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Question

List- IList-II(I)32301+x+x2+x3x+1+xdx(P) 1(II)Number of integer(s) in the range of the(Q) 2function f(x)=285+x2+x4 is(III) If limxπ2tan2x(4sin2x+sinx+3(R) 3 3sin2x+4sinx+1)=132k,then k=(IV)a,b and c are three vectors such(S) 4that there magnitudes are in the ratio1:2:3. The angle between each ofthem is π3 and |a+b+c|=10,then magnitude of a is(T) 5(U) 7

Which of the following is the only INCORRECT combination?

A
(I)(U)
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B
(II)(T)
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C
(IV)(R)
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D
(III)(S)
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Solution

The correct option is C (IV)(R)
32301+x+x2+x3x+1+xdx
=32301+x(1+x+x)x+1+xdx
=32[23(1+x)32]30
=7


The denominator of f(x) will always be positive and will lie in [5,)
At x=0, we get f(x)=285
As x, f(x)0.
So, the range of f(x) is (0,285]
Number of integers will be 5 i.e., {1,2,3,4,5}

limxπ2tan2x(4sin2x+sinx+33sin2x+4sinx+1)
=limxπ2sin2xcos2x(4sin2x+sinx+33sin2x4sinx14sin2x+sinx+3+3sin2x+4sinx+1)
=limxπ2sin2x(1sinx)(1+sinx)((sinx1)(sinx2)4sin2x+sinx+3+3sin2x+4sinx+1)
=limxπ2sin2x(1+sinx)((2sinx)4sin2x+sinx+3+3sin2x+4sinx+1)
=1×(21)(1+1)(8+8)
=182=1128=132k
k=4


|a+b+c|=10
|a+b+c|2=100
|a+b+c|2=|a|2+|b|2+|c|2 +2(a.b+b.c+c.a)

|a+b+c|2=|a|2+4|a|2+9|a|2 +2(2|a|2cos(π3)+6|a|2cos(π3)+3|a|2cos(π3))

25|a|2=100
|a|=2



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