Let y=f(x) be a function such that (d y/d x)=y-2 x+2, f(0)=1. The equation f(x)=k x(k ∈ R) has n distinct real roots then which of the following is/are correct?

Question

Let y=f(x) be a function such that (d y/d x)=y-2 x+2, f(0)=1. The equation f(x)=k x(k ∈ R) has n distinct real roots then which of the following is/are correct? (A) if k=e+2 then n=1 (B)  if k > e +2 then n =2 (C) if 2< k< e+2 then n=0 (D) if k< 2 then n=1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/6bf6841a6ada9b7340b2285b89a06c2c-.png" /> y 4+(-1) y =2-2 x ⇒ ye - x =∫(2-2 x ) e - x dx f ( x )=2 x + ce x text but f (0)=1 ⇒ f ( x )=2 x + e x ( e x / x )= k -2 Now g(x)=(ex/x) now interpret from the graph.

Q. Let $y = f (x)$ be a function such that $\frac{d y}{d x} = y - 2 x + 2, f (0) = 1$ . The equation $f (x) = k x (k \in$ R) has $n$ distinct real roots then which of the following is/are correct?

if $k = e + 2$ then $n = 1$

$i f$ k > e +2 $t h e n$ n =2$

if $2 < k < e + 2$ then $n = 0$

if $k < 2$ then $n = 1$

$y^{4} + (- 1) y = 2 - 2 x$
$\Rightarrow y e^{- x} = \int (2 - 2 x) e^{- x} d x$
$f (x) = 2 x + c e^{x} but f (0) = 1$
$\Rightarrow f (x) = 2 x + e^{x}$
$\frac{e ^{x}}{x} = k - 2$
Now $g (x) = \frac{e ^{x}}{x}$ now interpret from the graph.

Q. Let y=f(x) be a function such that dxdy​=y−2x+2,f(0)=1. The equation f(x)=kx(k∈ R) has n distinct real roots then which of the following is/are correct?

if k=e+2 then n=1

ifk > e +2thenn =2$

if 2<k<e+2 then n=0

if k<2 then n=1