Let f be an odd function defined on the real numbers such that f(x) = 3 sin x + 4 cos x, for x ≥ 0, then f(x) for x

Question

Let f be an odd function defined on the real numbers such that f(x) = 3 sin x + 4 cos x, for x ≥ 0, then f(x) for x < 0 is : (A) 3 sin x - 4 cos x  (B) -3 sin x + 4 cos x  (C)  - 3 sin x - 4 cos x  (D) 3 sin x + 4 cos x

Tardigrade Admin · Accepted Answer

Given, f(x)=3 sin x+4 cos x When, X< 0 ∴ -X> 0 ∴ f(-x)=3 sin (-x)+4 cos (-x) ...(i) But, f(x) is odd function ∴ f(-X) =-f(x), when x< 0 ⇒ f(x) =-f(-X) =-[3 sin (-x)+4 cos (-x)] [From Eq. (i)] =3 sin x-4 cos x

Q. Let $f$ be an odd function defined on the real numbers such that $f (x) = 3 sin x + 4 cos x,$ for $x \geq 0,$ then $f (x)$ for $x < 0$ is :

$3 sin x - 4 cos x$

$- 3 sin x + 4 cos x$

$- 3 sin x - 4 cos x$

$3 sin x + 4 cos x$

Given, $f (x) = 3 sin x + 4 cos x$
When, $X < 0$
$∴ - X > 0$
$∴ f (- x) = 3 sin (- x) + 4 cos (- x) ... (i)$
But, $f (x)$ is odd function
$∴ f (- X) = - f (x)$ , when $x < 0$
$\Rightarrow f (x) = - f (- X)$
$= - [3 sin (- x) + 4 cos (- x)]$ [From Eq. (i)]
$= 3 sin x - 4 cos x$

Q. Let f be an odd function defined on the real numbers such that f(x)=3sinx+4cosx, for x≥0, then f(x) for x<0 is :

3sinx−4cosx

−3sinx+4cosx

−3sinx−4cosx

3sinx+4cosx