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List - 1List - 2(I)If f(x)=e[x] and g(x)=x24x+3x22x+3,(P)    0then number of integer(s) in the range of (fg)(x) is(where [.] represents the greatest integer function)(II)40z18117n=0zn dz(Q)    1(III)    In a ΔXYZ,  y2sin(2Z)+z2sin(2Y)=2yz,(R)    2where y=15,z=8. Then the length of inradius is(IV)The number of integers in the range of the function(S)    3f(x)=sin1xcos1x+tan1xcot1x  is (T)    4(U)    5

Which of the following has CORRECT pair of combination? 
  1. (I)(T), (III)(P)
  2. (I)(Q), (II)(T)
  3. (II)(U), (IV)(R)
  4. (III)(S), (IV)(Q)


Solution

The correct option is B (I)(Q), (II)(T)
(I)
(fg)(x)=ex24x+3x22x+3
Let y=x24x+3x22x+3
(y1)x2+(42y)x+3y3=0
It is a quadratic equation and its discriminant should be 0
(42y)24(y1)(3y3)0
y[132,1+32]
Now, for y[132,1+32]

Range of (fg)(x) ={e1,e0,e1}
Only integer in the range of (fg)(x) is e0=1
(I)(Q)

(II)
40z18117n=0zndz=40(z1) dz
40z18117n=0zndz=4224=4
(II)(T)

(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 r=Δs=15×815+8+17=12040=3
(III)(S)

(IV)
sin1xcos1x0
2sin1xπ2
x[12,1]

tan1xcot1x0
2tan1xπ2
x[1,]

Therefore, domain of f(x) is [12,1][1,]={1}

At x=1
sin11cos11=π2
and tan11cot11=0
f(1)=π2
(IV)(P) 

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