%%

Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x$$-$$ 1 and it leaves remainder 6 when divided by x + 1; then :

JEE · Math · previous-year question

  1. A.p(2) = 11
  2. B.p(2) = 19
  3. C.p($$-$$ 2) = 19correct
  4. D.p($$-$$ 2) = 11

Answer

C. p($$-$$ 2) = 19

Explanation

Let, P(x) = ax2 + bx + c As, P(0) = 1, $$\therefore\,\,\,$$ a(0)2 + b(0) + c = 1 $$ \Rightarrow $$$$\,\,\,$$ c = 1 $$\therefore\,\,\,$$ P(x) = ax2 + bx + 1 If P(x) is divided by x $$-$$ 1, remainder = 4 $$ \Rightarrow $$$$\,\,\,$$ P$$\left( 1 \right) = 4$$ $$\therefore\,\,\,$$ a + b + 1 = 4 . . . . . (1) If P(x) is divided by x + 1, remainder = 6 $$ \Rightarrow $$$$\,\,\,$$ P($$-$$ 1) = 6 $$\therefore\,\,\,$$ a $$-$$ b + 1 = 6 . . . .(2) By solving (1) and (2) we get, a = 4, and b = $$-$$1 $$\therefore\,\,\,$$ P(x) = 4x2 $$-$$ x + 1 P(2) = 4(2)2 $$-$$ 2 + 1 = 15 P($$-$$ 2) = 4 ($$-$$2)2 $$-$$ ($$-$$ 2) + 1 = 19

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions