Directions: |
The following questions consist of two statements, one labelled as "Assertion [A] and the other labelled as Reason [R]". You are to examine these two statements carefully and decide if Assertion [A] and Reason [R] are individually true and if so, whether the Reason [R] is the correct explanation for the given Assertion [A]. Select your answer from following options. |
Let \[f\left( x \right)=2+cos\text{ }x\]for all real x |
Statement-1 : For each real 't', then exist a point C in \[\left[ t,t+\pi \right]\] such that\[f'\left( C \right)\text{ }=\text{ }0\]. |
Statement-2 : \[f\left( t \right)=f\left( t+2\pi \right)\]for each real t |
Assertion [A]: One root of \[{{x}^{3}}-2{{x}^{2}}-1=0\] and lies between 2 and 3. |
Reason [R]: If \[f\left( x \right)\] is continuous function and f [a], f[b] have opposite signs then at least one or odd number of roots of \[f\left( x \right)=0\] lies between a and b. |
[A] Statement-1 is true, Statement-2 is true; statement-I 2 is a correct explanation of statement-1. |
[B] Statement-1 is true, Statement-2 is true, statement- 2 is not correct explanation of Statement-1 |
[C] Statement-1 is true, Statement-2 is false |
[D] Statement-1 is false, Statement-2 is True. |
Assertion [A]: \[f\left( x \right)\text{ }=\text{ }\left[ x \right]\]is not differentiable |
Reason [R]: \[f\left( x \right)\text{ }=\text{ }\left[ x \right]\]is not continuous at\[x=0\]. |
Assertion [A]: \[f\left( x \right)=x\text{ }sin\left( \frac{1}{x} \right)\]is differentiable at x = 0. |
Reason [R]: f(x) is continuous at x = 0. |
Assertion [A]: If then \[f'\left( x \right)\]does not exist |
Reason [R]: \[f'\left( x \right)\]is not continuous at x =2. |
Assertion [A]: \[\frac{d}{dx}\left( {{x}^{{{x}^{x}}}} \right)={{x}^{{{x}^{x}}}}\,.\,x\left( 1+2\,\,\log \,x \right)\] |
Reason [R]: \[{{\left( {{x}^{x}} \right)}^{x}}={{x}^{{{x}^{2}}}}={{e}^{{{x}^{2}}}}={{e}^{{{x}^{2}}}}\log \,x\] |
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