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Rational Function

A quadratc fu...

Question

A quadratc function in x has the value 10 when x = 1 and has minimum value 1 when x = -2 the function is

A

2 x^{2} + 3 x + 5

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B

3 x^{2} + 2 x + 5

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C

x^{2} + 3 x + 6

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D

x^{2} + 4 x + 5

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Solution

The correct option is D $x^{2} + 4 x + 5$
Let quadratic $f (x) = a x^{2} + b x + c$
$f (1) = a + b + c = 10$
$f (- 2) = 4 a - 2 b + c = 1$
$f^{'} (x) = 2 a x + b$ (at extremum $f^{'} (x) = 0$ )
$f^{'} (- 2) = - 4 a + b = 0$
$b = 4 a$
So $a = 1, b = 4, c = 5$

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Similar questions

Q. Match the following.

	Column - I		Column - II
(P)	$(x^{2} + 5) (x^{3} + 3) + 5$	(1)	${- x}^{3} - 3 x^{2} + 3 x + 2$
(Q)	$(\frac{- 10}{3} x y^{3}) \times (\frac{6}{5} x^{3} y)$	(2)	$- x^{3} + x^{2} + 3 x - 6$
(R)	$(x^{3} - x^{2} - x - 2) - (2 x^{3} + 2 x^{2} - 4 x - 4)$	(3)	$x^{5} + 5 x^{3} + 3 x^{2} + 20$
(S)	$(x^{3} - x^{2} - x - 2) + (2 x^{2} - 2 x^{3} + 4 x - 4)$	(4)	$- 4 x^{4} y^{4}$

Find the value of -

(x^{2} - 3 x + 5) + (2 x^{2} - x - 2) - (3 x^{2} + 7 x - 3)

Solve graphically.

(1)

(2)

(3)

(4)

(5)

(6)

f (x) = x^{2} - 3 x + 4

has minimum value of

x

Q. A cubic function of x has maximum value 10 and minimum

\frac{- 5}{2}

x = - 3, x = 2

then the function is

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