Match the statements/expressions in Column I with the values given in Column II: .tg border-collapse:collapse;border-spacing:0; .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-c3owborder-color:inherit;text-align:center;vertical-align:top .tg .tg-7bttborder-color:inherit;font-weight:bold;text-align:center;vertical-align:top .tg .tg-0pkyborder-color:inherit;text-align:left;vertical-align:top .tg .tg-fymrborder-color:inherit;font-weight:bold;text-align:left;vertical-align:top Column I Column II (A) The number of solutions of the equation xe^\sin x - \cos x = 0 in the interval ( 0,\pi \over 2 ) (P) 1 (B) Value(s) of k for which the planes kx + 4y + z = 0,4x + ky + 2z = 0 and 2x + 2y + z = 0 intersect in a straight line (Q) 2 (C) Value(s) of k for which |x - 1| + |x - 2| + |x + 1| + |x + 2| = 4k has integer solution(s) (R) 3 (D) If y' = y + 1 and y(0) = 1 then value(s) of y(\ln 2) (S) 4 (T) 5

(A) \to (T); (B) \to (Q), (S); (C) \to (Q), (S), (T); (D) \to (Q)

(A) \to (S); (B) \to (Q), (S); (C) \to (P), (R), (S), (T); (D) \to (R)

(A) \to (P); (B) \to (Q), (S); (C) \to (Q), (R), (T); (D) \to (S)

Differential Equations PYQ — JEE Advanced 2009 · Shift 2 · Q22 with Solution

Match the statements/expressions in Column I with the values given in Column II:

	Column I		Column II
(A)	The number of solutions of the equation $x{e^{\sin x}} - \cos x = 0$ in the interval $\left( {0,{\pi \over 2}} \right)$	(P)	1
(B)	Value(s) of $k$ for which the planes $kx + 4y + z = 0,4x + ky + 2z = 0$ and $2x + 2y + z = 0$ intersect in a straight line	(Q)	2
(C)	Value(s) of $k$ for which $\|x - 1\| + \|x - 2\| + \|x + 1\| + \|x + 2\| = 4k$ has integer solution(s)	(R)	3
(D)	If $y' = y + 1$ and $y(0) = 1$ then value(s) of $y(\ln 2)$	(S)	4
		(T)	5

A
(A) $\to$ (P); (B) $\to$ (Q), (S); (C) $\to$ (Q), (R), (S), (T); (D) $\to$ (R)
B
(A) $\to$ (T); (B) $\to$ (Q), (S); (C) $\to$ (Q), (S), (T); (D) $\to$ (Q)
C
(A) $\to$ (S); (B) $\to$ (Q), (S); (C) $\to$ (P), (R), (S), (T); (D) $\to$ (R)
D
(A) $\to$ (P); (B) $\to$ (Q), (S); (C) $\to$ (Q), (R), (T); (D) $\to$ (S)

View written solutionFree

Correct answer: A

We solve each item in Column I and match with Column II.

1. Part (A)

We need the number of solutions of $x e^{\sin x}-\cos x=0, \qquad x\in \left(0,\frac{\pi}{2}\right).$

This is equivalent to $x e^{\sin x}=\cos x.$ Define $f(x)=x e^{\sin x}-\cos x.$

Now check monotonicity: $f'(x)=e^{\sin x}+x e^{\sin x}\cos x+\sin x.$ Since for $x\in\left(0,\frac{\pi}{2}\right)$ , $e^{\sin x}>0,\quad x e^{\sin x}\cos x>0,\quad \sin x>0,$ we get $f'(x)>0.$ So $f$ is strictly increasing.

Now evaluate at endpoints: $f(0)=0\cdot e^0-1=-1<0,$ and $f\left(\frac{\pi}{2}\right)=\frac{\pi}{2}e^{1}-0>0.$ By continuity and strict monotonicity, there is exactly one solution.

Thus, $(A)\to (P).$

2. Part (B)

We need value(s) of $k$ for which the planes $kx+4y+z=0,$ $4x+ky+2z=0,$ $2x+2y+z=0$ intersect in a straight line.

For three planes through the origin to intersect in a straight line, the system must have rank $2$ , i.e. determinant $=0$ but not all three dependent to rank $1$ .

Coefficient matrix:

k&4&1\\ 4&k&2\\ 2&2&1 \end{pmatrix}.$$ Compute determinant: $$\det A= \begin{vmatrix} k&4&1\\ 4&k&2\\ 2&2&1 \end{vmatrix}.$$ Expanding: $$\det A=k\begin{vmatrix}k&2\\2&1\end{vmatrix}-4\begin{vmatrix}4&2\\2&1\end{vmatrix}+1\begin{vmatrix}4&k\\2&2\end{vmatrix}.$$ So $$\det A=k(k-4)-4(4-4)+(8-2k)=k^2-4k+8-2k=k^2-6k+8.$$ Hence $$\det A=(k-2)(k-4).$$ Therefore determinant is zero for $$k=2,4.$$ Check rank is $2$ in each case. ### For $k=2$: Matrix becomes $$\begin{pmatrix} 2&4&1\\ 4&2&2\\ 2&2&1 \end{pmatrix}.$$ Rows are not all proportional, so rank is not $1$. Since determinant is $0$, rank $=2$. Thus planes intersect in a line. ### For $k=4$: Matrix becomes $$\begin{pmatrix} 4&4&1\\ 4&4&2\\ 2&2&1 \end{pmatrix}.$$ Again rows are not all proportional, so rank is not $1$. Hence rank $=2$. Thus planes intersect in a line. So there are two values: $2,4$. Thus, $$(B)\to (Q),(S).$$ --- ## 3. Part (C) We need value(s) of $k$ for which $$|x-1|+|x-2|+|x+1|+|x+2|=4k$$ has integer solution(s). Let $$g(x)=|x-1|+|x-2|+|x+1|+|x+2|.$$ We analyze piecewise. ### Case 1: $x\ge 2$ $$g(x)=(x-1)+(x-2)+(x+1)+(x+2)=4x,$$ so $$4k=4x\implies k=x.$$ For integer solution $x$, any integer $x\ge 2$ works, giving integer $k\ge 2$. Among Column II values, that gives $$(Q)=2,\ (R)=3,\ (S)=4,\ (T)=5.$$ ### Case 2: $1\le x\le 2$ $$g(x)=(x-1)+(2-x)+(x+1)+(x+2)=2x+4,$$ so $$4k=2x+4 \implies x=2k-2.$$ For integer $x\in[1,2]$, possible integer $x$ are $1,2$, giving - $x=1 \Rightarrow 4k=6 \Rightarrow k=\frac32$ (not in Column II), - $x=2 \Rightarrow 4k=8 \Rightarrow k=2$. ### Case 3: $-1\le x\le 1$ $$g(x)=(1-x)+(2-x)+(x+1)+(x+2)=6,$$ so $$4k=6\Rightarrow k=\frac32,$$ not in Column II. ### Case 4: $-2\le x\le -1$ $$g(x)=(1-x)+(2-x)+(-x-1)+(x+2)=4-2x.$$ For integer $x=-2,-1$, values are $8,6$, so $k=2,\frac32$. Thus from Column II only $k=2$. ### Case 5: $x\le -2$ $$g(x)=(1-x)+(2-x)+(-x-1)+(-x-2)=-4x,$$ so $$4k=-4x\implies k=-x.$$ For integer $x\le -2$, any integer $k\ge 2$ occurs. Hence the equation has integer solution(s) exactly for integer $k\ge 2$. From Column II, these are $$2,3,4,5,$$ i.e. $$(Q),(R),(S),(T).$$ Thus, $$(C)\to (Q),(R),(S),(T).$$ --- ## 4. Part (D) Given $$y'=y+1, \qquad y(0)=1.$$ We need $y(\ln 2)$. Solve the differential equation: $$\frac{dy}{dx}=y+1.$$ Rewrite: $$\frac{dy}{y+1}=dx.$$ Integrate: $$\ln|y+1|=x+C.$$ So $$y+1=Ce^x.$$ Using $y(0)=1$: $$1+1=C\cdot 1 \Rightarrow C=2.$$ Hence $$y=2e^x-1.$$ Now at $x=\ln 2$, $$y(\ln 2)=2e^{\ln 2}-1=2\cdot 2-1=3.$$ Thus, $$(D)\to (R).$$ --- ## 5. Final matching We get: - $(A)\to(P)$ - $(B)\to(Q),(S)$ - $(C)\to(Q),(R),(S),(T)$ - $(D)\to(R)$ This matches **Option A**. --- ## 6. Comparison with stored answer Stored correct answer: **A** Our derived answer: **A** So the stored answer is correct.

More from Differential Equations

1. Part (A)

We need the number of solutions of

x e^{\sin x}-\cos x=0, \qquad x\in \left(0,\frac{\pi}{2}\right).

This is equivalent to

x e^{\sin x}=\cos x.

Define

f(x)=x e^{\sin x}-\cos x.

Now check monotonicity:

f'(x)=e^{\sin x}+x e^{\sin x}\cos x+\sin x.

Since for

x\in\left(0,\frac{\pi}{2}\right)

e^{\sin x}>0,\quad x e^{\sin x}\cos x>0,\quad \sin x>0,

we get

f'(x)>0.

f

is strictly increasing.

Now evaluate at endpoints:

f(0)=0\cdot e^0-1=-1<0,

and

f\left(\frac{\pi}{2}\right)=\frac{\pi}{2}e^{1}-0>0.

By continuity and strict monotonicity, there is exactly one solution.

Thus,

(A)\to (P).

2. Part (B)

We need value(s) of

k

for which the planes

kx+4y+z=0,

4x+ky+2z=0,

2x+2y+z=0

intersect in a straight line.

For three planes through the origin to intersect in a straight line, the system must have rank

2

, i.e. determinant

=0

but not all three dependent to rank

1

Coefficient matrix:

k&4&1\\ 4&k&2\\ 2&2&1 \end{pmatrix}.$$ Compute determinant: $$\det A= \begin{vmatrix} k&4&1\\ 4&k&2\\ 2&2&1 \end{vmatrix}.$$ Expanding: $$\det A=k\begin{vmatrix}k&2\\2&1\end{vmatrix}-4\begin{vmatrix}4&2\\2&1\end{vmatrix}+1\begin{vmatrix}4&k\\2&2\end{vmatrix}.$$ So $$\det A=k(k-4)-4(4-4)+(8-2k)=k^2-4k+8-2k=k^2-6k+8.$$ Hence $$\det A=(k-2)(k-4).$$ Therefore determinant is zero for $$k=2,4.$$ Check rank is $2$ in each case. ### For $k=2$: Matrix becomes $$\begin{pmatrix} 2&4&1\\ 4&2&2\\ 2&2&1 \end{pmatrix}.$$ Rows are not all proportional, so rank is not $1$. Since determinant is $0$, rank $=2$. Thus planes intersect in a line. ### For $k=4$: Matrix becomes $$\begin{pmatrix} 4&4&1\\ 4&4&2\\ 2&2&1 \end{pmatrix}.$$ Again rows are not all proportional, so rank is not $1$. Hence rank $=2$. Thus planes intersect in a line. So there are two values: $2,4$. Thus, $$(B)\to (Q),(S).$$ --- ## 3. Part (C) We need value(s) of $k$ for which $$|x-1|+|x-2|+|x+1|+|x+2|=4k$$ has integer solution(s). Let $$g(x)=|x-1|+|x-2|+|x+1|+|x+2|.$$ We analyze piecewise. ### Case 1: $x\ge 2$ $$g(x)=(x-1)+(x-2)+(x+1)+(x+2)=4x,$$ so $$4k=4x\implies k=x.$$ For integer solution $x$, any integer $x\ge 2$ works, giving integer $k\ge 2$. Among Column II values, that gives $$(Q)=2,\ (R)=3,\ (S)=4,\ (T)=5.$$ ### Case 2: $1\le x\le 2$ $$g(x)=(x-1)+(2-x)+(x+1)+(x+2)=2x+4,$$ so $$4k=2x+4 \implies x=2k-2.$$ For integer $x\in[1,2]$, possible integer $x$ are $1,2$, giving - $x=1 \Rightarrow 4k=6 \Rightarrow k=\frac32$ (not in Column II), - $x=2 \Rightarrow 4k=8 \Rightarrow k=2$. ### Case 3: $-1\le x\le 1$ $$g(x)=(1-x)+(2-x)+(x+1)+(x+2)=6,$$ so $$4k=6\Rightarrow k=\frac32,$$ not in Column II. ### Case 4: $-2\le x\le -1$ $$g(x)=(1-x)+(2-x)+(-x-1)+(x+2)=4-2x.$$ For integer $x=-2,-1$, values are $8,6$, so $k=2,\frac32$. Thus from Column II only $k=2$. ### Case 5: $x\le -2$ $$g(x)=(1-x)+(2-x)+(-x-1)+(-x-2)=-4x,$$ so $$4k=-4x\implies k=-x.$$ For integer $x\le -2$, any integer $k\ge 2$ occurs. Hence the equation has integer solution(s) exactly for integer $k\ge 2$. From Column II, these are $$2,3,4,5,$$ i.e. $$(Q),(R),(S),(T).$$ Thus, $$(C)\to (Q),(R),(S),(T).$$ --- ## 4. Part (D) Given $$y'=y+1, \qquad y(0)=1.$$ We need $y(\ln 2)$. Solve the differential equation: $$\frac{dy}{dx}=y+1.$$ Rewrite: $$\frac{dy}{y+1}=dx.$$ Integrate: $$\ln|y+1|=x+C.$$ So $$y+1=Ce^x.$$ Using $y(0)=1$: $$1+1=C\cdot 1 \Rightarrow C=2.$$ Hence $$y=2e^x-1.$$ Now at $x=\ln 2$, $$y(\ln 2)=2e^{\ln 2}-1=2\cdot 2-1=3.$$ Thus, $$(D)\to (R).$$ --- ## 5. Final matching We get: - $(A)\to(P)$ - $(B)\to(Q),(S)$ - $(C)\to(Q),(R),(S),(T)$ - $(D)\to(R)$ This matches **Option A**. --- ## 6. Comparison with stored answer Stored correct answer: **A** Our derived answer: **A** So the stored answer is correct.

Differential Equations question

Differential Equations question

1. Part (A)

2. Part (B)

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2. Part (B)

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