Let the function f(x) =x2 +x + sin x- cos x + log(1 + | x |). be defined over the interval [0, 1]. The odd extension of f(x) to the interval [- 1, 1] is

Question

Let the function f(x) =x2 +x + sin x- cos x + log(1 + | x |). be defined over the interval [0, 1]. The odd extension of f(x) to the interval [- 1, 1] is (A) x2 + x + sin x + cos x - log (1 + | x |) (B) -x2 + x + sin x + cos x + log (1 + | x |) (C) -x2 + x + sin x - cos x + log (1 + | x |) (D) none of these

Tardigrade Admin · Accepted Answer

Let the function in (a), (b), (c) be denoted by fa,fb,fc Then fa(-x) ≠ - f(x) ;fb (-x)=-f(x); fc(-x) ≠ -f(x) ∴ fb is odd in (-1,0) Hence fb is odd extension of f(x) in [-1, 1].

Q. Let the function $f (x) = x^{2} + x + sin x - cos x + lo g (1 + ∣ x ∣) .$ be defined over the interval [0, 1]. The odd extension of $f (x)$ to the interval [- 1, 1] is

$x^{2} + x + sin x + cos x - l o g (1 + ∣ x ∣)$

$- x^{2} + x + sin x + cos x + l o g (1 + ∣ x ∣)$

$- x^{2} + x + sin x - cos x + l o g (1 + ∣ x ∣)$

none of these

Let the function in (a), (b), (c) be denoted by $f_{a}, f_{b}, f_{c}$
Then $f_{a} (- x) \neq = - f (x); f_{b} (- x) = - f (x);$
$f_{c} (- x) \neq = - f (x) ∴ f_{b}$ is odd in $(- 1, 0)$
Hence $f_{b}$ is odd extension of $f (x)$ in $[- 1, 1]$ .

Q. Let the function f(x)=x2+x+sinx−cosx+log(1+∣x∣). be defined over the interval [0, 1]. The odd extension of f(x) to the interval [- 1, 1] is

x2+x+sinx+cosx−log(1+∣x∣)

−x2+x+sinx+cosx+log(1+∣x∣)

−x2+x+sinx−cosx+log(1+∣x∣)