SATHEE: Functions 2 Question 10

Functions 2 Question 10

11.

If $f (x) = \cos (\log x)$ , then $f (x) \cdot f (y) - \frac{1}{2} f \frac{x}{y} + f (x y)$ has the value

(1983, 1M)

(a) -1

(b) $\frac{1}{2}$

(c) -2

(d) None of these

Show Answer

Answer:

Correct Answer: 11. (d)

Solution:

Given, $f (x) = \cos (\log x)$

$∴ f (x) \cdot f (y) - \frac{1}{2} [f \frac{x}{y} + f (x y)]$

$= \cos (\log x) \cdot \cos (\log y) - \frac{1}{2} [\cos (\log x - \log y)$

$+ \cos (\log x + \log y)]$

$= \cos (\log x) \cdot \cos (\log y) - \frac{1}{2} [(2 \cos (\log x) \cdot \cos (\log y)]$

$= \cos (\log x) \cdot \cos (\log y) - \cos (\log x) \cdot \cos (\log y) = 0$

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