CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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 ax,x,a+x(Q)5are in A.P., then least possible value of a is(C)If 3a+2b+4c=0,a,b,cR 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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
CS;DT
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
CT;DR
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
CR;DP
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D CR;DP
(A) x=6+6+6+up to
x=6+xx2x6=0x=2,3
As x>0, so x=3
AS

(B) ax,x,a+x are in A.P., then
2x=ax+a+x
Squaring both the sides, we get
4x2=2a+2(a2x2)4x=2a+2a2x22xa=a2x2
Again squaring both the sides, we get
4x2+a24xa=a2x25x2=4xax(5x4a)=0
As x and a are positive integers, so x=4a5
Hence, the minimum value of a=5
BS

(C) 3a+2b+4c=0
b=(3a+4c)2
Now, ax+by+c=0
ax(3a+4c)y2+c=0a(x3y2)+c(12y)=0(x3y2)+ca(12y)=0
The fixed point is (p,q)=(34,12)
Therefore, 2p+q=2
CR

(D)
k(sin18+cos36)2=5k(514+5+14)2=5k=4
DP

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon