Subject

Mathematics

Class

CBSE Class 12

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 Multiple Choice QuestionsShort Answer Type

1. If space space straight f open parentheses straight x close parentheses equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin open parentheses straight a plus 1 close parentheses plus 2 sinx over denominator straight x end fraction comma space straight x less than 0 end cell row cell 2 space space space space space space space space space space space space space space space space space space space space space space comma space x equals 0 end cell row cell fraction numerator square root of 1 plus b x end root minus 1 over denominator straight x end fraction space space space space space space comma space straight x greater than 0 end cell end table close
is continuous at x = 0, then find the values of a and b.

Given space that space straight f space is space continous space at space straight x space equals space 0

If space space straight f open parentheses straight x close parentheses equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin open parentheses straight a plus 1 close parentheses plus 2 sinx over denominator straight x end fraction comma space straight x space less than space 0 end cell row cell 2 space space space space space space space space space space space space space space space space space space space space space space comma space x equals 0 end cell row cell fraction numerator square root of 1 plus b x end root minus 1 over denominator straight x end fraction space space space space space space comma space straight x space greater than space 0 end cell end table close

Since space straight f left parenthesis straight x right parenthesis space is space continous space at space straight x equals 0 comma space limit as straight x rightwards arrow 0 of straight f space left parenthesis straight x right parenthesis space equals limit as straight x rightwards arrow 0 of straight f space left parenthesis 0 right parenthesis
Thus space straight R. straight H. straight L space equals space limit as straight x rightwards arrow 0 of straight f space left parenthesis straight x right parenthesis

space equals space limit as straight x rightwards arrow 0 of straight f space left parenthesis 0 plus straight h right parenthesis

equals space limit as straight h rightwards arrow 0 of space space fraction numerator square root of 1 plus bh end root minus 1 over denominator straight h end fraction

equals limit as straight h rightwards arrow 0 of space fraction numerator square root of 1 plus bh end root minus 1 over denominator straight h end fraction space straight x space fraction numerator square root of 1 plus bh end root plus 1 over denominator square root of 1 plus bh end root plus 1 end fraction

equals space limit as straight h rightwards arrow 0 of space fraction numerator 1 plus bh minus 1 over denominator straight h left parenthesis square root of 1 plus bh end root plus 1 right parenthesis end fraction

equals space limit as straight h rightwards arrow 0 of space fraction numerator straight b over denominator square root of 1 plus bh end root plus 1 end fraction

equals straight b over 2

Given space that space straight f space left parenthesis straight x right parenthesis space equals space 2

rightwards double arrow space limit as straight x rightwards arrow 0 of straight f space left parenthesis straight x right parenthesis space space equals space straight f left parenthesis 0 right parenthesis

rightwards double arrow space straight b over 2 space space equals space 2 space

rightwards double arrow space straight b equals space 4

Similarly comma space

straight L. straight H. straight L space equals space limit as straight x rightwards arrow 0 of straight f space left parenthesis straight x right parenthesis

space equals space limit as straight x rightwards arrow 0 of straight f space left parenthesis 0 minus straight h right parenthesis

equals space limit as straight h rightwards arrow 0 of fraction numerator sin space left parenthesis straight a plus 1 right parenthesis left parenthesis 0 minus straight h right parenthesis plus 2 sin left parenthesis 0 minus straight h right parenthesis over denominator 0 minus straight h end fraction

equals space limit as straight h rightwards arrow 0 of fraction numerator negative sin left parenthesis straight a plus 1 right parenthesis straight h minus 2 space sin space straight h over denominator negative straight h end fraction

equals space limit as straight h rightwards arrow 0 of fraction numerator negative sin space left parenthesis straight a plus 1 right parenthesis straight h over denominator negative straight h end fraction space plus space limit as straight h rightwards arrow 0 of fraction numerator negative 2 sin space straight h over denominator negative straight h end fraction

equals space limit as straight h rightwards arrow 0 of fraction numerator sin left parenthesis straight a plus 1 right parenthesis straight h over denominator straight h end fraction fraction numerator left parenthesis straight a plus 1 right parenthesis over denominator left parenthesis straight a plus 1 right parenthesis end fraction space plus space 2 stack space lim with straight h rightwards arrow 0 below fraction numerator sin space straight h over denominator straight h end fraction

equals space straight a plus 1 plus 2 space space space space space space open square brackets therefore space lim space fraction numerator sin space straight theta over denominator straight theta end fraction equals 1 space close square brackets

Given space that space straight f space left parenthesis straight x right parenthesis equals 2

rightwards double arrow stack space lim with straight x space rightwards arrow 0 below space straight f space left parenthesis straight x right parenthesis space equals space straight f space left parenthesis 0 right parenthesis
rightwards double arrow space straight a plus 1 plus space 2 space space equals space 2
rightwards double arrow space straight a space equals negative 1
1266 Views

2. if space cos space left parenthesis straight a plus straight y right parenthesis space equals space cos space straight y space then space prove space that space dy over dx space equals fraction numerator cos squared left parenthesis straight a plus straight y right parenthesis over denominator sin space straight a end fraction
Hence space show space that space sin space straight a space fraction numerator straight d squared straight y over denominator dx squared end fraction plus sin space 2 space left parenthesis straight a plus straight y right parenthesis dy over dx equals 0

Given space that comma

straight x space cos space left parenthesis straight a plus straight y right parenthesis space equals space cos space straight y space.... space left parenthesis straight i right parenthesis

rightwards double arrow space straight x space equals space fraction numerator cos space straight y over denominator cos space left parenthesis straight a space plus space straight y right parenthesis end fraction space... space left parenthesis ii right parenthesis

Differentiating space both space sides space of space the space equation space left parenthesis straight i right parenthesis comma space we space have comma

straight x space left parenthesis negative sin space left parenthesis straight a space plus space straight y right parenthesis right parenthesis space dy over dx equals space minus space cos space left parenthesis straight a plus straight y right parenthesis space equals negative sin space straight y space dy over dx

rightwards double arrow space space open square brackets sin space straight y minus space straight x space sin space left parenthesis straight a space plus space straight y right parenthesis close square brackets dy over dx space equals space minus cos space left parenthesis straight a space plus space straight y right parenthesis

rightwards double arrow space open square brackets sin space straight y space minus fraction numerator cos space straight y over denominator cos space left parenthesis straight a plus straight y right parenthesis end fraction sin space left parenthesis straight a space plus space straight y right parenthesis close square brackets dy over dx space space equals negative cos space left parenthesis straight a space plus space straight y right parenthesis

rightwards double arrow space open square brackets fraction numerator cos space left parenthesis straight a space plus space straight y right parenthesis space space straight x space space sin space straight y space minus space cos space straight y space sin space left parenthesis straight a space plus space straight y right parenthesis over denominator cos space left parenthesis straight a plus straight y right parenthesis end fraction close square brackets dy over dx space equals space minus space cos left parenthesis straight a space plus space straight y right parenthesis space

rightwards double arrow space open square brackets cos space left parenthesis straight a space plus space straight y right parenthesis space straight x space sin space straight y space minus space cosy space space straight x space space sin space left parenthesis straight a space plus space straight y right parenthesis close square brackets dy over dx equals negative cos left parenthesis straight a space plus space straight y right parenthesis space space straight x space space cos space left parenthesis straight a space plus space straight y right parenthesis

rightwards double arrow space space open square brackets sin space left parenthesis straight a space plus space straight y space minus space straight y right parenthesis close square brackets dy over dx space space equals space minus cos squared left parenthesis straight a space plus space straight y right parenthesis space space space space space open square brackets sin left parenthesis straight A minus straight B right parenthesis space equals space sinA space cos space straight B space minus space cosA space sinB close square brackets

rightwards double arrow left square bracket sina space right square bracket space dy over dx equals negative cos squared left parenthesis straight a space plus space straight y right parenthesis

rightwards double arrow space dy over dx space equals fraction numerator negative cos squared left parenthesis straight a space plus space straight y right parenthesis over denominator sin space straight a space end fraction space space space space.. space left parenthesis iii right parenthesis

differentiating space once space again space with space respect space to space straight x space comma we space have comma

sin space straight a space fraction numerator straight d squared straight y over denominator dx squared end fraction space plus space 2 cos left parenthesis straight a space plus space straight y right parenthesis space sin space left parenthesis straight a space plus space straight y right parenthesis dy over dx

rightwards double arrow space sin space straight a space fraction numerator straight d squared straight y over denominator dx squared end fraction space plus space 2 space cos space left parenthesis straight a space plus space straight y right parenthesis space sin space left parenthesis straight a space plus space straight y right parenthesis dy over dx equals 0

rightwards double arrow space sin space straight a space fraction numerator straight d squared straight y over denominator dx squared end fraction space plus space sin space 2 space left parenthesis straight a space plus space straight y right parenthesis dy over dx equals 0

Hence space proved.
1206 Views

3. Prove space that space tan to the power of negative 1 end exponent space open parentheses fraction numerator 6 straight x minus 8 straight x cubed over denominator 1 minus 12 straight x squared end fraction close parentheses minus tan to the power of negative 1 end exponent space open parentheses fraction numerator 4 straight x over denominator 1 minus 4 straight x squared end fraction close parentheses space equals space tan to the power of negative 1 end exponent space 2 straight x semicolon space vertical line 2 straight x vertical line less than fraction numerator 1 over denominator square root of 3 end fraction

Taking L.H.S,
tan to the power of negative 1 end exponent open parentheses fraction numerator 6 straight x minus 8 straight x cubed over denominator 1 minus 12 straight x squared end fraction close parentheses space minus space tan to the power of negative 1 end exponent open parentheses fraction numerator 4 straight x over denominator 1 minus 4 straight x squared end fraction close parentheses

We space know space that comma
tan to the power of negative 1 end exponent space left parenthesis straight A right parenthesis space minus space tan to the power of negative 1 end exponent space left parenthesis straight B right parenthesis space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator straight A plus straight B over denominator 1 plus AB end fraction close parentheses

Thus comma space straight L. straight H. straight S space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator 6 straight x minus 8 straight x cubed over denominator 1 minus 12 straight x squared end fraction end style minus begin display style fraction numerator 4 straight x over denominator 1 minus 4 straight x squared end fraction end style over denominator 1 plus open parentheses begin display style fraction numerator 6 straight x minus 8 straight x cubed over denominator 1 minus 12 straight x squared end fraction end style close parentheses open parentheses begin display style fraction numerator 4 straight x over denominator 1 minus 4 straight x squared end fraction end style close parentheses end fraction close parentheses

equals space tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator left parenthesis 6 straight x minus 8 straight x cubed right parenthesis left parenthesis 1 minus 4 straight x squared right parenthesis minus 4 straight x left parenthesis 1 minus 12 straight x squared right parenthesis over denominator left parenthesis 1 minus 12 straight x squared right parenthesis left parenthesis 1 minus 4 straight x right parenthesis squared end fraction end style over denominator begin display style fraction numerator 4 straight x left parenthesis 6 straight x minus 8 straight x cubed right parenthesis over denominator left parenthesis 1 minus 12 straight x squared right parenthesis left parenthesis 1 minus 4 straight x squared right parenthesis end fraction end style end fraction close parentheses space

equals space tan to the power of negative 1 end exponent open parentheses fraction numerator left parenthesis 6 straight x minus 24 straight x cubed minus 8 straight x cubed plus 32 straight x to the power of 5 minus 4 straight x plus 48 straight x cubed over denominator 1 minus 4 straight x squared minus 12 straight x squared plus 48 straight x to the power of 4 plus 24 straight x squared minus 32 straight x to the power of 4 end fraction close parentheses

equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 32 straight x to the power of 5 plus 16 straight x to the power of 6 plus 2 straight x over denominator 16 straight x to the power of 4 plus 8 straight x squared plus 1 end fraction close parentheses

equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x left parenthesis 16 straight x to the power of 4 plus 2 straight x over denominator 16 straight x to the power of 4 plus 8 straight x squared plus 1 end fraction close parentheses

equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x left parenthesis 16 straight x to the power of 4 plus 8 straight x squared plus 1 over denominator 16 straight x to the power of 4 plus 8 straight x squared plus 1 end fraction close parentheses

equals space tan to the power of negative 1 end exponent space 2 straight x

Thus comma space straight L. straight H. straight S space equals space straight R. straight H. straight S

676 Views

4.

Find the equation of tangents to the curve y= x3+2x-4, which are perpendicular to line x+14y+3 =0.


Taking the given equation,
y = x3+2x-4
Differentiating the above function with respect to x, we have,
dy over dx equals space 3 straight x squared plus 2
rightwards double arrow space straight m subscript 1 equals space 3 straight x squared plus 2

Given that the tangents to the given curve are perpendicular to the line x+ 14y + 3 = 0
Slope of this line, m2=-1/14
Since the given line and the tangents to the curve are perpendicular, we have,

m1 x m2 =-1

rightwards double arrow space left parenthesis 3 straight x squared plus 2 right parenthesis open parentheses fraction numerator negative 1 over denominator 14 end fraction close parentheses space equals negative 1
rightwards double arrow space 3 straight x squared space plus 2 space equals space 14
rightwards double arrow space 3 straight x squared space equals space 12
rightwards double arrow space straight x squared space equals 4
rightwards double arrow space straight x space equals plus-or-minus 2
If space straight x equals 2 comma space straight y equals straight x cubed space plus 2 straight x minus 4
rightwards double arrow space straight y equals left parenthesis negative 2 right parenthesis cubed plus 2 straight x space left parenthesis negative 2 right parenthesis minus 4
rightwards double arrow straight y equals negative 16

Equation space of space the space tangent space having space slope space straight m space at space the space point space left parenthesis straight x subscript 1 comma straight y subscript 1 right parenthesis space is
left parenthesis straight y minus straight y subscript 1 right parenthesis space equals straight m left parenthesis straight x minus straight x subscript 1 right parenthesis

Equation space of space the space tangent space at space straight P space left parenthesis 2 comma 8 right parenthesis space with space slope space 14
left parenthesis straight y minus 8 right parenthesis equals 14 left parenthesis straight x minus 2 right parenthesis
rightwards double arrow space straight y minus 8 space equals space 14 space straight x minus 28
rightwards double arrow 14 straight x minus straight y equals 20

Equation space of space the space tangent space at space straight P left parenthesis negative 2 comma negative 16 right parenthesis space with space slope space 14
left parenthesis straight y minus 8 right parenthesis equals 14 left parenthesis straight x minus 2 right parenthesis
rightwards double arrow space straight y minus 8 space equals space 14 space straight x minus 28
rightwards double arrow 14 space straight x minus straight y space equals 20

Equation space of space the space tangent space at space straight P left parenthesis negative 2 comma negative 16 right parenthesis space with space slope space 1
left parenthesis straight y plus 16 right parenthesis equals 14 left parenthesis straight x plus 2 right parenthesis
rightwards double arrow straight y plus 16 space equals space 14 straight x plus 28
rightwards double arrow 14 space straight x minus straight y space equals space minus 12
thus equation of the tangent is 14 x- y =-12

2336 Views

5. Find space dy over dx space if space straight y equals sin to the power of negative 1 end exponent open square brackets fraction numerator 6 straight x minus 4 square root of 1 minus 4 straight x squared end root over denominator 5 end fraction close square brackets

Given space that
straight y space equals space sin to the power of negative 1 end exponent space open square brackets fraction numerator 6 straight x minus 4 square root of 1 space minus space 4 straight x squared end root over denominator 5 end fraction close square brackets

if space straight y space equals space sin to the power of negative 1 end exponent straight x comma space then space dy over dx space equals fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction

straight y space equals space sin to the power of negative 1 end exponent open square brackets fraction numerator 6 straight x minus 4 square root of 1 space minus space 4 straight x squared end root over denominator 5 end fraction close square brackets

rightwards double arrow space space straight y space equals space sin to the power of negative 1 end exponent open square brackets fraction numerator 6 straight x over denominator 5 end fraction minus fraction numerator 4 square root of 1 space minus space 4 straight x squared end root over denominator 5 end fraction close square brackets

rightwards double arrow space straight y space equals space sin to the power of negative 1 end exponent open square brackets fraction numerator 2 straight x.3 over denominator 3 end fraction space minus space fraction numerator 4 square root of 1 space minus space left parenthesis 2 straight x right parenthesis squared end root over denominator 5 end fraction close square brackets

rightwards double arrow space straight y space equals space sin to the power of negative 1 end exponent open square brackets 2 straight x.3 over 5 minus 4 over 5 square root of 1 space minus space left parenthesis 2 straight x right parenthesis squared end root close square brackets

rightwards double arrow space straight y space equals space sin to the power of negative 1 end exponent open square brackets 2 straight x space square root of 1 minus open parentheses 4 over 5 close parentheses squared end root minus space 4 over 5 square root of 1 space minus left parenthesis 2 straight x right parenthesis squared end root close square brackets

we space know space that comma

sin to the power of negative 1 end exponent straight p space minus space sin to the power of negative 1 end exponent straight q space equals space sin to the power of negative 1 end exponent space left parenthesis straight p square root of 1 space minus space straight q squared end root minus straight q-th root of 1 space minus space straight p squared end root right parenthesis

Here comma space straight p space equals space 2 straight x space space and space straight q space equals 4 over 5

Differentiating space the space above space functions space with space respect space straight x comma space space we space have comma space

dy over dx space equals space fraction numerator 1 over denominator square root of 1 minus left parenthesis 2 straight x right parenthesis squared end root end fraction space straight x space 2 minus 0

rightwards double arrow space space dy over dx space equals space fraction numerator 2 over denominator square root of 1 minus 4 straight x squared end root end fraction
space
783 Views

6.

For what values of k, the system of linear equations

x + y + z = 2
2x + y - z = 3
3x + 2y + kz = 4

has a unique solution?


For space unique space solution space vertical line straight A vertical line space not equal to 0

open vertical bar table row 1 1 1 row 2 1 cell negative 1 end cell row 3 2 straight k end table close vertical bar space not equal to space 0
straight C subscript 2 rightwards arrow straight C subscript 2 minus straight C subscript 1 semicolon space straight C subscript 3 rightwards arrow straight C subscript 3 minus straight C subscript 1

open vertical bar table row 1 0 0 row 2 cell negative 1 end cell cell negative 3 end cell row 3 cell negative 1 end cell cell straight k minus 3 end cell end table close vertical bar space not equal to space 0

expansion space along space straight R subscript 1
minus left parenthesis straight k minus 3 right parenthesis minus 3 space not equal to 0

minus straight k plus 3 minus 3 space not equal to 0

straight k not equal to 0
1308 Views

7. Solve space for space straight x space colon space tan to the power of negative 1 end exponent space left parenthesis straight x space plus 1 right parenthesis space plus space tan to the power of negative 1 end exponent straight x space plus tan to the power of negative 1 end exponent space left parenthesis straight x plus 1 right parenthesis space equals space tan to the power of negative 1 end exponent 3 straight x.

Given space that comma
tan to the power of negative 1 end exponent space left parenthesis straight x minus 1 right parenthesis space plus space tan to the power of negative 1 end exponent space straight x space plus space tan to the power of negative 1 end exponent space left parenthesis straight x plus 1 right parenthesis space equals space tan to the power of negative 1 end exponent 3 straight x

rightwards double arrow space space tan to the power of negative 1 end exponent space left parenthesis straight x minus 1 right parenthesis space plus space tan to the power of negative 1 end exponent space left parenthesis straight x plus 1 right parenthesis space equals space tan to the power of negative 1 end exponent space 3 straight x space minus tan to the power of negative 1 end exponent space straight x space.... space left parenthesis straight i right parenthesis
we space know space that comma space tan to the power of negative 1 end exponent space straight A space plus space tan to the power of negative 1 end exponent space straight B space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator straight A plus straight B over denominator 1 plus AB end fraction close parentheses
and comma space tan to the power of negative 1 end exponent straight A space minus space tan to the power of negative 1 end exponent straight B space equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator straight A minus straight B over denominator 1 plus AB end fraction close parentheses
Thus comma space tan to the power of negative 1 end exponent left parenthesis straight x minus 1 right parenthesis space plus space tan to the power of negative 1 end exponent space left parenthesis straight x plus 1 right parenthesis space equals space tan to the power of negative 1 end exponent space equals space open parentheses fraction numerator straight x minus 1 plus straight x plus 1 over denominator 1 minus left parenthesis straight x minus 1 right parenthesis left parenthesis straight x plus 1 right parenthesis end fraction close parentheses
equals space space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 1 minus left parenthesis straight x to the power of 2 minus end exponent 1 right parenthesis end fraction close parentheses

equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 2 minus straight x squared end fraction close parentheses space space.. space left parenthesis ii right parenthesis
Similarly comma space tan to the power of negative 1 end exponent 3 straight x space minus space tan to the power of negative 1 end exponent space straight x space space equals space space tan to the power of negative 1 end exponent space open parentheses fraction numerator 3 straight x minus straight x over denominator 1 plus 3 straight x left parenthesis straight x right parenthesis end fraction close parentheses
equals space space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus 3 straight x squared end fraction close parentheses space.. space left parenthesis iii right parenthesis

From space space equ. space left parenthesis straight i right parenthesis space comma space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis comma space we space have comma

tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 2 minus straight x squared end fraction close parentheses space space equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 1 plus 3 straight x squared end fraction close parentheses space

rightwards double arrow space fraction numerator 2 straight x over denominator 2 minus straight x squared end fraction space equals fraction numerator 2 straight x over denominator 1 plus 3 straight x squared end fraction

cross space multiply space the space above space equ.

rightwards double arrow space 2 minus straight x squared space equals space 1 plus 3 straight x squared

rightwards double arrow space 4 straight x squared minus 1

rightwards double arrow space 4 straight x squared space equals 1

rightwards double arrow space straight x space equals space plus for minus of space 1 half
1117 Views

8. If space straight A space equals space open parentheses table row cell cos space straight alpha end cell cell sin space straight alpha end cell row cell negative sin space straight alpha end cell cell cos space straight alpha end cell end table close parentheses comma space find space straight alpha space satisfying space 0 space less than space straight alpha space less than space straight pi over 2 space when space straight A space plus space straight A to the power of straight T space space equals space square root of 2 straight I subscript 2 semicolon space Where space straight A to the power of straight T space is space
transpose space of space straight A.


straight A space equals space open square brackets table row cell cos space straight alpha end cell cell sin space straight alpha end cell row cell negative sin space straight alpha end cell cell cos space straight alpha end cell end table close square brackets space space 0 space less than space straight alpha space less than space straight pi over 2

straight A space plus space straight A to the power of straight T space equals space square root of 2 straight I subscript 2 space

open square brackets table row cell cos space straight alpha end cell cell sin space straight alpha end cell row cell negative sin space straight alpha end cell cell cos space straight alpha end cell end table close square brackets space plus space open square brackets table row cell cos space straight alpha end cell cell negative sin space straight alpha end cell row cell sin space straight alpha end cell cell cos space straight alpha end cell end table close square brackets space space equals space square root of 2 space end root open square brackets table row 1 0 row 0 1 end table close square brackets

open square brackets table row cell 2 cos space straight alpha end cell 0 row 2 cell 2 cos space straight alpha end cell end table close square brackets space space equals space open square brackets table row cell square root of 2 end cell 0 row 0 cell square root of 2 end cell end table close square brackets

2 space cos space straight alpha space equals space square root of 2

cos space straight alpha space space equals space fraction numerator square root of 2 over denominator 2 end fraction space equals space fraction numerator 1 over denominator square root of 2 end fraction

straight alpha space equals space straight pi over 4
4254 Views

9.

A typist charges Rs. 145 for typing 10 English and 3 Hindi pages, while charges for typing 3 English and 10 Hindi pages are Rs. 180. Using matrices, find the charges of typing one English and one Hindi page separately. However, typist charged only Rs. 2 per page from a poor student Shyam for 5 Hindi pages. How much less was charged from this poor boy? Which values are reflected in this problem?


Let charges for typing one English page be Rs. x.

Let charges for typing one Hindi page be Rs.y.

Thus from the given statements, we have,

10x  + 3y = 145

3x + 10y = 180

Thus the above system can be written as,

open square brackets table row 10 3 row 3 10 end table close square brackets open square brackets table row straight x row straight y end table close square brackets space space equals space open square brackets table row 145 row 180 end table close square brackets
rightwards double arrow space AX space equals space straight B comma space where comma space straight A space equals space open square brackets table row 10 3 row 3 10 end table close square brackets comma space straight X space space equals space open square brackets table row straight x row straight y end table close square brackets space and space open square brackets table row 145 row 180 end table close square brackets
Multiply space straight A to the power of negative 1 end exponent space on space both space the space sides comma space we space have comma

straight A to the power of negative 1 end exponent space straight x space AX space equals space straight A to the power of negative 1 end exponent straight B

rightwards double arrow space IX space space equals space straight A to the power of negative 1 end exponent straight B

rightwards double arrow space straight X space equals space straight A to the power of negative 1 end exponent straight B

Thus comma space we space need space to space find space the space inverse space of space the space matrix space straight A.

We space know space that comma space if space space straight P space equals space open square brackets table row straight a straight b row straight c straight d end table close square brackets space then space straight p to the power of negative 1 end exponent space equals space fraction numerator 1 over denominator ad minus bc end fraction space open square brackets table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close square brackets

Thus comma space straight A to the power of negative 1 end exponent space equals space fraction numerator 1 over denominator 10 space straight x space 10 minus 3 space straight x space 3 end fraction open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets

equals fraction numerator 1 over denominator 100 minus 9 end fraction open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets

equals 1 over 91 open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets

therefore space space straight X space equals space 1 over 91 open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets space space open square brackets table row 145 row 180 end table close square brackets

equals 1 over 91 open square brackets table row cell 10 space straight x space 145 space minus end cell cell 3 straight x space 180 end cell row cell negative 3 space straight x space 145 space plus end cell cell 10 straight x 180 end cell end table close square brackets

space equals 1 over 91 open square brackets table row 910 row 1365 end table close square brackets

equals open square brackets table row 10 row 15 end table close square brackets

rightwards double arrow space open square brackets table row straight x row straight y end table close square brackets space equals space open square brackets table row 10 row 15 end table close square brackets

therefore comma

straight x space equals 10 space and space straight y space equals space 15

Amount space taken space from space shyam space equals 2 space straight x space 5 space equals Rs. space 10

Actual space rate space equals space 15 space straight x space 5 space equals Rs. space 75

Difference space amount space equals space 75 minus 10 space equals space 65

Rs. space 65 space was space less space charged space from space the space poor space boy space Shyam.
Humanity is reflected in this problem.

2328 Views

10.

If A is a 3 x 3 matrix |3A| = k|A|, then write the value of k.


|3A| = k |A|

|3A| = 27|A|

k = 27

920 Views