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Which of the following is strictly decreasing on $\mathbb R$ ?

Option 1)
$-2 + \sin x$

Option 2)
$-2x + \sin x$

Option 3)
$x^3 - x$

Option 4)
$x^2 + 1$

Answers (1)

best_answer

As we have learned

Strictly decreasing functions -

A function f is called strictly decreasing in an interval L

$if\:for\:\:x_{1}>x_{2}\Rightarrow f(x_{1})<f(x_{2})$

$or\:if\:for\:\:x_{1}<x_{2}\Rightarrow f(x_{1})>f(x_{2})$

$\therefore \:\:\;\frac{dy}{dx}=f'(x)<0\:\:\:x\epsilon (a,b)$

- wherein

$-2+\sin x$ has derivative $\cos x$

$x^{3}-x$ has derivative $3x^{2}-1$

$x^{2}+1$ has derivative $2x$

All above derivative changes sign on R, but $-2x+\sin x$ has derivative $-2+\cos x$ which will be always negative on R . so $-2x+\sin x$ is strictly decreasing

Option 1)

$-2 + \sin x$

Option 2)

$-2x + \sin x$

Option 3)

$x^3 - x$

Option 4)

$x^2 + 1$

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Himanshu

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