Plzz solve this question for me.. Plzz solve this question for me.. 30.1f 4b2 4c2 4b 1 1 f(-l) f(l) f(2) 3b2 +3b , f (x) is a quadratic function and its maximum value 3c2 + occurs at a point V. A is a point of intersection of y = f(x) with x-axis and point B is such that chord AB subtends a right angle at V. Find the area enclosed by f(x) and chord AB. 125 sq. units 3 115 (b) — sq. units 135 c — sq. units 3 (d) none of these Share with your friends Share 0 Zuhaib Ul Zamann answered this Dear student, we have4a2f(-1)+4af(1)+f(2)=3a2+3aie a2(4f(-1)-3)+a(4f(1)-3)+f(2)=0Similarlyb2(4f(-1)-3)+b(4f(1)-3)+f(2)=0andc2(4f(-1)-3)+c(4f(1)-3)+f(2)=0Hence a,b,c are three roots of quadraticx2(4f(-1)-3)+x(4f(1)-3)+f(2)=0 (1)But a quadratic can have at maximum two rootsHence 1 is an identity in xHence coefficient of x2=coefficient of x=constant term=0i.e4f(-1)-3=0f(-1)=34 (2)4f(1)-3=0f(1)=34 (3).andf(2)=0 (4)Since f(x) is quadratic we havef(x)=Ax2+Bx+CFrom 2 we haveA-B+C=34 (5)From 3 we haveA+B+C=34 (6)From 4 we have4A+2B+C=0 (7)Subtracting 5 from 6 we have2B=0B=0put in 5 we get A+C=34 (8)put in 7 we have4A+C=0 (9)Subtracting 8 from 9 we have3A=-34A=-14put in 9 we getC=1Hence f(x)=-x24+1Now max of f(x) occurs at x=-B2A=0.Hence V≡(0,1)Now let A≡(h,0)A lies on f we have-h24+1=0i.e h=±2Choose h=2 {Due to symmetry answers will be same from 2 or -2}Now B≡(k,r)so we haver=-k24+1Also AV is perpendicular to BVWe have0-12-0×r-1k=-r=2k+1Substituting we have2k+1=-k24+1i.ek=0 or k=-8k=0 is rejected because for k=0 V and B will coincideSo k=-8r=-15Now the area is to be found by integrationA=∫-82(-x24+1-32(x-2))dx=1253Regards -1 View Full Answer
