Let A(m) be the area bounded by the curves y=x2−3 and y=mx+2, then
A
The range of A(m) is [10√53,∞).
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B
The range of A(m) is [20√53,∞).
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C
A(m) is many-one function for m∈[−2,∞)
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D
The area is minimum when m=1.
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Solution
The correct option is CA(m) is many-one function for m∈[−2,∞)
The line y=mx+2 is always passes through (0,2).
The point of intersection of the line and the parabola is, mx+2=x2−3⇒x2−mx−5=0
Let the roots of the equation be a,b where a>0;b<0
Now, a+b=mab=−5(a−b)=√(a+b)2−4ab=√m2+20
Now the area will be, A(m)=a∫bmx+2−x2+3dx=[−x33+mx22+5x]ab=(a−b)[−(a2+b2+ab3)+m(a+b)2+5]=√m2+20[−13((a+b)2−ab)+m22+5]=√m2+20[−13(m2+5)+m22+5]⇒A(m)=(m2+20)3/26
Area will me minimum when m=0
So the minimum area will be, A(0)=203/26=20√53 A(m) will be many one function in the given domain.