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The acute angle between F(1)(x)=int(2)^...

The acute angle between `F_(1)(x)=int_(2)^(x) (2t-5)dt and F_(2)(x)=int_(0)^(x)2t dt," is , if "tantheta=a/b` where a, b are co-prime numbers then b – a is:

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The correct Answer is:
6
Let `F_(1) (x)=y_(1)= int_(2)^(x) (2t-5) dt and F_(2)(x)=y_(2)=int_(2)^(x) 2tdt`
Now point of intersection means those point at which `y_(1)=y_(2)=y rArr y_(1)=x^(2)-5x+6 and y_(2)=x^(2)`
On solving, we get `x^(2)=x^(2)-5x+6 rArr x=6/5 and y=x^(2)=(36/(25)`
Thus point of intersection is `6/5,(36)/(25))` slope of tangents are `12/5 and (-13)/(5)`
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