JEE Advance - Mathematics (2009 - Paper 1 Offline - No. 16)
Match the statements/expressions in Column I with the open intervals in Column II :
| Column I | Column II | ||
|---|---|---|---|
| (A) | Interval contained in the domain of definition of non-zero solutions of the differential equation $${(x - 3)^2}y' + y = 0$$ | (P) | $$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$ |
| (B) | Interval containing the value of the integral $$\int\limits_1^5 {(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)dx} $$ | (Q) | $$\left( {0,{\pi \over 2}} \right)$$ |
| (C) | Interval in which at least one of the points of local maximum of $${\cos ^2}x + \sin x$$ lies | (R) | $$\left( {{\pi \over 8},{{5\pi } \over 4}} \right)$$ |
| (D) | Interval in which $${\tan ^{ - 1}}(\sin x + \cos x)$$ is increasing | (S) | $$\left( {0,{\pi \over 8}} \right)$$ |
| (T) | $$( - \pi ,\pi )$$ |
(A)$$\to$$(P), (Q), (S); (B)$$\to$$(P), (T), (S); (C)$$\to$$(P), (Q), (R), (T); (D)$$\to$$(S)
(A)$$\to$$(P), (Q), (S); (B)$$\to$$(P), (T), (R); (C)$$\to$$(P), (Q), (R), (T); (D)$$\to$$(R)
(A)$$\to$$(P), (Q), (S); (B)$$\to$$(P), (T), (S); (C)$$\to$$(S), (Q), (R), (T); (D)$$\to$$(S)
(A)$$\to$$(P), (T), (S); (B)$$\to$$(P), (T), (S); (C)$$\to$$(P), (Q), (R), (T); (D)$$\to$$(S)
Explanation
(A) We have $${(x - 3)^2}{{dy} \over {dx}} + y = 0$$
$$\int {{{dx} \over {{{(x - 3)}^2}}} = - \int {{{dy} \over y}} } $$
$$ \Rightarrow {1 \over {x - 3}} = \ln |y| + c$$
So the domain is $$R \to \{ 3\} $$.
(B) On substituting $$x = t + 3$$, we get
$$\int\limits_{ - 2}^2 {(t + 2)(t + 1)t(t - 1)(t - 2)dt = \int\limits_{ - 2}^2 {t({t^2} - 1)({t^2} - 4)dt = 0} } $$ (being odd function)
(C) $$f(x) = {5 \over 4} - {\left( {\sin x - {1 \over 2}} \right)^2}$$
The maximum value occurs when $$\sin x = 1/2$$.
(D) $$f'(x) > 0$$ if $$\cos x > \sin x$$.
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