Let A= {−4, −2, −1, 0, 3, 5} and f: A→R be defined by f(x)={3x−1 for x>3x2+1for −3≤x≤32x−3for x < −3Then the range of f is

A= {−4, −2, −1,0,3,5}f(x)={3x−1forx> 3x2+1for−3≤x≤32x−3forx< −3Range = f(A)={f(−4), f(−2), f(−1),f(0), f(3), f(5)}f(−4)=2(−4)−3=−11f(−2)=(−2)2+1=5f(−1)=(−1)2+1=2f(0)=02+1=1f(3)=3(3)−1=8f(5)=3(5)−1=14∴ Range = {−11, 5, 2,1,8,14}

Let A= {−4, −2, −1, 0, 3, 5} and f: A→R be defined by f(x)={3x−1 for x>3x2+1for −3≤x≤32x−3for x < −3Then the range of f is

A= {−4, −2, −1,0,3,5}f(x)={3x−1forx> 3x2+1for−3≤x≤32x−3forx< −3Range = f(A)={f(−4), f(−2), f(−1),f(0), f(3), f(5)}f(−4)=2(−4)−3=−11f(−2)=(−2)2+1=5f(−1)=(−1)2+1=2f(0)=02+1=1f(3)=3(3)−1=8f(5)=3(5)−1=14∴ Range = {−11, 5, 2,1,8,14}

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$\begin{array}{l} L e t A = {- 4, - 2, - 1, 0, 3, 5} a n d f : A \to R b e d e f i n e d b y f (x) = {\begin{cases} 3 x - 1 f o r x > 3 \\ x^{2} + 1 f o r - 3 \leq x \leq 3 \\ 2 x - 3 f o r x < - 3 \end{cases} \\ T h e n t h e r a n g e o f f i s \end{array}$

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${- 11, 5, 2, 1, 8, 14}$

${- 11, - 7, 2, 1, 8, 14}$

${- 11, - 7, - 5, 1, 10, 14}$

${- 11, 5, 2, 1, 10, 14}$

answer is B.

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Detailed Solution

$\begin{array}{l} A = {- 4, - 2, - 1, 0, 3, 5} \\ f (x) = {\begin{cases} 3 x - 1 f o r x > 3 \\ x^{2} + 1 f o r - 3 \leq x \leq 3 \\ 2 x - 3 f o r x < - 3 \end{cases} \\ R a n g e = f (A) = {f (- 4), f (- 2), f (- 1), f (0), f (3), f (5)} \\ f (- 4) = 2 (- 4) - 3 = - 11 \\ f (- 2) = {(- 2)}^{2} + 1 = 5 \\ f (- 1) = {(- 1)}^{2} + 1 = 2 \\ f (0) = 0^{2} + 1 = 1 \\ f (3) = 3 (3) - 1 = 8 \\ f (5) = 3 (5) - 1 = 14 \\ ∴ R a n g e = {- 11, 5, 2, 1, 8, 14} \end{array}$