-
question_answer1)
For which interval, the function \[\frac{{{x}^{2}}-3x}{x-1}\] satisfies all the conditions of Rolle's theorem [MP PET 1993]
A)
[0, 3] done
clear
B)
[? 3, 0] done
clear
C)
[1.5, 3] done
clear
D)
For no interval done
clear
View Solution play_arrow
-
question_answer2)
For the function\[f(x)={{e}^{x}},a=0,b=1\], the value of c in mean value theorem will be [DCE 2002]
A)
log x done
clear
B)
\[\log (e-1)\] done
clear
C)
0 done
clear
D)
1 done
clear
View Solution play_arrow
-
question_answer3)
Rolle's theorem is not applicable to the function \[f(x)=|x|\] defined on [?1, 1] because [AISSE 1986; MP PET 1994, 95]
A)
f is not continuous on [ ?1, 1] done
clear
B)
f is not differentiable on (?1,1) done
clear
C)
\[f(-1)\ne f(1)\] done
clear
D)
\[f(-1)=f(1)\ne 0\] done
clear
View Solution play_arrow
-
question_answer4)
If \[f(x)=\cos x,0\le x\le \frac{\pi }{2}\], then the real number ?c? of the mean value theorem is [MP PET 1994]
A)
\[\frac{\pi }{6}\] done
clear
B)
\[\frac{\pi }{4}\] done
clear
C)
\[{{\sin }^{-1}}\left( \frac{2}{\pi } \right)\] done
clear
D)
\[{{\cos }^{-1}}\left( \frac{2}{\pi } \right)\] done
clear
View Solution play_arrow
-
question_answer5)
From mean value theorem \[f(b)-f(a)=\] \[(b-a)f'({{x}_{1}});\] \[a<{{x}_{1}}<b\]if \[f(x)=\frac{1}{x}\], then \[{{x}_{1}}=\] [MP PET 1996]
A)
\[\sqrt{ab}\] done
clear
B)
\[\frac{a+b}{2}\] done
clear
C)
\[\frac{2ab}{a+b}\] done
clear
D)
\[\frac{b-a}{b+a}\] done
clear
View Solution play_arrow
-
question_answer6)
The function \[f(x)=x(x+3){{e}^{-(1/2)x}}\] satisfies all the conditions of Rolle's theorem in [?3, 0]. The value of c is
A)
0 done
clear
B)
?1 done
clear
C)
? 2 done
clear
D)
? 3 done
clear
View Solution play_arrow
-
question_answer7)
Rolle's theorem is true for the function \[f(x)={{x}^{2}}-4\]in the interval
A)
[?2, 0] done
clear
B)
[?2, 2] done
clear
C)
\[\left[ 0,\,\frac{1}{2} \right]\] done
clear
D)
\[[0,\,\,2]\] done
clear
View Solution play_arrow
-
question_answer8)
For the function\[x+\frac{1}{x},x\in [1,\,3]\], the value of c for the mean value theorem is [MP PET 1997]
A)
1 done
clear
B)
\[\sqrt{3}\] done
clear
C)
2 done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer9)
If from mean value theorem, \[f'({{x}_{1}})=\frac{f(b)-f(a)}{b-a}\], then [MP PET 1999]
A)
\[a<{{x}_{1}}\le b\] done
clear
B)
\[a\le {{x}_{1}}<b\] done
clear
C)
\[a<{{x}_{1}}<b\] done
clear
D)
\[a\le {{x}_{1}}\le b\] done
clear
View Solution play_arrow
-
question_answer10)
Let \[f(x)\] satisfy all the conditions of mean value theorem in [0, 2]. If f (0) = 0 and \[|f'(x)|\,\le \frac{1}{2}\] for all x, in [0, 2] then
A)
\[f(x)\le 2\] done
clear
B)
\[|f(x)|\le 1\] done
clear
C)
\[f(x)=2x\] done
clear
D)
\[f(x)=3\]for at least one x in [0, 2] done
clear
View Solution play_arrow
-
question_answer11)
The function \[f(x)={{x}^{3}}-6{{x}^{2}}+ax+b\] satisfy the conditions of Rolle's theorem in [1, 3]. The values of a and b are
A)
11, ? 6 done
clear
B)
? 6, 11 done
clear
C)
?11, 6 done
clear
D)
6, ?11 done
clear
View Solution play_arrow
-
question_answer12)
Consider the function \[f(x)={{e}^{-2x}}\]sin 2x over the interval \[\left( 0,\frac{\pi }{2} \right)\]. A real number \[c\in \left( 0,\frac{\pi }{2} \right)\,,\] as guaranteed by Rolle?s theorem, such that \[{f}'\,(c)=0\] is [AMU 1999]
A)
\[\pi /8\] done
clear
B)
\[\pi /6\] done
clear
C)
\[\pi /4\] done
clear
D)
\[\pi /3\] done
clear
View Solution play_arrow
-
question_answer13)
Let \[f(x)=\sqrt{x-1}+\sqrt{x+24-10\sqrt{x-1};}\] \[1<x<26\] be real valued function. Then \[f\,'(x)\] for \[1<x<26\] is [MP PET 2000]
A)
0 done
clear
B)
\[\frac{1}{\sqrt{x-1}}\] done
clear
C)
\[2\sqrt{x-1}-5\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer14)
If \[f(x)\] satisfies the conditions of Rolle?s theorem in \[[1,\,2]\] and \[f(x)\] is continuous in \[[1,\,2]\] then \[\int_{1}^{2}{f'(x)dx}\] is equal to [DCE 2002]
A)
3 done
clear
B)
0 done
clear
C)
1 done
clear
D)
2 done
clear
View Solution play_arrow
-
question_answer15)
If the function \[f(x)={{x}^{3}}-6{{x}^{2}}+ax+b\] satisfies Rolle?s theorem in the interval \[[1,\,3]\] and \[f'\left( \frac{2\sqrt{3}+1}{\sqrt{3}} \right)=0\], then [MP PET 2002]
A)
\[a=-11\] done
clear
B)
\[a=-6\] done
clear
C)
\[a=6\] done
clear
D)
\[a=11\] done
clear
View Solution play_arrow
-
question_answer16)
In the Mean-Value theorem \[\frac{f(b)-f(a)}{b-a}=f'(c),\] if \[a=0,b=\frac{1}{2}\] and \[f(x)=x(x-1)(x-2),\]the value of c is [MP PET 2003]
A)
\[1-\frac{\sqrt{15}}{6}\] done
clear
B)
\[1+\sqrt{15}\] done
clear
C)
\[1-\frac{\sqrt{21}}{6}\] done
clear
D)
\[1+\sqrt{21}\] done
clear
View Solution play_arrow
-
question_answer17)
The abscissa of the points of the curve \[y={{x}^{3}}\]in the interval [?2, 2], where the slope of the tangents can be obtained by mean value theorem for the interval [?2, 2], are [MP PET 1993]
A)
\[\pm \frac{2}{\sqrt{3}}\] done
clear
B)
\[\pm \sqrt{3}\] done
clear
C)
\[\pm \frac{\sqrt{3}}{2}\] done
clear
D)
0 done
clear
View Solution play_arrow
-
question_answer18)
The function \[f(x)={{(x-3)}^{2}}\] satisfies all the conditions of mean value theorem in [3, 4]. A point on\[y={{(x-3)}^{2}}\], where the tangent is parallel to the chord joining (3, 0) and (4, 1) is
A)
\[\left( \frac{7}{2},\frac{1}{2} \right)\] done
clear
B)
\[\left( \frac{7}{2},\frac{1}{4} \right)\] done
clear
C)
(1, 4) done
clear
D)
(4, 1) done
clear
View Solution play_arrow
-
question_answer19)
In the mean value theorem, \[f(b)-f(a)=(b-a)f'(c)\]if \[a=4\], \[b=9\] and \[f(x)=\sqrt{x}\] then the value of c is [J & K 2005]
A)
8.00 done
clear
B)
5.25 done
clear
C)
4.00 done
clear
D)
6.25 done
clear
View Solution play_arrow