Let A=⎡⎢⎣−3021x5−20x2⎤⎥⎦,B=⎡⎢⎣2b−1⎤⎥⎦ and C=[351]. If tr denotes the trace of a matrix, then the number of integral values of b for which tr(ABC)≤−18 for all x∈R, is
A
3
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B
4
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C
5
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D
6
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Solution
The correct option is C5 ABC=⎡⎢⎣−3021x5−20x2⎤⎥⎦⎡⎢⎣2b−1⎤⎥⎦[351]
=⎡⎢⎣−8bx−3−x2−4⎤⎥⎦[351]
=⎡⎢⎣−24−40−83bx−95bx−15bx−3−3x2−12−5x2−20−x2−4⎤⎥⎦
tr(ABC)=−x2+5bx−43≤−18
i.e., −x2+5bx−25≤0
As the coefficient of x2 is negative, so for this inequality D≤0 ⇒25b2−100≤0 ⇒b2−4≤0 ⇒b∈[−2,2]