10 ∘sin50 ∘sin70 ∘ =72sin(60 ∘−10 ∘)sin(10 ∘)sin(60 ∘+10 ∘) =72×14sin(3×10 ∘) =9

D→T

(A) Let x1,x2,x3 be the roots of the equation x3−ax2+bx−2=0, then

A.M.≥H.M.⇒x1+x2+x33≥3x1x2x3x1x2+x2x3+x3x1⇒a3≥3×2b∴ab≥18

Hence, the minimum value of ab is 18.

A→R

(B) Number of quadrilateral having two adjacent sides common

= 8C1× 3C1=8×3=24

Number of quadrilateral not having two adjacent sides common

=12( 8C1× 3C1)=12

Hence, total number of quadrilateral =24+12=36

(C) 2nC4, 2nC5 and 2nC6 are in A.P., so

2× 2nC5= 2nC4+ 2nC6⇒2= 2nC4 2nC5+ 2nC6 2nC5⇒2=52n−4+2n−56⇒2n2−21n+49=0⇒(n−7)(2n−7)=0∴n=7 (∵n∈N)

Hence, 2n=14

C→S

(D) 72sinπ18sin5π18sin7π18 =72sin10 ∘sin50 ∘sin70 ∘ =72sin(60 ∘−10 ∘)sin(10 ∘)sin(60 ∘+10 ∘) =72×14sin(3×10 ∘) =9

D→T

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