Let fx - 2 - cosx for all real x- Statement I- For each real t- there exists a point c in -t- t - - such that f-c - 0 because - Statement II-ft - f t - 2-for each realt

Explanation for the correct option:Given that: f(x)=2+cosx0ex0exf'(x)= sinx0ex0exf'(c)= sincNow we have, f^'(c)=00ex0ex⇒sinc=00ex0ex⇒ c=nπtherefore, there ...

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Standard VII

Mathematics

Equal Angles Subtend Equal Sides

Let fx = 2 + ...

Question

Let $f (x) = 2 + \cos x$ for all real $x$ .
Statement I: For each real $t$ , there exists a point $c$ in $[t, t + π]$ such that $f' (c) = 0$ because .
Statement II: $f (t) = f (t + 2 π)$ for each real $t$

Statement I is true, Statement II is true; statement II is a correct explanation for statement I.

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Statement I is true, statement II is true; statement II is not a correct explanation for statement I

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Statement I is true, statement II is false

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Statement I is false, statement II is true

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Solution

The correct option is B
Statement I is true, statement II is true; statement II is not a correct explanation for statement I

Explanation for the correct option:
Given that:
$f (x) = 2 + cosx f' (x) = - sinx f' (c) = - sinc$
Now we have,
$f^{'} (c) = 0 \Rightarrow \sin c = 0 \Rightarrow c = n π$
therefore, there exists a point $c$ in $[t, t + π]$ for $t \in R$
Hence, statement I is true.
Also $f (x)$ being periodic of period $2 π$ , hence statement II is true, but statement II is not a correct explanation of statement I
Hence, the correct answer is option (B).

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Let f(x)=2+cosx for all real x. Statement I: For each real t, there exists a point c in [t,t+π] such that f'(c)=0 because . Statement II:f(t)=f(t+2π)for each realt

Let $f (x) = 2 + \cos x$ for all real $x$ .
Statement I: For each real $t$ , there exists a point $c$ in $[t, t + π]$ such that $f' (c) = 0$ because .
Statement II: $f (t) = f (t + 2 π)$ for each real $t$