JEE MAIN - Mathematics (2017 - 8th April Morning Slot - No. 4)
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 :
p(2) = 11
p(2) = 19
p($$-$$ 2) = 19
p($$-$$ 2) = 11
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
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
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