If the mapping f(x)=ax+b,a < 0 maps [-1,1]onto [0,2] then-Turito

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Question

If the mapping $f left parenthesis x right parenthesis equals a x plus b comma a less than 0$ maps [-1,1] onto [0,2] then for all values of $theta$ , $A equals cos squared space theta plus sin to the power of 4 space theta$ is such that

$f (\frac{1}{4}) \leq A \leq f (0)$
$f (0) \leq A \leq f (- 1)$
$f (\frac{1}{3}) \leq A \leq f (0)$
$f (1) < A \leq f (- 1)$

Hint:

In the question we will simplify the equation to make it a quadratic after that we can easily find the minimum and maximum value.

The correct answer is: $f open parentheses 1 fourth close parentheses less or equal than A less or equal than f left parenthesis 0 right parenthesis$

$A equals cos squared space theta plus sin to the power of 4 space theta equals cos squared theta plus open parentheses 1 minus cos squared theta close parentheses squared equals cos squared theta plus 1 plus cos to the power of 4 theta minus 2 cos squared theta equals cos to the power of 4 theta minus cos squared theta plus 1 l e t space cos squared theta space b e space t. S o comma space A equals t squared minus t plus 1 a s space cos squared theta space equals t space s o comma space 0 less or equal than t less or equal than 1 m i n i m u m space v a l u e space equals fraction numerator negative open parentheses b squared minus 4 a c close parentheses over denominator 4 end fraction equals fraction numerator negative open parentheses open parentheses negative 1 close parentheses squared minus 4.1.1 close parentheses over denominator 4 end fraction equals fraction numerator negative open parentheses 1 minus 4 close parentheses over denominator 4 end fraction equals 3 over 4 m a x i m u m space v a l u e equals 0 minus 0 plus 1 space equals 1 t h e r e f o r e comma space r a n g e open square brackets 3 over 4 comma 1 close square brackets a s space f open parentheses 1 fourth close parentheses equals 3 over 4 space a n d space f open parentheses 0 close parentheses equals 1 s o comma space t h e space r e q u i r e d space r a n g e space i s space f open parentheses 1 fourth close parentheses less or equal than A less or equal than f open parentheses 0 close parentheses$

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Question

If the mapping $f left parenthesis x right parenthesis equals a x plus b comma a less than 0$ maps [-1,1] onto [0,2] then for all values of $theta$ , $A equals cos squared space theta plus sin to the power of 4 space theta$ is such that

$f (\frac{1}{4}) \leq A \leq f (0)$
$f (0) \leq A \leq f (- 1)$
$f (\frac{1}{3}) \leq A \leq f (0)$
$f (1) < A \leq f (- 1)$

Hint:

Answer

The correct answer is: $f open parentheses 1 fourth close parentheses less or equal than A less or equal than f left parenthesis 0 right parenthesis$

Solution

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If the mapping maps [-1,1] onto [0,2] then for all values of , is such that

Hint:

Answer

The correct answer is:

Solution

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