If tangents are drawn to the parabola $y = x^{2} + b x + c$ (for b and c fixed real number) at the points $(i, y_{i})$ for i = 1, 2, …, 10. Let $l_{1}, l_{2,} l_{3} . . . . . . l_{9}$ be the point of intersection of tangents at $(i, y_{i})$ and $(i + 1, y_{i + 1})$ then the least degree polynomial whose graph passes through all nine points, is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$y = x^{2} + b x + c - \frac{1}{2}$

$y = x^{2} + b x + c - \frac{1}{4}$

$y = x^{2} + b x + c$

$y = x^{2} + b x + c - \frac{1}{8}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} Equation of tangent at (i, y_{i}) \\ y = (2 i + b) x - i^{2} + c \to (1) (s i n c e \frac{d y}{d x} = 2 x + b = 2 i + b a t (i, y_{i})) \\ Also at (i + 1, y_{i + 1}) \\ y = (2 i + 2 + b) x - {(i + 1)}^{2} + c \to (2) \\ s o l v e (1), (2) t h e n x = \frac{2 i + 1}{2} \Rightarrow i = \frac{2 x - 1}{2} \\ Put this i in any tangent, we get y = x^{2} + b x + c - \frac{1}{4} \end{array}$