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(A) Let $f(x) = x e^{\sin x} - \cos x$. Then $f'(x) = e^{\sin x} + x e^{\sin x} \cos x + \sin x > 0$ for all $x \in (0, \frac{\pi}{2})$.Since $f(0) =

Vedclass · Accepted Answer

$A-p, B-q, s, C-q, r, s, t, D-r$

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(A) Let $f(x) = x e^{\sin x} - \cos x$. Then $f'(x) = e^{\sin x} + x e^{\sin x} \cos x + \sin x > 0$ for all $x \in (0, \frac{\pi}{2})$.Since $f(0) = -1  0$,there is exactly $1$ solution.$(B)$ The planes intersect in a line if the determinant of the coefficients is $0$ and they are not parallel. The determinant is $k(k-4) - 4(4-4) + 1(8-2k) = k^2 - 4k + 8 - 2k = k^2 - 6k + 8 = 0$,giving $k=2, 4$. For $k=2$,the planes are $2x+4y+z=0, 4x+2y+2z=0, 2x+2y+z=0$. The first and third are not parallel,so they intersect in a line.$(C)$ Let $f(x) = |x+2|+|x+1|+|x-1|+|x-2|$. The minimum value of $f(x)$ is $6$ (for $x \in [-1, 1]$). For $f(x) = 4k$ to have integer solutions,$4k \ge 6$,so $k \ge 1.5$. For $k=2, 3, 4, 5$,$4k$ is $8, 12, 16, 20$,all of which allow integer solutions for $x$.$(D)$ $\frac{dy}{y+1} = dx \implies \ln|y+1| = x + C$. Since $y(0)=1$,$\ln 2 = C$. Thus $y+1 = 2e^x$,so $y = 2e^x - 1$. Then $y(\ln 2) = 2(2) - 1 = 3$.

Column $I$	Column $II$
$(A)$ The number of solutions of the equation $x e^{\sin x}-\cos x=0$ in the interval $(0, \frac{\pi}{2})$	$(p)$ $1$
$(B)$ Value$(s)$ of $k$ for which the planes $k x+4 y+z=0, 4 x+k y+2 z=0$ and $2 x+2 y+z=0$ intersect in a straight line	$(q)$ $2$
$(C)$ Value$(s)$ of $k$ for which $\|x-1\|+\|x-2\|+\|x+1\|+\|x+2\|=4 k$ has integer solution$(s)$	$(r)$ $3$
$(D)$ If $y^{\prime}=y+1$ and $y(0)=1$,then value$(s)$ of $y(\ln 2)$	$(s)$ $4$
	$(t)$ $5$

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