f(x) is a quadratic polynomial with maximum value of 3 at x = 2. If f (0) = 2. Find f (6)

Option 1 -

-13

Option 2 -

Option 3 -

Option 4 -

-6

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1 Answer

Answered by

Payal Gupta | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

Let f (x) = ax² + bx + c

f (0)= 2

2 = a (0)² + b (0) + c

c = 2

f (– 2) = 3

a (–2)² +b (–2) + 2 = 3

4a – 2b = 1 ………. (1)

The given equation has a maximum value at x = –2

$\frac{- b}{2 a} = - 2$

b = 4a

put in equation 1

we get, $a = \frac{- 1}{4}, b = - 1$

f (6)= $\frac{- 1}{4} {(6)}^{2} + (- 1) 6 + 2$

= –13

Let f (x) = ax2 + bx + c f (0)= 22 = a (0)2 + b (0) + cc = 2f (– 2) = 3a (–2)2 +b (–2) + 2 = 34a – 2b = 1                    ………. (1)The given equation has a maximum value at x = –2 <math> <mrow> <mfrac> <mrow> <mo>−</mo> <mi>b</mi> </mrow> <mrow> <mn>2</mn> <mi>a</mi> </mrow> </mfrac> <mo>=</mo> <mo>−</mo> <mn>2</mn> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1816433319"> </o:OLEObject></xml><! [endif]->b = 4aput in equation 1 we get, <math> <mrow> <mi>a</mi> <mo>=</mo> <mfrac> <mrow> <mo>−</mo> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>, </mo> <mi>b</mi> <mo>=</mo> <mo>−</mo> <mn>1</mn> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1816433320"> </o:OLEObject></xml><! [endif]->f (6)= <math> <mrow> <mfrac> <mrow> <mo>−</mo> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <msup> <mrow> <mo stretchy="false"> (</mo> <mn>6</mn> <mo stretchy="false">)</mo> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mo stretchy="false"> (</mo> <mo>−</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mn>6</mn> <mo>+</mo> <mn>2</mn> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1816433321"> </o:OLEObject></xml><! [endif]->= –13

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