For the function f ( x ) = ( cos x ) - x + 1 , ? x ? R , between the following two statements (S1) f ( x ) = 0 for only one value of x in. (S2) f ( x ) is decreasing in and increasing in [2024]

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Solution

Q.
For the function $f (x) = (\cos x) - x + 1, x \in R,$ between the following two statements [2024]

(S1) $f (x) = 0$ for only one value of $x$ in $[0, π] .$

(S2) $f (x)$ is decreasing in $[0, \frac{π}{2}]$ and increasing in $[\frac{π}{2}, π .]$

1 Only (S2) is correct.

2 Both (S1) and (S2) are incorrect.

3 Only (S1) is correct.

4 Both (S1) and (S2) are correct.

Ans.
(3)

We have, $f (x) = \cos x - x + 1, x \in R$

$\Rightarrow f' (x) = - \sin x - 1 < 0 \forall x \in R$

$\Rightarrow f (x) is decreasing \forall x \in R$

For $f (x) = 0$

Now, $f (0) = 2$

$f (π) = - π$

$f (0) f (π) < 0$

So, $f$ is strictly decreasing in $[0, π]$ .

So, $f (x) = 0$ has one solution in $[0, π]$

(S1) is correct but (S2) is incorrect.

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