IIT JEE 1983 Maths - MCQ Question 14

Geometry Level 3

If f ( x ) = cos ⁡ ( ln ⁡ x ) f(x)=\cos (\ln x) , then f ( x ) f ( y ) − 1 2 [ f ( x y ) + f ( x y ) ] f(x)f(y)-\frac12\left[f\left(\frac{x}{y}\right)+f(xy)\right] simplifies to which of the following numbers?

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− 2 -2 1 2 \frac12 − 1 -1 None of the others

Shubhamkar Ayare by Shubhamkar Ayare , 22, India

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1 solution

f ( x ) f ( y ) = cos ⁡ ( ln ⁡ x ) cos ⁡ ( ln ⁡ y ) As cos ⁡ A cos ⁡ B = 1 2 [ cos ⁡ ( A − B ) + cos ⁡ ( A + B ) ] = 1 2 [ cos ⁡ ( ln ⁡ x − ln ⁡ y ) + cos ⁡ ( ln ⁡ x + ln ⁡ y ) ] = 1 2 [ cos ⁡ ( ln ⁡ ( x y ) ) + cos ⁡ ( ln ⁡ ( x y ) ) ] = 1 2 [ f ( x y ) + f ( x y ) ] \begin{aligned} f(x)f(y) & = \cos(\ln x) \cos (\ln y) & \small \color{#3D99F6} \text{As } \cos A \cos B = \frac 12[\cos(A-B)+\cos(A+B)] \\ & = \frac 12[\cos(\ln x-\ln y)+\cos(\ln x + \ln y)] \\ & = \frac 12 \left[\cos \left(\ln \left(\frac xy \right) \right)+\cos (\ln (xy))\right] \\ & = \frac 12 \left[f \left( \frac xy \right)+ f(xy) \right] \end{aligned}

⟹ f ( x ) f ( y ) − 1 2 [ f ( x y ) + f ( x y ) ] = 0 \begin{aligned} \implies f(x)f(y) - \frac 12 \left[f \left( \frac xy \right)+ f(xy) \right] & = 0 \end{aligned}

Therefore, the answer is None of the others \boxed{\text{None of the others}} .

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