We have, curve y=x3−2x2+3x−4 intersects the line ‌y=−2‌ at ‌P(h,k) ‌⇒‌‌k=−2‌ and ‌−2=h3−2h2+3h−4 ‌⇒‌‌h3−2h2+3h−2=0 ‌(h−1)(h2−h+2)=0⇒h=1 So, point P≡(1,−2) ‌
dy
dx
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|‌at ‌P=3x2−4x+3=2 Now, equation of tangent at P(1,−2) is ‌y+2=2(x−1) ⇒‌‌2x−y−4=0 ∵ Tangent at P meets X-axis at (x1,y1) ⇒‌‌y1=0 From Eq. (i), we get 2x1−0−4=0⇒x1=‌
