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Page 1
Edurev123
1. Function of a Real Variable
1.1 Suppose that ?? ''
is continuous on [?? ,?? ] and that ?? has three zeros in the
interval (?? ,?? ) . Shov that ?? ''
has at least one zero in the interval (?? ,?? ) .
(2009 : 12 . Marks)
Solution:
Insight: This question uses the fact that continuity of any derivative of a function ensures
continuity and differentiability of lower order derivatives and the Rolle's theorem.
?? ''
is continuous on [1,2]
??? is continuous and differentiable on [1,2]
??? is continuous and differentiable on [1,2]
?? has three zeros in (1,2) . Let them be ?? 1
,?? 2
,?? 3
with ?? 1
<?? 2
<?? 3
.
Inflection Point
In the interval [?? 1
,?? 2
], applying Rolle's theorem.
?? is continuous on [?? 1
,?? 2
].
?? is differentiable on (?? 1
,?? 2
) .
?? (?? 1
)=?? (?? 2
)=0
???? ?(?? 1
,?? 2
) such that ?? (?? )=0 by Rolle's theorem.
Page 2
Edurev123
1. Function of a Real Variable
1.1 Suppose that ?? ''
is continuous on [?? ,?? ] and that ?? has three zeros in the
interval (?? ,?? ) . Shov that ?? ''
has at least one zero in the interval (?? ,?? ) .
(2009 : 12 . Marks)
Solution:
Insight: This question uses the fact that continuity of any derivative of a function ensures
continuity and differentiability of lower order derivatives and the Rolle's theorem.
?? ''
is continuous on [1,2]
??? is continuous and differentiable on [1,2]
??? is continuous and differentiable on [1,2]
?? has three zeros in (1,2) . Let them be ?? 1
,?? 2
,?? 3
with ?? 1
<?? 2
<?? 3
.
Inflection Point
In the interval [?? 1
,?? 2
], applying Rolle's theorem.
?? is continuous on [?? 1
,?? 2
].
?? is differentiable on (?? 1
,?? 2
) .
?? (?? 1
)=?? (?? 2
)=0
???? ?(?? 1
,?? 2
) such that ?? (?? )=0 by Rolle's theorem.
Similarly, applying Rolle's theorem in interval [?? 2
,?? 3
],??? 2
?(?? 2
,?? 3
) such that ?? (?? 2
)=0.
As ?? 1
<?? 2
and ?? 2
>?? 2
??? 1
<?? 2
.
Applying Rolle's theorem on ?? in (?? 1
,?? 2
?
) .
?? is continuous on [?? 1
,?? 2
].
?? is differentiable on (?? 1
,?? 2
) as ?? '
is continuous on that interval.
?? (?? 1
)=?? (?? 2
)=0
???? ?(?? 1
,?? 2
) so that ?? '
(?? )=0 by Rolle's theorem.
Also, (?? 1
,?? 2
)?(1,2)??? ?(1,2)
1.2 If ?? is the derivative of some function defined on [?? ,?? ] prove that there exists a
number ?? ?[?? ,?? ] such that
? ?
?? ?? =?? ?? (?? )???? =?? (?? )(?? -?? )
(2009 : 12 Marks)
Solution:
Insight: This is the mean value theorem of integral calculus with the difference that ?? is
not given as continuous but as derivative of some function. We use 2nd Fundamental
Theorem of Calculus.
?? is derivative on some function
? ?? (?? )=?? (?? ) on [?? ,?? ]
i.e., ?? (?? ) has an anti-derivative ?? (?? ) defined on [?? ,?? ].
By 2nd Fundamental Theorem of Algebra, for any ?? 1
,?? 2
?[?? ,?? ]
? ?
?? 2
?? 1
?? (?? )???? =?? (?? 2
)-?? (?? 1
)
Proof of 2nd Fundamental Theorem: ?? (?? ) is an anti-derivative of ?? (?? ) .
Let
?? (?? )=? ?
?? ?? 1
?? (?? )???? where ?? 1
?(?? ,?? )
Then ?? (?? ) is an anti-derivative.
As two anti-derivatives differ by a constant
Page 3
Edurev123
1. Function of a Real Variable
1.1 Suppose that ?? ''
is continuous on [?? ,?? ] and that ?? has three zeros in the
interval (?? ,?? ) . Shov that ?? ''
has at least one zero in the interval (?? ,?? ) .
(2009 : 12 . Marks)
Solution:
Insight: This question uses the fact that continuity of any derivative of a function ensures
continuity and differentiability of lower order derivatives and the Rolle's theorem.
?? ''
is continuous on [1,2]
??? is continuous and differentiable on [1,2]
??? is continuous and differentiable on [1,2]
?? has three zeros in (1,2) . Let them be ?? 1
,?? 2
,?? 3
with ?? 1
<?? 2
<?? 3
.
Inflection Point
In the interval [?? 1
,?? 2
], applying Rolle's theorem.
?? is continuous on [?? 1
,?? 2
].
?? is differentiable on (?? 1
,?? 2
) .
?? (?? 1
)=?? (?? 2
)=0
???? ?(?? 1
,?? 2
) such that ?? (?? )=0 by Rolle's theorem.
Similarly, applying Rolle's theorem in interval [?? 2
,?? 3
],??? 2
?(?? 2
,?? 3
) such that ?? (?? 2
)=0.
As ?? 1
<?? 2
and ?? 2
>?? 2
??? 1
<?? 2
.
Applying Rolle's theorem on ?? in (?? 1
,?? 2
?
) .
?? is continuous on [?? 1
,?? 2
].
?? is differentiable on (?? 1
,?? 2
) as ?? '
is continuous on that interval.
?? (?? 1
)=?? (?? 2
)=0
???? ?(?? 1
,?? 2
) so that ?? '
(?? )=0 by Rolle's theorem.
Also, (?? 1
,?? 2
)?(1,2)??? ?(1,2)
1.2 If ?? is the derivative of some function defined on [?? ,?? ] prove that there exists a
number ?? ?[?? ,?? ] such that
? ?
?? ?? =?? ?? (?? )???? =?? (?? )(?? -?? )
(2009 : 12 Marks)
Solution:
Insight: This is the mean value theorem of integral calculus with the difference that ?? is
not given as continuous but as derivative of some function. We use 2nd Fundamental
Theorem of Calculus.
?? is derivative on some function
? ?? (?? )=?? (?? ) on [?? ,?? ]
i.e., ?? (?? ) has an anti-derivative ?? (?? ) defined on [?? ,?? ].
By 2nd Fundamental Theorem of Algebra, for any ?? 1
,?? 2
?[?? ,?? ]
? ?
?? 2
?? 1
?? (?? )???? =?? (?? 2
)-?? (?? 1
)
Proof of 2nd Fundamental Theorem: ?? (?? ) is an anti-derivative of ?? (?? ) .
Let
?? (?? )=? ?
?? ?? 1
?? (?? )???? where ?? 1
?(?? ,?? )
Then ?? (?? ) is an anti-derivative.
As two anti-derivatives differ by a constant
?? (?? )=?? (?? )+??
? ?? (?? 1
) =?? (?? 1
)+??
and ? ?
?? 1
?? 1
??? (?? )=?? (?? 1
)=0
? ?? (?? 1
)+?? =0? ?? =-?? (?? 1
)
? ?? (?? )=?? (?? )-?? (?? 1
)
? ? ?
?? ?? 1
??? (?? )???? =?? (?? )-?? (?? 1
)
? ? ?
?? ?? 2
??? (?? )???? =?? (?? 2
)-?? (?? 1
)
Now, ?? '
(?? )=?? (?? ) on [?? ,?? ] so ?? (?? ) is continuous on [?? ,?? ] as every differentiable
function is continuous and ?? (?? ) is differentiable on (?? ,?? ) .
So by Mean Value Theorem, ??? ?(?? ,?? ) such that
?? (?? )=
?? (?? )-?? (?? )
?? -?? ? ?? (?? )-?? (?? )=(?? -?? )?? (?? )
? ? ?
?? ?? ??? (?? )???? =(?? -?? )?? (?? )
1.2 A twice differentiable function ?? (?? ) is such that ?? (?? )=?? =?? (?? ) and ?? (?? )>?? for
?? <?? <?? . Prove that there is at least one point ?? ,?? <?? <?? , for which ?? ''
(?? )<?? .
(2010 : 12 Marks)
Solution:
Given ?? (?? )=?? (?? )=0 and ?? ?(?? ,?? ) such that ?? (?? )>0.
By Lagrange's Mean Value Theorem (LMVT), ??? ?(?? ,?? ) and ?? ?(?? ,?? ) :
?? (?? )=
?? (?? )-?? (?? )
?? -?? and ?? (?? )=
?? (?? )-?? (?? )
?? -?? (1)
Now, let ?? ?(?? ,?? )
? ?? ?(?? ,?? ) or ?? <?? <??
By Lagrange's Mean Value Theorem (LMVT)
?? '
(?? )=
?? '
(?? )-?? '
(?? )
?? -??
Page 4
Edurev123
1. Function of a Real Variable
1.1 Suppose that ?? ''
is continuous on [?? ,?? ] and that ?? has three zeros in the
interval (?? ,?? ) . Shov that ?? ''
has at least one zero in the interval (?? ,?? ) .
(2009 : 12 . Marks)
Solution:
Insight: This question uses the fact that continuity of any derivative of a function ensures
continuity and differentiability of lower order derivatives and the Rolle's theorem.
?? ''
is continuous on [1,2]
??? is continuous and differentiable on [1,2]
??? is continuous and differentiable on [1,2]
?? has three zeros in (1,2) . Let them be ?? 1
,?? 2
,?? 3
with ?? 1
<?? 2
<?? 3
.
Inflection Point
In the interval [?? 1
,?? 2
], applying Rolle's theorem.
?? is continuous on [?? 1
,?? 2
].
?? is differentiable on (?? 1
,?? 2
) .
?? (?? 1
)=?? (?? 2
)=0
???? ?(?? 1
,?? 2
) such that ?? (?? )=0 by Rolle's theorem.
Similarly, applying Rolle's theorem in interval [?? 2
,?? 3
],??? 2
?(?? 2
,?? 3
) such that ?? (?? 2
)=0.
As ?? 1
<?? 2
and ?? 2
>?? 2
??? 1
<?? 2
.
Applying Rolle's theorem on ?? in (?? 1
,?? 2
?
) .
?? is continuous on [?? 1
,?? 2
].
?? is differentiable on (?? 1
,?? 2
) as ?? '
is continuous on that interval.
?? (?? 1
)=?? (?? 2
)=0
???? ?(?? 1
,?? 2
) so that ?? '
(?? )=0 by Rolle's theorem.
Also, (?? 1
,?? 2
)?(1,2)??? ?(1,2)
1.2 If ?? is the derivative of some function defined on [?? ,?? ] prove that there exists a
number ?? ?[?? ,?? ] such that
? ?
?? ?? =?? ?? (?? )???? =?? (?? )(?? -?? )
(2009 : 12 Marks)
Solution:
Insight: This is the mean value theorem of integral calculus with the difference that ?? is
not given as continuous but as derivative of some function. We use 2nd Fundamental
Theorem of Calculus.
?? is derivative on some function
? ?? (?? )=?? (?? ) on [?? ,?? ]
i.e., ?? (?? ) has an anti-derivative ?? (?? ) defined on [?? ,?? ].
By 2nd Fundamental Theorem of Algebra, for any ?? 1
,?? 2
?[?? ,?? ]
? ?
?? 2
?? 1
?? (?? )???? =?? (?? 2
)-?? (?? 1
)
Proof of 2nd Fundamental Theorem: ?? (?? ) is an anti-derivative of ?? (?? ) .
Let
?? (?? )=? ?
?? ?? 1
?? (?? )???? where ?? 1
?(?? ,?? )
Then ?? (?? ) is an anti-derivative.
As two anti-derivatives differ by a constant
?? (?? )=?? (?? )+??
? ?? (?? 1
) =?? (?? 1
)+??
and ? ?
?? 1
?? 1
??? (?? )=?? (?? 1
)=0
? ?? (?? 1
)+?? =0? ?? =-?? (?? 1
)
? ?? (?? )=?? (?? )-?? (?? 1
)
? ? ?
?? ?? 1
??? (?? )???? =?? (?? )-?? (?? 1
)
? ? ?
?? ?? 2
??? (?? )???? =?? (?? 2
)-?? (?? 1
)
Now, ?? '
(?? )=?? (?? ) on [?? ,?? ] so ?? (?? ) is continuous on [?? ,?? ] as every differentiable
function is continuous and ?? (?? ) is differentiable on (?? ,?? ) .
So by Mean Value Theorem, ??? ?(?? ,?? ) such that
?? (?? )=
?? (?? )-?? (?? )
?? -?? ? ?? (?? )-?? (?? )=(?? -?? )?? (?? )
? ? ?
?? ?? ??? (?? )???? =(?? -?? )?? (?? )
1.2 A twice differentiable function ?? (?? ) is such that ?? (?? )=?? =?? (?? ) and ?? (?? )>?? for
?? <?? <?? . Prove that there is at least one point ?? ,?? <?? <?? , for which ?? ''
(?? )<?? .
(2010 : 12 Marks)
Solution:
Given ?? (?? )=?? (?? )=0 and ?? ?(?? ,?? ) such that ?? (?? )>0.
By Lagrange's Mean Value Theorem (LMVT), ??? ?(?? ,?? ) and ?? ?(?? ,?? ) :
?? (?? )=
?? (?? )-?? (?? )
?? -?? and ?? (?? )=
?? (?? )-?? (?? )
?? -?? (1)
Now, let ?? ?(?? ,?? )
? ?? ?(?? ,?? ) or ?? <?? <??
By Lagrange's Mean Value Theorem (LMVT)
?? '
(?? )=
?? '
(?? )-?? '
(?? )
?? -??
Using values in eqn. (1), we get
?? '
(?? )=
?? (?? )-?? (?? )
?? -?? -
?? (?? )-?? (?? )
?? -?? ?? -?? ? ?? ''
(?? )=
-
?? (?? )
?? -?? -
?? (?? )
?? -?? ?? -?? [?? (?? )=?? (?? )=0]
?? '
(?? )=-[
?? (?? )
?? -?? +
?? '
(?? )
?? -?? ?? -?? ]<0 as ?? >?? ,?? >?? ,?? >??
1.4 Show that a box (rectangular parallelopiped) of maximum volume ?? with
prescribed surface area is a cube.
(2010 : 20 Marks)
Solution:
Let ?? ,?? ,?? be length of edges of given rectangular parallelopiped. Its surface area be ?? ,
volume be ?? .(?? ,?? , ?? ?0) .
? ?? =2???? +2???? +2???? ,?? =??????
Let ?? be the Lagrange's multiplier (?? ?0)
So,
?? =?????? +?? (2???? +2???? +2???? -?? )
? At extremum,
?? ?? =0????? +?? (2?? +2?? )=0 (1)
?? ?? =0????? +?? (2?? +2?? )=0 (2)
?? ?? =0????? +?? (2?? +???? )=0 (3)
Multiplying (1) by ?? and (2) by ?? and subtracting then, we get
?????? +?? (2???? +2???? )-?????? -?? (2???? +2???? )=0
Page 5
Edurev123
1. Function of a Real Variable
1.1 Suppose that ?? ''
is continuous on [?? ,?? ] and that ?? has three zeros in the
interval (?? ,?? ) . Shov that ?? ''
has at least one zero in the interval (?? ,?? ) .
(2009 : 12 . Marks)
Solution:
Insight: This question uses the fact that continuity of any derivative of a function ensures
continuity and differentiability of lower order derivatives and the Rolle's theorem.
?? ''
is continuous on [1,2]
??? is continuous and differentiable on [1,2]
??? is continuous and differentiable on [1,2]
?? has three zeros in (1,2) . Let them be ?? 1
,?? 2
,?? 3
with ?? 1
<?? 2
<?? 3
.
Inflection Point
In the interval [?? 1
,?? 2
], applying Rolle's theorem.
?? is continuous on [?? 1
,?? 2
].
?? is differentiable on (?? 1
,?? 2
) .
?? (?? 1
)=?? (?? 2
)=0
???? ?(?? 1
,?? 2
) such that ?? (?? )=0 by Rolle's theorem.
Similarly, applying Rolle's theorem in interval [?? 2
,?? 3
],??? 2
?(?? 2
,?? 3
) such that ?? (?? 2
)=0.
As ?? 1
<?? 2
and ?? 2
>?? 2
??? 1
<?? 2
.
Applying Rolle's theorem on ?? in (?? 1
,?? 2
?
) .
?? is continuous on [?? 1
,?? 2
].
?? is differentiable on (?? 1
,?? 2
) as ?? '
is continuous on that interval.
?? (?? 1
)=?? (?? 2
)=0
???? ?(?? 1
,?? 2
) so that ?? '
(?? )=0 by Rolle's theorem.
Also, (?? 1
,?? 2
)?(1,2)??? ?(1,2)
1.2 If ?? is the derivative of some function defined on [?? ,?? ] prove that there exists a
number ?? ?[?? ,?? ] such that
? ?
?? ?? =?? ?? (?? )???? =?? (?? )(?? -?? )
(2009 : 12 Marks)
Solution:
Insight: This is the mean value theorem of integral calculus with the difference that ?? is
not given as continuous but as derivative of some function. We use 2nd Fundamental
Theorem of Calculus.
?? is derivative on some function
? ?? (?? )=?? (?? ) on [?? ,?? ]
i.e., ?? (?? ) has an anti-derivative ?? (?? ) defined on [?? ,?? ].
By 2nd Fundamental Theorem of Algebra, for any ?? 1
,?? 2
?[?? ,?? ]
? ?
?? 2
?? 1
?? (?? )???? =?? (?? 2
)-?? (?? 1
)
Proof of 2nd Fundamental Theorem: ?? (?? ) is an anti-derivative of ?? (?? ) .
Let
?? (?? )=? ?
?? ?? 1
?? (?? )???? where ?? 1
?(?? ,?? )
Then ?? (?? ) is an anti-derivative.
As two anti-derivatives differ by a constant
?? (?? )=?? (?? )+??
? ?? (?? 1
) =?? (?? 1
)+??
and ? ?
?? 1
?? 1
??? (?? )=?? (?? 1
)=0
? ?? (?? 1
)+?? =0? ?? =-?? (?? 1
)
? ?? (?? )=?? (?? )-?? (?? 1
)
? ? ?
?? ?? 1
??? (?? )???? =?? (?? )-?? (?? 1
)
? ? ?
?? ?? 2
??? (?? )???? =?? (?? 2
)-?? (?? 1
)
Now, ?? '
(?? )=?? (?? ) on [?? ,?? ] so ?? (?? ) is continuous on [?? ,?? ] as every differentiable
function is continuous and ?? (?? ) is differentiable on (?? ,?? ) .
So by Mean Value Theorem, ??? ?(?? ,?? ) such that
?? (?? )=
?? (?? )-?? (?? )
?? -?? ? ?? (?? )-?? (?? )=(?? -?? )?? (?? )
? ? ?
?? ?? ??? (?? )???? =(?? -?? )?? (?? )
1.2 A twice differentiable function ?? (?? ) is such that ?? (?? )=?? =?? (?? ) and ?? (?? )>?? for
?? <?? <?? . Prove that there is at least one point ?? ,?? <?? <?? , for which ?? ''
(?? )<?? .
(2010 : 12 Marks)
Solution:
Given ?? (?? )=?? (?? )=0 and ?? ?(?? ,?? ) such that ?? (?? )>0.
By Lagrange's Mean Value Theorem (LMVT), ??? ?(?? ,?? ) and ?? ?(?? ,?? ) :
?? (?? )=
?? (?? )-?? (?? )
?? -?? and ?? (?? )=
?? (?? )-?? (?? )
?? -?? (1)
Now, let ?? ?(?? ,?? )
? ?? ?(?? ,?? ) or ?? <?? <??
By Lagrange's Mean Value Theorem (LMVT)
?? '
(?? )=
?? '
(?? )-?? '
(?? )
?? -??
Using values in eqn. (1), we get
?? '
(?? )=
?? (?? )-?? (?? )
?? -?? -
?? (?? )-?? (?? )
?? -?? ?? -?? ? ?? ''
(?? )=
-
?? (?? )
?? -?? -
?? (?? )
?? -?? ?? -?? [?? (?? )=?? (?? )=0]
?? '
(?? )=-[
?? (?? )
?? -?? +
?? '
(?? )
?? -?? ?? -?? ]<0 as ?? >?? ,?? >?? ,?? >??
1.4 Show that a box (rectangular parallelopiped) of maximum volume ?? with
prescribed surface area is a cube.
(2010 : 20 Marks)
Solution:
Let ?? ,?? ,?? be length of edges of given rectangular parallelopiped. Its surface area be ?? ,
volume be ?? .(?? ,?? , ?? ?0) .
? ?? =2???? +2???? +2???? ,?? =??????
Let ?? be the Lagrange's multiplier (?? ?0)
So,
?? =?????? +?? (2???? +2???? +2???? -?? )
? At extremum,
?? ?? =0????? +?? (2?? +2?? )=0 (1)
?? ?? =0????? +?? (2?? +2?? )=0 (2)
?? ?? =0????? +?? (2?? +???? )=0 (3)
Multiplying (1) by ?? and (2) by ?? and subtracting then, we get
?????? +?? (2???? +2???? )-?????? -?? (2???? +2???? )=0
? ?? (2???? -2???? )=0
? 2???? -2???? =0 (?? ?0)
? ?? =?? as ?? ?0 (4)
Similarly, multiplying ( 2 ) by ?? and (3) by ?? and subtracting, we get
?????? +?? (2???? +2?? ?? )-?????? -?? (2???? +2???? )=0
? ?? (2???? -2???? )=0
? ?? =?? (?? ?0) (5)
? from (4) and (5), it can be concluded ?? =?? =?? which is a cube.
1.5 Let ?? be a function defined on R such that ?? (?? )=-?? and ?? '
(?? )=?? for all
values of ?? in R. How long can ?? (?? ) possibly be?
(2011: 10 marks)
Solution :
?? (?? ) =?? ??? ???
Let ?? (?? )
=5+ where
?? (?? )=0
? ?? (?? )
=5?? +?? (?? )+?? 1
,h(?? )=??? (?? )???? is a
decreasing function
? ?? (0) =?? (0)+?? 1
=-3
? ?? 1
=-3-?? (0)
? ?? (?? ) =5?? +?? (?? )-3-?? (0)
? ?? (2) =10+?? (2)-3-?? (0)
=7+(?? (2)-?? (0))
=7 (??? is a decreasing function, ?? (2)-?? (0)<
0)
? (2) =7
?? '
(?? )=5??? ???
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