If area bounded by the curve y = cos[x], x–axis and the lines given by $x^{2} - 3 x + 2 = 0$ is A, then (where [x] represents greatest integer function)

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$A = | \int_{1}^{π / 2} \cos [x] d x | + \int_{π / 2}^{2} \cos [x] d x$

$A = \int_{1}^{π / 2} \cos [x] d x + | \int_{π / 2}^{2} \cos [x] d x |$

A < sin1

A = 2cos1

answer is A, C, D.

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Detailed Solution

y = cos[x], x–axis, lines by $x^{2} - 3 x + 2 = 0$ (x = 1 and x = 2)
As $x \in (1, 2), [x] = 1$ and cos[x] = cos1
Question Image
Area = A = cos1
(A) $A = \int_{1}^{π / 2} \cos [x] d x + | \int_{π / 2}^{2} \cos [x] d x | = \cos 1 (\frac{π}{2} - 1) + \cos 1 (2 - \frac{π}{2}) = \cos 1$
(C) $A = | \int_{1}^{π / 2} \cos [x] d x | + \int_{π / 2}^{2} \cos [x] d x = \cos 1 (\frac{π}{2} - 1) + \cos 1 (2 - \frac{π}{2}) = \cos 1$
(D) As $(\frac{π}{4}, \frac{π}{2})$ , cosx < sinx
$\Rightarrow$ cos1 < sin1
Hence, ACD is correct.