IIT-JEE 2009 · Paper 1 · Mathematics Q39

MathematicsMatrix match+8Question 39 of 60

Match the statements/expressions in Column I with the open intervals in Column II.

Column I

A
Interval contained in the domain of definition of non-zero solutions of the differential equation $(x-3)^2y'+y=0$
B
Interval containing the value of the integral $\displaystyle\int_1^5(x-1)(x-2)(x-3)(x-4)(x-5)\,dx$
C
Interval in which at least one of the points of local maximum of $\cos^2x+\sin x$ lies
D
Interval in which $\tan^{-1}(\sin x+\cos x)$ is increasing

Column II

P
$\left(-\frac\pi2,\frac\pi2\right)$
Q
$\left(0,\frac\pi2\right)$
R
$\left(\frac\pi8,\frac{5\pi}4\right)$
S
$\left(0,\frac\pi8\right)$
T
$(-\pi,\pi)$

Answer and worked solution

Official answer: A → P, Q, S; B → P, T; C → P, Q, R, T; D → S

A: $\dfrac{dy}y=-\dfrac{dx}{(x-3)^2}$ gives $y=Ce^{1/(x-3)}$ , defined for $x\ne3$ . The intervals not containing 3 are P, Q and S ( $\frac{5\pi}4>3$ and $\pi>3$ , so R and T contain 3). → P, Q, S.
B: With $t=x-3$ the integrand $t(t^2-1)(t^2-4)$ is odd on $[-2,2]$ , so the integral is 0. Intervals containing 0: P and T (Q and S are open at 0). → P, T.
C: $f=\cos^2x+\sin x$ , $f'=\cos x(1-2\sin x)$ . Local maxima at $\sin x=\frac12$ : $x=\frac\pi6,\frac{5\pi}6$ (plus $2k\pi$ ). $\frac\pi6\approx0.52$ lies in P, Q, R and T but not in S ( $\frac\pi8\approx0.39$ ). → P, Q, R, T.
D: $\dfrac d{dx}\tan^{-1}(\sin x+\cos x)=\dfrac{\cos x-\sin x}{1+(\sin x+\cos x)^2}>0$ for $-\frac{3\pi}4<x<\frac\pi4$ . Only S lies inside this. → S.

Answer: A → P, Q, S; B → P, T; C → P, Q, R, T; D → S

IIT-JEE 2009 · Paper 1, Mathematics question 39