Suppose y=y(x) be the solution curve to the differential equation… - Vidyadip

MCQ

Suppose $y=y(x)$ be the solution curve to the differential equation $\frac{d y}{d x}-y=2-e^{-x}$ such that $\lim _{x \rightarrow \infty} y(x)$ is finite. If $a$ and $b$ are respectively the $x-$ and $y$-intercepts of the tangent to the curve at $x=0$, then the value of $a-4 b$ is equal to$....$

A
$6$
B
$2$
✓
$3$
D
$0$

✓

Answer

Correct option: C.

$3$

c
$\frac{d y}{d x}-y=2-e^{-x}$

$\text { I.F. }=e^{-\int d x}=e^{-x}$

$\therefore \text { solution of D.E }$

$y \cdot e^{-x}=\int\left(2 e^{-x}-e^{-2 x}\right) d x$

$\Rightarrow y=-2+\frac{e^{-x}}{2}+C \cdot e^{x}$

$\because \lim _{x \rightarrow \infty} y$ is finite

$\therefore \lim _{x \rightarrow \infty}\left(-2+\frac{ e ^{- x }}{2}+ C ^{ x }\right) \rightarrow \text { finite }$

This is possible only when $C =0$

$\therefore y = y ( x )=-2+\frac{ e ^{- x }}{2}$

$\frac{ dy }{ dx }=-\frac{1}{2} e ^{- x }$

$\left.\frac{ dy }{ dx }\right|_{ x =0}=-\frac{1}{2}= m , y (0)=-2+\frac{1}{2}=\frac{-3}{2}$

$\therefore$ equation of tangent

$y +\frac{3}{2}=-\frac{1}{2}( x -0)$

$\Rightarrow x +2 y =-3$

$a =-3, b =\frac{-3}{2}$

$a -4 b =-3+6=3$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If the components of $\overrightarrow{ a }=\alpha \hat{ i }+\beta \hat{ j }+\gamma \hat{ k }$ along and perpendiculer to $\overrightarrow{ b }=3 \hat{ i }+\hat{ j }-\hat{ k }$ respectively, are $\frac{16}{11}(3 \hat{ i }+\hat{ j }-\hat{ k })$ and $\frac{1}{11}(-4 \hat{ i }-5 \hat{ j }-17 \hat{ k })$, then $\alpha^2+\beta^2+\gamma^2$ is equal to :

If $\frac{{\cos x}}{a} = \frac{{\cos (x + \theta )}}{b} = \frac{{\cos (x + 2\theta )}}{c} = \frac{{\cos (x + 3\theta )}}{d} \, ,$ then $\left( {\frac{{a + c}}{{b + d}}} \right)$ is equal to :-

Let $R$ be the set of all real numbers and $f(x)=\sin ^{10} x\left(\cos ^8 x+\cos ^4 x+\cos ^2 x+1\right)$ $x \in R$. Let $S=\{\lambda \in R$ there exists a point $c \in(0,2 \pi)$ with $\left.f^{\prime}(c)=\lambda f(c)\right\}$ Then,

The sum of the digits in the unit place of all numbers formed with the help of $3, 4, 5, 6$ taken all at a time is

If the length of the latus rectum of the ellipse $x^{2}+$ $4 y^{2}+2 x+8 y-\lambda=0$ is $4$ , and $l$ is the length of its major axis, then $\lambda+l$ is equal to$......$

A company has $10$ employes. The company has decided to form a team including atleast three employes and also excluding atleast three employes. Then the number of ways of forming the team is

If $n(U)$ = $600$ , $n(A)$ = $100$ , $n(B)$ = $200$ and $n(A \cap B )$ = $50$, then $n(\bar A \cap \bar B )$ is

($U$ is universal set and $A$ and $B$ are subsets of $U$)

If $f(x)$ = $x\sqrt {1 - {{\left[ x \right]}^2}} $ then (where $[.]$ denotes greatest integer function)

The distance of the point $Q(0,2,-2)$ form the line passing through the point $\mathrm{P}(5,-4,3)$ and perpendicular to the lines $\overrightarrow{\mathrm{r}}=(-3 \hat{\mathrm{i}}+2 \hat{\mathrm{k}})$ $\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}), \quad \lambda \in \mathbb{R} \quad$ and $\quad \overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+$ $\mu(-\hat{i}+3 \hat{J}+2 \hat{K}), \mu \in \mathbb{R}$ is

The function $f(x)=x^{3}-6 x^{2}+a x+b$ is such that $f(2)=f(4)=0$. Consider two statements.

$(S_1)$ there exists $\mathrm{x}_{1}, \mathrm{x}_{2} \in(2,4), \mathrm{x}_{1}<\mathrm{x}_{2}$, such that $f^{\prime}\left(x_{1}\right)=-1$ and $f^{\prime}\left(x_{2}\right)=0$

$(S_2)$ there exists $\mathrm{x}_{3}, \mathrm{x}_{4} \in(2,4), \mathrm{x}_{3}<\mathrm{x}_{4}$, such that $f$ is decreasing in $\left(2, x_{4}\right)$, increasing in $\left(x_{4}, 4\right)$ and $2 f^{\prime}\left(x_{3}\right)=\sqrt{3} f\left(x_{4}\right)$.

Then