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List - IList - II(I)If A and B are two sets with elements 4 and 3,(P) 1respectively. Then the number of onto functionfrom A to B is(II)The number of points in the interval [13,13](Q) 3at which f(x)=sin(x2)+cos(x2) attains itsmaximum value, is(III) In a ΔXYZ, y2sin(2Z)+z2sin(2Y)=2yz,(R) 4where y=15,z=8. Then the length of inradius is(IV)logx+1(4x3+9x2+6x+1)+log4x+1(x2+2x+1)(S) 8=5, then the number of solution is(T) 24(U) 36

Which of the following has INCORRECT pair of combination?

A
(I)(U)
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B
(II)(S)
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C
(III)(Q)
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D
(IV)(P)
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Solution

The correct option is B (II)(S)
(I)
The number of onto functions from A to B is :
=3k=0(1)k nCk(3k)4
=34 3C1×(31)4+ 3C2×(32)4
=8148+3=36

(II)
f(x)=sin(x2)+cos(x2)
x2[0,13]
Maximum value of f(x)=2 and its attans at
sin(x2)=cos(x2)
tan(x2)=1
x2=π4,9π4<13x=±π2,±3π2
Hence, there are 4 points in x[13,13]

(III)
y2sin(2Z)+z2sin(2Y)=2yz
y2(2sinZcosZ)+z2(2sinYcosY)=2yz
Multiplying both sides by R where R is the circumradius.
y2(zcosZ)+z2(ycosY)=2Ryz
ycosZ+zcosY=2R
x=2R
X=90

x2=y2+z2 =225+64 =289
x=17

Now, inradius is, r=Δs=15×815+8+17=12040=3

(IV)
logx+1(4x3+9x2+6x+1)+log4x+1(x2+2x+1)=5
x+1>0,x0 and4x+1>0,x0 (1)

Now,
logx+1[(x+1)2(4x+1)]+log4x+1(x+1)2=5
2logx+1(x+1)+logx+1(4x+1)+2log4x+1(x+1)=5
logx+1(4x+1)+2logx+1(4x+1)=3

Put logx+1(4x+1)=t
t+2t=3
t=1,2

When
t=1logx+1(4x+1)=1
4x+1=x+1x=0
Which is not possible [from eqn(1)]

When
t=2logx+1(4x+1)=2
4x+1=(x+1)2x=0,2 but x0
x=2 is the only solution.

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