Subject

Mathematics

Class

CBSE Class 12

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

1.

Show that:
2 sin to the power of negative 1 end exponent open parentheses 3 over 5 close parentheses minus tan to the power of negative 1 end exponent open parentheses 17 over 31 close parentheses equals space straight pi over 4


2 sin to the power of negative 1 end exponent open parentheses 3 over 5 close parentheses minus tan to the power of negative 1 end exponent open parentheses 17 over 31 close parentheses space equals straight pi over 4
straight L. straight H. straight S. comma
space space space equals cos to the power of negative 1 end exponent open parentheses 1 minus 2 cross times 9 over 25 close parentheses minus tan to the power of negative 1 end exponent open parentheses 17 over 31 close parentheses
space space space equals cos to the power of negative 1 end exponent open parentheses 7 over 25 close parentheses minus tan to the power of negative 1 end exponent open parentheses 17 over 31 close parentheses
space space space space equals tan to the power of negative 1 end exponent open parentheses 24 over 7 close parentheses minus tan to the power of negative 1 end exponent open parentheses 17 over 31 close parentheses
space space space space space equals tan to the power of negative 1 end exponent open parentheses 24 over 7 close parentheses minus tan to the power of negative 1 end exponent open parentheses 17 over 31 close parentheses
space space space space space equals tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 24 over 7 end style minus begin display style 17 over 31 end style over denominator 1 plus begin display style 24 over 7 end style cross times begin display style 17 over 31 end style end fraction close parentheses
space space space space space equals tan to the power of negative 1 end exponent open parentheses fraction numerator 24 cross times 31 minus 17 cross times 7 over denominator 31 cross times 7 plus 24 cross times 17 end fraction close parentheses
space space space space space space equals tan to the power of negative 1 end exponent open parentheses 625 over 625 close parentheses
space space space space space space equals tan to the power of negative 1 end exponent 1
space space space space space space equals straight pi over 4
space space space space space equals straight R. straight H. straight S space space
space space space Hence space Proved
638 Views

 Multiple Choice QuestionsLong Answer Type

2. Let space straight f colon straight W rightwards arrow straight W space be space defined space as
straight f left parenthesis straight n right parenthesis space equals space open curly brackets table row cell straight n minus 1 comma space space if space straight n space is space odd end cell row cell straight n plus 1 comma space if space straight n space is space even end cell end table close curly brackets
Show that f is invertible and find the inverse of f. Here, W is the set of all whole numbers. 

Let f: W→W be defined as
straight f left parenthesis straight n right parenthesis space equals space open curly brackets table row cell straight n minus 1 comma space if space straight n space is space odd end cell row cell straight n plus 1 comma space space if space straight n space is space even end cell end table close curly brackets
We need to prove that 'f' is invertible.
In order to prove that 'f' is invertible it is sufficient to prove that f is a bijection.
A function f: A→B is a one-one function or an injection, if
f(x) = f(y) ⇒ x = y for all x, y ∈ A.
Case i:
If x and y are odd.
Let f(x) = f(y)
⇒x − 1 = y − 1
⇒x = y
Case ii:
If x and y are even,
Let f(x) = f(y)
⇒x + 1 = y + 1
⇒x = y
Thus, in both the cases, we have,
f(x) = f(y) ⇒ x = y for all x, y ∈ W.
Hence f is an injection. 

Let n be an arbitrary element of W.
If n is an odd whole number, there exists an even whole number n − 1 ∈ W such that
f(n − 1) = n − 1 + 1 = n.
If n is an even whole number, then there exists an odd whole number n + 1 ∈ W such that f(n + 1) = n + 1 − 1 = n. Also, f(1) = 0 and f(0) = 1
Thus, every element of W (co-domain) has its pre-image in W (domain).
So f is an onto function.

Thus, it is proved that f is an invertible function.
Thus, a function g: B→A which associates each element y ∈ B to a unique element x ∈ A such that f(x) = y is called the inverse of f.
That is, f(x) = y ⇔ g(y) = x.
The inverse of f is generally denoted by f -1.

Now let us find the inverse of f.
Let x, y ∈ W such that f(x) = y
⇒x + 1 = y, if x is even
And
straight x minus 1 space equals space straight y comma space if space straight x space is space odd
rightwards double arrow space space straight x space equals space open curly brackets table row cell straight y minus 1 comma space if space straight y space is space odd end cell row cell straight y plus 1 comma if space straight y space is space even end cell end table close curly brackets
rightwards double arrow space straight f to the power of negative 1 end exponent left parenthesis straight y right parenthesis space equals open curly brackets table row cell straight y minus 1 comma space if space straight y space is space odd end cell row cell straight y plus 1 comma space if space straight y space straight i space even end cell end table close curly brackets
Interchange comma space straight x space and space straight y comma space we space have comma space
rightwards double arrow straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis space equals open curly brackets table row cell straight x minus 1 comma space if space straight x space is space odd end cell row cell straight x plus 1 comma space if space straight x space is space even end cell end table close curly brackets
Re space writing space the space above space we space have comma
space space rightwards double arrow straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis space equals space open curly brackets table row cell straight x plus 1 comma space if space straight x space is space even space end cell row cell straight x minus 1 comma space if space straight x space is space odd end cell end table close curly brackets
Thus comma space straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis space equals space straight f left parenthesis straight x right parenthesis

579 Views

 Multiple Choice QuestionsShort Answer Type

3. If space straight A space equals space open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets comma space then space show space that space straight A squared minus 4 straight A minus 5 straight I space equals 0 comma space and space hence space find space straight A to the power of negative 1 end exponent
space space space space space space space space space space space space space space space space space space space space space space space space

straight A space equals space open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets
straight A squared equals open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets space open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets
space space space equals space open square brackets table row cell 1 cross times 1 plus 2 cross times 2 plus 2 cross times 2 end cell cell space space 1 cross times 2 plus 2 cross times 1 plus 2 cross times 2 end cell cell space space space space space 1 cross times 2 plus 2 cross times 2 plus 2 cross times 1 end cell row cell 2 cross times 1 plus 1 cross times 2 plus 2 cross times 2 end cell cell space 2 cross times 2 plus 1 cross times 1 plus 2 cross times 2 end cell cell space space space space 2 cross times 2 plus 1 cross times 2 plus 2 cross times 1 end cell row cell 2 cross times 1 plus 2 cross times 2 plus 1 cross times 2 end cell cell space 2 cross times 2 plus 2 cross times 1 plus 1 cross times 2 end cell cell space space space space 2 cross times 2 plus 2 cross times 2 plus 1 cross times 1 end cell end table close square brackets
space equals space open square brackets table row cell 1 plus 4 plus 4 end cell cell space space space 2 plus 2 plus 4 end cell cell space space 2 plus 4 plus 2 end cell row cell 2 plus 2 plus 4 end cell cell space space 4 plus 1 plus 4 end cell cell space space 4 plus 2 plus 2 end cell row cell 2 plus 4 plus 2 end cell cell space 4 plus 2 plus 2 end cell cell space 4 plus 4 plus 1 end cell end table close square brackets
equals space open square brackets table row 9 8 8 row 8 9 8 row 8 8 9 end table close square brackets

Conisder space straight A squared minus 4 straight A minus 5 straight I

space equals space open square brackets table row 9 8 8 row 8 9 8 row 8 8 9 end table close square brackets space minus space 4 open square brackets table row 1 cell space 2 end cell cell space 2 end cell row 2 1 2 row 2 cell space 2 end cell 1 end table close square brackets space minus space 5 open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets
space equals open square brackets table row 9 cell space 8 end cell cell space 8 end cell row 8 cell space 9 end cell cell space 8 end cell row 8 cell space 8 end cell cell space 9 end cell end table close square brackets space minus space open square brackets table row 4 cell space 8 end cell cell space 8 end cell row 8 cell space 4 end cell cell space 8 end cell row 8 cell space 8 end cell cell space 4 end cell end table close square brackets space minus space open square brackets table row 5 cell space 0 end cell cell space 0 end cell row 0 cell space 5 end cell cell space 0 end cell row 0 0 cell space 5 end cell end table close square brackets
space equals space open square brackets table row cell 9 minus 9 end cell cell space 8 minus 8 end cell cell space space 8 minus 8 end cell row cell 8 minus 8 end cell cell space 9 minus 9 end cell cell space 8 minus 8 end cell row cell 8 minus 8 end cell cell space 8 minus 8 end cell cell space 9 minus 9 end cell end table close square brackets
equals space open square brackets table row 0 0 0 row 0 0 0 row 0 0 0 end table close square brackets

Now
straight A squared minus 4 straight A minus 5 straight I space equals space 0
straight A squared minus 4 straight A space equals space 5 straight I
straight A squared straight A to the power of negative 1 end exponent space minus 4 AA to the power of negative 1 end exponent space equals space 5 IA to the power of negative 1 end exponent left parenthesis Postmultiply space by space straight A to the power of negative 1 end exponent right parenthesis
straight A minus 4 straight I space equals space 5 straight A to the power of negative 1 end exponent
open square brackets table row 1 cell space 2 end cell cell space 2 end cell row 2 1 cell space 2 end cell row 2 cell space 2 end cell cell space 1 end cell end table close square brackets space minus space open square brackets table row 4 cell space 0 end cell cell space 0 end cell row 0 cell space 4 end cell cell space 0 end cell row 0 cell space 0 end cell cell space 4 end cell end table close square brackets space equals space 5 straight A to the power of negative 1 end exponent
open square brackets table row cell negative 3 end cell cell space space space 2 end cell cell space space space 2 end cell row 2 cell negative 3 end cell cell space space space 2 end cell row 2 cell space space 2 end cell cell negative 3 end cell end table close square brackets space equals space 5 straight A to the power of negative 1 end exponent
straight A to the power of negative 1 end exponent space equals space open square brackets table row cell fraction numerator negative 3 over denominator 5 end fraction end cell cell space 2 over 5 end cell cell space 2 over 5 end cell row cell 2 over 5 end cell cell fraction numerator negative 3 over denominator 5 end fraction end cell cell 2 over 5 end cell row cell 2 over 5 end cell cell 2 over 5 end cell cell fraction numerator negative 3 over denominator 5 end fraction end cell end table close square brackets



616 Views

4. If space straight A space equals space open vertical bar table row 2 cell space space 0 end cell cell negative 1 end cell row 5 cell space 1 end cell cell space 0 end cell row 0 cell space 1 end cell cell space 3 end cell end table close vertical bar comma space then space find thin space straight A to the power of negative 1 end exponent space using space elementary space row space operations.Applying space straight R subscript 3 space rightwards arrow space straight R subscript 3 plus left parenthesis negative 1 right parenthesis straight R subscript 2
straight A to the power of negative 1 end exponent open square brackets table row 1 cell space 0 end cell cell negative 1 half end cell row 0 cell space 1 end cell cell space 5 over 2 end cell row 0 cell space 1 end cell cell 1 half end cell end table close square brackets space equals space open square brackets table row cell 1 half end cell cell space space 0 end cell cell space 0 end cell row cell negative 5 over 2 end cell 1 cell space 0 end cell row cell 5 over 2 end cell cell negative 1 end cell cell space 1 end cell end table close square brackets
Applying space straight R subscript 3 space rightwards arrow space left parenthesis 2 right parenthesis straight R subscript 3
straight A to the power of negative 1 end exponent open square brackets table row 1 cell space 0 end cell cell negative 1 half end cell row 0 1 cell space 5 over 2 end cell row 0 cell space 0 end cell 1 end table close square brackets space equals space open square brackets table row cell 1 half end cell 0 cell space space space 0 end cell row cell negative 5 over 2 end cell 1 cell space space 0 end cell row 5 cell negative 2 end cell cell space space 2 end cell end table close square brackets

open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space 0 end cell cell space minus 1 end cell row 5 cell space space space 1 end cell cell space space space 0 end cell row 0 cell space space space 1 end cell cell space space space 3 end cell end table close vertical bar
space space equals space 2 left parenthesis 3 minus 0 right parenthesis minus 0 left parenthesis 15 minus 0 right parenthesis minus 1 left parenthesis 5 minus 0 right parenthesis
space space equals 6 minus 0 minus 5
space space equals space 1
space space not equal to 0
space
Hence space straight A to the power of negative 1 end exponent space exists.
straight A to the power of negative 1 end exponent straight A space equals space 1
straight A to the power of negative 1 end exponent open square brackets table row 2 cell space space 0 end cell cell space minus 1 end cell row cell space 5 space end cell cell space 1 end cell cell space space space 0 end cell row 0 cell space 1 end cell cell space space space 3 end cell end table close square brackets space equals space open square brackets table row 1 cell space 0 end cell cell space 0 end cell row 0 cell space 1 end cell cell space 0 end cell row 0 cell space 0 end cell cell space 1 end cell end table close square brackets
Applying space straight R subscript 1 space rightwards arrow space open parentheses 1 half close parentheses straight R subscript 1
straight A to the power of negative 1 end exponent open square brackets table row 1 cell space 0 end cell cell negative 1 half end cell row 5 1 cell space space 0 end cell row 0 1 cell space 3 end cell end table close square brackets space equals space open square brackets table row cell 1 half end cell cell space 0 end cell cell space space space 0 end cell row 0 cell space 1 end cell cell space space space 0 end cell row 0 cell space 0 end cell cell space space space 1 end cell end table close square brackets
Applying space straight R subscript 2 space rightwards arrow space straight R subscript 2 space plus left parenthesis negative 5 right parenthesis straight R subscript 1
straight A to the power of negative 1 end exponent open square brackets table row 1 cell space 0 end cell cell negative 1 half end cell row 0 1 cell space space 5 over 2 end cell row 0 1 cell space 3 end cell end table close square brackets space equals space open square brackets table row cell 1 half end cell cell space space 0 end cell cell space space space 0 end cell row cell negative 5 over 2 end cell cell space 1 end cell cell space space 0 end cell row 0 cell space 0 end cell cell space space space 1 end cell end table close square brackets
Applying space straight R subscript 3 space rightwards arrow straight R subscript 3 plus left parenthesis negative 1 right parenthesis straight R subscript 2
straight A to the power of negative 1 end exponent open square brackets table row 1 cell space space 0 end cell cell space space minus 1 half end cell row 0 cell space space 1 end cell cell space space space space 5 over 2 end cell row 0 cell space space 0 end cell cell space space space space 1 half end cell end table close square brackets space equals space open square brackets table row cell 1 half end cell cell space 0 end cell cell space 0 end cell row cell negative 5 over 2 end cell 1 cell space 0 end cell row cell 5 over 2 end cell cell negative 1 end cell 1 end table close square brackets
Applying space straight R subscript 3 space rightwards arrow left parenthesis 2 right parenthesis straight R subscript 3
straight A to the power of negative 1 end exponent open square brackets table row 1 cell space 0 end cell cell space minus 1 half end cell row 0 cell space 1 end cell cell space space 5 over 2 end cell row 0 cell space 0 end cell cell space 1 end cell end table close square brackets space equals open square brackets table row cell 1 half end cell cell space 0 end cell 0 row cell negative 5 over 2 end cell 1 0 row 5 cell negative 2 end cell cell space 2 end cell end table close square brackets
Applying space straight R subscript 1 space rightwards arrow space straight R subscript 1 space plus space open parentheses 1 half close parentheses straight R subscript 3
space space space space space space space space space space space space straight R subscript 2 space rightwards arrow space straight R subscript 2 space plus space open parentheses negative 5 over 2 close parentheses straight R subscript 3
space space straight A to the power of negative 1 end exponent open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets space equals space open square brackets table row 3 cell space minus 1 end cell cell space space space 1 end cell row cell negative 15 end cell cell space space space 6 end cell cell space minus 5 end cell row 5 cell space minus 2 end cell cell space space space 2 end cell end table close square brackets
straight A to the power of negative 1 end exponent space equals space open square brackets table row 3 cell space minus 1 end cell cell space space space space 1 end cell row cell negative 15 end cell cell space space space 6 end cell cell space minus 5 end cell row 5 cell space minus 2 end cell cell space space 2 end cell end table close square brackets




460 Views

5.

Solve the following for x:

sin to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis minus 2 sin to the power of negative 1 end exponent straight x equals straight pi over 2


sin to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis minus 2 sin to the power of negative 1 end exponent straight x space equals space straight pi over 2
rightwards double arrow space space sin to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis space equals space straight pi over 2 plus 2 sin to the power of negative 1 end exponent straight x
rightwards double arrow space space left parenthesis 1 minus straight x right parenthesis space equals space sin left parenthesis straight pi over 2 plus 2 sin to the power of negative 1 end exponent straight x right parenthesis
space rightwards double arrow space space left parenthesis 1 minus straight x right parenthesis space equals space cos left parenthesis 2 sin to the power of negative 1 end exponent straight x right parenthesis
space rightwards double arrow space space left parenthesis 1 minus straight x right parenthesis space equals space cos left parenthesis cos to the power of negative 1 end exponent left parenthesis 1 minus 2 straight x squared right parenthesis right parenthesis
space rightwards double arrow space left parenthesis 1 minus straight x right parenthesis equals space left parenthesis 1 minus 2 straight x squared right parenthesis
space rightwards double arrow space 1 minus straight x equals 1 minus 2 straight x squared
space space rightwards double arrow 2 straight x squared minus straight x space equals space 0
therefore straight x space equals 0 comma space space straight x space equals space 1 half
space space space space space
536 Views

 Multiple Choice QuestionsLong Answer Type

6.

Let A = Q × Q, where Q is the set of all rational numbers, and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b+ad) for (a, b), (c, d) element of A. Then find
(i) The identify element of * in A.
(ii) Invertible elements of A, and write the inverse of elements (5, 3) and open parentheses 1 half comma space 4 close parentheses.


Let A = Q x Q, where Q is the set of rational numbers.
Given that * is the binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for 
(a, b), (c, d) ∈ A.
(i)
We need to find the identity element of the operation * in A.
Let (x, y) be the identity element in A.
Thus,
(a, b) * (x, y) = (x, y) * (a, b) = (a, b), for all (a, b) ∈ A
⇒(ax, b + ay) = (a, b)
⇒ ax = a and b + ay =b
⇒ y = 0 and x = 1
Therefore, (1, 0) ∈ A is the identity element in A with respect to the operation *.

(ii) We need to find the invertible elements of A.
Let (p, q) be the inverse of the element (a, b)
Thus,
left parenthesis straight a comma space straight b right parenthesis asterisk times left parenthesis straight p comma space straight q right parenthesis space equals space left parenthesis 1 comma space 0 right parenthesis
rightwards double arrow space left parenthesis ap comma space straight b plus aq right parenthesis space equals left parenthesis 1 comma space 0 right parenthesis
rightwards double arrow space ap space equals space 1 space and space straight b plus aq space equals space 0
rightwards double arrow space straight p space equals 1 over straight a space space and space straight q equals negative straight b over straight a
space space Thus space the space inverse space elements space of space left parenthesis straight a comma space straight b right parenthesis space is space open parentheses 1 over straight a comma space minus straight b over straight a close parentheses
space space Now space let space us space find space the space inverse space of space left parenthesis 5 comma space 3 right parenthesis space and space open parentheses 1 half comma space 4 close parentheses
space Hence comma space inverse space of space left parenthesis 5 comma space 3 right parenthesis space is space open parentheses 1 fifth comma space minus 3 over 5 close parentheses
And space inverse space of space open parentheses 1 half comma space 4 close parentheses space is space open parentheses 2 comma space fraction numerator negative 4 over denominator begin display style 1 half end style end fraction close parentheses space equals space space left parenthesis 2 comma space minus 8 right parenthesis

1348 Views

 Multiple Choice QuestionsShort Answer Type

7.

Using the properties of determinants, solve the following for x:

open vertical bar table row cell straight x plus 2 end cell cell space space straight x plus 6 end cell cell space straight x minus 1 end cell row cell straight x plus 6 end cell cell straight x minus 1 end cell cell straight x plus 2 end cell row cell straight x minus 1 end cell cell straight x plus 2 end cell cell straight x plus 6 end cell end table close vertical bar space equals space 0


Let space increment space equals space open vertical bar table row cell straight x plus 2 end cell cell space space straight x plus 6 end cell cell space straight x minus 1 end cell row cell straight x plus 6 end cell cell straight x minus 1 end cell cell straight x plus 2 end cell row cell straight x minus 1 end cell cell straight x plus 2 end cell cell straight x plus 6 end cell end table close vertical bar space equals space 0
Applying space straight C subscript 2 rightwards arrow straight C subscript 2 minus straight C subscript 1 space and space straight C subscript 3 space rightwards arrow space straight C subscript 3 space minus straight C subscript 1
increment space equals space open vertical bar table row cell straight x plus 2 end cell cell space space 4 end cell cell space space minus 3 end cell row cell straight x plus 6 end cell cell space minus 7 end cell cell space minus 4 end cell row cell straight x minus 1 end cell cell space 3 end cell cell space space space 7 end cell end table close vertical bar
Applying space straight R subscript 2 space rightwards arrow straight R subscript 2 minus straight R subscript 1 space and space straight R subscript 3 space rightwards arrow straight R subscript 3 minus straight R subscript 1
space space space increment equals space open vertical bar table row cell straight x plus 2 end cell cell space space 4 end cell cell space space minus 3 end cell row 4 cell space minus 11 end cell cell space minus 1 end cell row cell negative 3 end cell cell space minus 1 end cell cell space space 10 end cell end table close vertical bar
space Applying space straight R subscript 2 rightwards arrow straight R subscript 2 space plus space straight R subscript 3
increment space equals space open vertical bar table row cell straight x plus 2 end cell cell space space 4 end cell cell space space minus 3 end cell row 1 cell space minus 12 end cell cell space space space 9 end cell row cell negative 3 end cell cell space minus 1 end cell cell space 10 end cell end table close vertical bar
Applying space straight R subscript 3 space rightwards arrow straight R subscript 3 plus left parenthesis 3 right parenthesis straight R subscript 2
space increment space equals space open vertical bar table row cell straight x plus 2 end cell cell space 4 end cell cell space minus 3 end cell row 1 cell negative 12 end cell cell space space 9 end cell row 0 cell negative 37 end cell cell space 37 end cell end table close vertical bar
Expanding space along space straight C subscript 1

increment space equals space left parenthesis straight x plus 2 right parenthesis space open vertical bar table row cell negative 12 end cell cell space space 9 end cell row cell negative 37 end cell cell space 37 end cell end table close vertical bar minus 1 open vertical bar table row 4 cell space minus 3 end cell row cell negative 37 end cell cell space space 37 end cell end table close vertical bar
increment equals left parenthesis straight x plus 2 right parenthesis thin space left parenthesis negative 444 plus 333 right parenthesis minus 1 left parenthesis 148 minus 111 right parenthesis
increment equals space left parenthesis straight x plus 2 right parenthesis thin space left parenthesis negative 111 right parenthesis minus 1 left parenthesis 37 right parenthesis
therefore space increment space equals space 0 space equals space minus 111 straight x minus 259
therefore space space straight x space equals space minus 259 over 111 equals negative 7 over 3
751 Views

 Multiple Choice QuestionsLong Answer Type

8.

Find the absolute maximum and absolute minimum values of the function f given by
straight f left parenthesis straight x right parenthesis space equals sin squared straight x minus cosx comma space straight x space element of space left parenthesis 0 comma space straight pi right parenthesis


straight f left parenthesis straight x right parenthesis space equals space sin squared straight x minus cosx comma
straight f apostrophe left parenthesis straight x right parenthesis equals 2 sinx. cosx space plus space sinx
equals space sin space straight x left parenthesis 2 cosx plus 1 right parenthesis
Equating space straight f apostrophe left parenthesis straight x right parenthesis space to space zero.
straight f apostrophe left parenthesis straight x right parenthesis space equals space 0
sinx left parenthesis 2 cosx plus 1 right parenthesis space equals space 0
sinx space equals space 0
therefore space straight x space equals space 0 comma space straight pi
2 cosx plus 1 space equals 0
rightwards double arrow cosx space equals space minus 1 half
therefore straight x space equals space fraction numerator 5 straight pi over denominator 6 end fraction
straight f left parenthesis 0 right parenthesis space equals space sin squared 0 minus cos 0 equals space minus 1
straight f open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses equals sin squared open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses minus cos open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses
equals sin squared straight pi over 6 plus cos straight pi over 6
equals 1 fourth minus fraction numerator square root of 3 over denominator 2 end fraction
equals open parentheses fraction numerator 1 minus 2 square root of 3 over denominator 4 end fraction close parentheses
straight f left parenthesis straight pi right parenthesis equals sin squared straight pi minus cosπ space equals 1
Of these values, the maximum value is 1, and the minimum value is −1.
Thus, the absolute maximum and absolute minimum values of f(x) are 1 and −1, which it attains at x = 0 and straight x equals straight pi
619 Views

 Multiple Choice QuestionsShort Answer Type

9.

If straight a with rightwards arrow on top space equals space 2 straight i with hat on top plus straight j with hat on top plus 3 straight k with hat on top space space space and space straight b with rightwards arrow on top space equals space 3 straight i with hat on top space plus space 5 straight j with hat on top space minus space 2 straight k with hat on top comma space then space find space open vertical bar straight a with rightwards arrow on top cross times space straight b with rightwards arrow on top close vertical bar.


Given space that space straight a with rightwards arrow on top space equals space 2 straight i with hat on top space plus space straight j with hat on top space plus space 3 straight k with hat on top space space and space straight b with rightwards arrow on top space equals space 3 straight i with hat on top space plus space 5 straight j with hat on top space minus space 2 straight k with hat on top
We space need space to space find space open vertical bar straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top close vertical bar
straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space equals space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 1 3 row 3 5 cell negative 2 end cell end table close vertical bar
space space space equals space straight i with hat on top left parenthesis negative 2 minus 15 right parenthesis space minus space straight j with hat on top left parenthesis negative 4 minus 9 right parenthesis space plus space straight k with hat on top left parenthesis 10 minus 3 right parenthesis
space space space space equals negative 17 straight i with hat on top space plus space 13 straight j with hat on top space plus space 7 straight k with hat on top
Hence comma space open vertical bar straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top close vertical bar space equals space square root of 17 squared plus 13 squared plus 7 squared end root
space space space space space space space space equals open vertical bar straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top close vertical bar space equals space square root of 507
1702 Views

10.

Write the element straight a subscript 12 of the matrix straight A space equals space open square brackets straight a subscript ij close square brackets subscript 2 cross times 2 end subscript comma whose elements straight a subscript ij are given by aij space equals space straight e to the power of 2 ix end exponent space sin space jx.


 Given space that space of space straight a subscript ij space equals space straight e to the power of 2 ix end exponent sin left parenthesis jx right parenthesis
Substitute space straight i space equals space 1 space and space straight j space equals space 2
Thus comma space straight a subscript 12 space equals straight e to the power of 2 cross times 1 cross times straight x end exponent space sin left parenthesis 2 cross times straight x right parenthesis space equals space straight e to the power of 2 straight x end exponent sin left parenthesis 2 straight x right parenthesis

1662 Views