Assertion (A) & Reason (B) MCQ

Question 11 Mark

Assertion (A) : $y=a \sin x+b \cos x$ is a general solution of $y^{\prime \prime}+y=0$.
Reason (R): $y=a \sin x+b \cos x$ is a trigonometric function.

Answer

(b) : $\because y=a \sin x+b \cos x$ ...(i)
$
\begin{array}{ll}
\therefore & y^{\prime}=a \cos x-b \sin x \\
\Rightarrow & y^{\prime \prime}=-a \sin x-b \cos x=-y [Using (i)] \\
\Rightarrow & y^{\prime \prime}+y=0
\end{array}
$

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Question 21 Mark

Assertion (A) : A differential equation of the form $y f(x y) d x+x g(x y) d y=0$ can be converted into homogeneous differential equation by substituting $x y=t$
Reason (R) : A differential equation is called homogeneous if $f(\lambda x, \lambda y)=\lambda^0 f(x, y)$.

Answer

(d) : Clearly, reason is true.
As, $x y=t$
$
\therefore \quad x d y / d x+y=\frac{d t}{d x} \Rightarrow \frac{x d y}{d x}=\frac{d t}{d x}-\frac{t}{x}
$
Now, given differential equation is
$
\begin{aligned}
& y f(x y) d x+x g(x y) d y=0 \\
\Rightarrow & \frac{t}{x} f(t)+g(t)\left(\frac{d t}{d x}-\frac{t}{x}\right)=0 \\
\Rightarrow & \frac{t}{x} f(t)+g(t) \frac{d t}{d x}-\frac{t}{x} g(t)=0 \\
\Rightarrow & t\left[\frac{f(t)-g(t)}{x}\right]+g(t) \frac{d t}{d x}=0 \\
\Rightarrow & \frac{d x}{x}+\frac{g(t)}{t[f(t)-g(t)]} d t=0,
\end{aligned}
$
which is variable separable form
$\therefore \quad$ It is not homogeneous differential equation.
$\therefore \quad$ Assertion is false; Reason is true.

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Question 31 Mark

Assertion (A) : ' $x$ ' is not an integrating factor for the differential equation $x \frac{d y}{d x}+2 y=e^x$.
Reason (R) : $x\left(x \frac{d y}{d x}+2 y\right)=\frac{d}{d x}\left(x^2 y\right)$.

Answer

(b) : $\frac{d y}{d x}+\frac{2}{x} y=\frac{e^x}{x}$
$
\text { I.F. }=e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log x^2}=x^2
$
$\Rightarrow$ Assertion is correct.
Now, $\frac{d}{d x}\left(x^2 y\right)=x^2 \frac{d y}{d x}+y \cdot 2 x=x\left(x \frac{d y}{d x}+2 y\right)$
$\Rightarrow$ Reason is correct.

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Question 41 Mark

Assertion (A) : Order of the differential equation whose solution is $y=c_1 e^{x+c_2}+c_3 e^{x+c_4}$ is 4.
Reason (R) : Order of the differential equation is equal to the number of independent arbitrary constants mentioned in the solution of the differential equation.

Answer

(d): $\because y=\left(c_1 e^{c_2}+c_3 e^{c_4}\right) e^x=c e^x$ ...(i)
Solution of differential equation containing the arbitrary constant.
$\therefore \quad$ Order is 1.

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Question 51 Mark

Assertion $(A) :$ If $\frac{d y}{d x}+x y=x^3 y^3, x>0, y \geq 0$ and $y(0)=1$, then $y(1)=\frac{1}{\sqrt{2}}$.
Reason $(R) :$ The differential equation is linear with integrating factor $e^x$.

Answer

$\text { (c) : } \frac{1}{y^3} \frac{d y}{d x}+\frac{x}{y^2}=x^3$
$\text { Put } \frac{1}{y^2}=z \Rightarrow-\frac{2}{y^3} d y=d z$
$\therefore \frac{d z}{d x}-2 x z=-2 x^3,$
which is a linear differential equation with I.F. $=e^{-x^2}$
$\therefore \text { Solution, } z e^{-x^2}=-\int e^{-x^2} 2 x^3 d x+C$
$\Rightarrow z e^{-x^2}=\left(x^2+1\right) e^{-x^2}+C$
$\Rightarrow z=x^2+1+C e^{x^2}$
$\therefore \frac{1}{y^2}=x^2+1+C e^{x^2}$
$\because y(0)=1 \Rightarrow C=0$
$\therefore y^2=\frac{1}{x^2+1} \Rightarrow y=\frac{1}{\sqrt{x^2+1}}$
$\Rightarrow y(1)=\frac{1}{\sqrt{2}}$

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Question 61 Mark

Assertion $(A) : x \sin x \frac{d y}{d x}+(x+x \cos x+\sin x)$ $y=\sin x, y\left(\frac{\pi}{2}\right)=1-\frac{2}{\pi} \Rightarrow y=\frac{x-\sin x}{x(1-\cos x)}$
Reason $(R) :$ The differential equation is linear with integrating factor $x(1-\cos x)$.

Answer

$\text { (a) : } \frac{d y}{d x}+\left(\frac{1}{\sin x}+\cot x+\frac{1}{x}\right) y=\frac{1}{x z}$
$\text { I.F. }=\exp \int\left(\frac{1}{\sin x}+\cot x+\frac{1}{x}\right) d x=\exp \ln x(1-\cos x)$
$=x(1-\cos x)$
$ \therefore \text { Solution is, } y x(1-\cos x)=\int \frac{1}{x} \cdot x(1-\cos x) d x+C$
$y\left(\frac{\pi}{2}\right)=1-\frac{2}{\pi} \Rightarrow c=0 x-\sin x+c$
$\therefore y=\frac{x-\sin x}{x(1-\cos x)}$

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Question 71 Mark

Assertion (A) : Integrating factor of $\left(1+x^2\right) \frac{d y}{d x}+x y=\frac{1}{2 x}$ is given by $\sqrt{1+x^2}$.
Reason (R) : Integrating factor of $\frac{d y}{d x}+P(x) \cdot y=Q(x)$ is $e^{\int P(x) d x}$.

Answer

(a) : Clearly, reason is true.
Now, $\frac{d y}{d x}\left(1+x^2\right)+x y=\frac{1}{2 x}$
$\Rightarrow \quad \frac{d y}{d x}+\frac{x}{1+x^2} y=\frac{1}{2 x\left(1+x^2\right)}$ which is of the form
$\frac{d y}{d x}+P(x) y=Q(x)$, where $P(x)=\frac{x}{1+x^2}, Q(x)=\frac{1}{2 x\left(1+x^2\right)}$
I.F. $=e^{\int P(x) d x}=e^{\int \frac{x}{1+x^2} d x}=e^{\int \frac{1}{2}\left(\frac{2 x}{1+x^2}\right) d x}=e^{\frac{1}{2} \log \left(1+x^2\right)}$
$=e^{\log \left(1+x^2\right)^{1 / 2}}=\sqrt{1+x^2}$
$\therefore \quad$ Assertion is true; Reason is true.

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MATHS Assertion (A) & Reason (B) MCQ

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