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NCERT Class XI Mathematics - Relations and Functions - Solutions

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Question : 23 of 36
Marks: +1, -0
The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t (C) = 9C5+32\frac{9C}{5}+32
Find
(i) t (0)
(ii) t (28)
(iii) t (–10)
(iv) The value of C, when t (C) = 212.
Solution:  
We are given t (C) = 9C5+32\frac{9C}{5}+32
(i) t (0) = 9(0)5\frac{9(0)}{5} + 32 = 0 + 32 = 32
(ii) t (28) = 9(28)5\frac{9(28)}{5} + 32 = 2525\frac{252}{5} + 32 = 252+1605\frac{252+160}{5} = 4125\frac{412}{5} = 82.4
(iii) t (- 10) = 9(−10)5\frac{9(-10)}{5} + 32 = - 18 + 32 = 14
(iv) t(C) = 212
⇒ 9C5\frac{9C}{5} + 32 = 212 ⇒ 9C5\frac{9C}{5} = 212 - 32 ⇒ 9C5\frac{9C}{5} = 180 ⇒ C = 180×59\frac{180 \times 5}{9} = 100
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