Subject

Mathematics

Class

CBSE Class 12

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 Multiple Choice QuestionsShort Answer Type

1.

Using properties of determinants, prove that 

open vertical bar table row cell straight x space end cell cell straight x plus straight y space end cell cell straight x plus 2 straight y space end cell row cell straight x plus 2 straight y space end cell cell straight x space end cell cell straight x plus straight y end cell row cell straight x plus straight y space end cell cell straight x plus 2 straight y space end cell cell straight x space end cell end table close vertical bar space equals space 9 straight y squared left parenthesis straight x plus straight y right parenthesis


open vertical bar table row cell straight x space end cell cell straight x plus straight y space end cell cell straight x plus 2 straight y space end cell row cell straight x plus 2 straight y space end cell cell straight x space end cell cell straight x plus straight y end cell row cell straight x plus straight y space end cell cell straight x plus 2 straight y space end cell cell straight x space end cell end table close vertical bar
Applying space straight R subscript 1 rightwards arrow straight R subscript 1 space plus straight R subscript 2 space plus space straight R subscript 3 space we space get colon
open vertical bar table row cell 3 left parenthesis straight x plus straight y right parenthesis end cell cell 3 left parenthesis straight x plus straight y right parenthesis end cell cell 3 left parenthesis straight x plus straight y right parenthesis end cell row cell straight x plus 2 straight y end cell straight x cell straight x plus straight y end cell row cell straight x plus straight y end cell cell straight x plus 2 straight y end cell straight x end table close vertical bar
space equals space 3 left parenthesis straight x plus straight y right parenthesis space open vertical bar table row 1 1 1 row cell straight x plus 2 straight y end cell straight x cell straight x plus straight y end cell row cell straight x plus straight y end cell cell straight x plus 2 straight y end cell straight x end table close vertical bar
Applying space straight C subscript 1 rightwards arrow straight C subscript 1 minus straight C subscript 2 space and space straight C subscript 2 space rightwards arrow straight C subscript 2 minus straight C subscript 3 space we space get colon
3 left parenthesis straight x plus straight y right parenthesis space open vertical bar table row 0 0 1 row cell 2 straight y end cell cell negative straight y end cell cell straight x plus straight y end cell row cell negative straight y end cell cell 2 straight y end cell straight x end table close vertical bar
equals space 3 left parenthesis straight x plus straight y right parenthesis open vertical bar table row cell 2 straight y end cell cell negative straight y end cell row cell negative straight y end cell cell 2 straight y end cell end table close vertical bar space equals space 3 left parenthesis straight x plus straight y right parenthesis left parenthesis 4 straight y squared minus straight y squared right parenthesis space equals space 9 straight y squared space left parenthesis straight x plus straight y right parenthesis
space RHS equals space LHS
Hence space Proved
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2.

Show that all the diagonal elements of a skew symmetric matrix are zero.


Let A [aij] be a skew symmetric matrix.
so,
aij =-aji for all i,j
⇒aii -aii for all values of i
⇒2aii =0
⇒aii =0 for all values of i
⇒a11 = a22 = a33 =..... ann =0

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3.

The volume of a sphere is increasing at the rate of 3 cubic centimetres per second. Find the rate of increase of its surface area, when the radius is 2 cm.


straight V space equals 4 over 3 πr cubed
rightwards double arrow space dV over dt space equals space 4 over 3 straight pi.3 straight r squared dr over dt
rightwards double arrow space dV over dt space equals space 4 πr squared dr over dt
rightwards double arrow dr over dt space equals space fraction numerator 1 over denominator 4 πr squared end fraction dV over dt
rightwards double arrow dr over dt space equals space fraction numerator 3 over denominator 4 straight pi left parenthesis 2 right parenthesis squared end fraction
open square brackets straight r equals space 2 space cm space and space dV over dt space equals space 3 space cm squared divided by sec close square brackets

rightwards double arrow dr over dt space space equals fraction numerator 3 over denominator 16 space straight pi end fraction cm divided by sec

Now, let S be the surface area of the sphere at any time t. then,
S = 4πr2
rightwards double arrow space dS over dt space equals space 8 πr dr over dt
rightwards double arrow space dS over dt space equals space 8 straight pi space left parenthesis 2 right parenthesis space straight x fraction numerator 3 over denominator 16 straight pi end fraction
open square brackets straight r space equals 2 space cm space and space dr over dt space equals space fraction numerator 3 over denominator 16 straight pi end fraction space cm divided by sec close square brackets
rightwards double arrow space dS over dt space equals space 3 space cm squared divided by sec



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4.

Prove that tan space open curly brackets straight pi over 4 plus 1 half cos to the power of negative 1 end exponent straight a over straight b close curly brackets space plus tan space open curly brackets straight pi over 4 minus 1 half cos to the power of negative 1 end exponent straight a over straight b close curly brackets space equals space fraction numerator 2 straight b over denominator straight a end fraction


Let space cos to the power of negative 1 end exponent space equals space open parentheses straight a over straight b close parentheses space equals space straight x
Then comma space cos space straight x space equals space straight a over straight b
LHS colon
tan open curly brackets straight pi over 4 plus 1 half cos to the power of negative 1 end exponent straight a over straight b close curly brackets plus space tan space open parentheses straight pi over 4 minus 1 half cos to the power of negative 1 end exponent straight a over straight b close parentheses
equals space tan open parentheses straight pi over 4 space plus straight x over 2 close parentheses plus space tan space open parentheses straight pi over 4 minus straight x over 2 close parentheses
fraction numerator 1 plus space tan begin display style straight x over 2 end style over denominator 1 minus tan begin display style straight x over 2 end style end fraction space plus fraction numerator 1 minus space tan begin display style straight x over 2 end style over denominator 1 plus tan begin display style straight x over 2 end style end fraction
fraction numerator open parentheses 1 plus tan begin display style straight x over 2 end style close parentheses squared plus open parentheses 1 minus tan begin display style straight x over 2 end style close parentheses squared over denominator 1 minus tan squared begin display style straight x over 2 end style end fraction
space equals space 2 open parentheses fraction numerator 1 plus tan squared begin display style straight x over 2 end style over denominator 1 minus tan squared begin display style straight x over 2 end style end fraction close parentheses
space equals space fraction numerator 2 over denominator cos space straight x end fraction space open square brackets because space cos space 2 straight x space equals space fraction numerator 1 minus tan squared straight x over denominator 1 plus space tan squared space straight x end fraction close square brackets
space equals space fraction numerator 2 straight b over denominator straight a end fraction space equals space RHS
Hence space proved
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5.

If A is a 3 × 3 invertible matrix, then what will be the value of k if det(A–1) = (det A)k.


straight A to the power of negative 1 end exponent space equals space fraction numerator Adj space straight A over denominator vertical line straight A vertical line end fraction
therefore space vertical line straight A to the power of negative 1 end exponent vertical line space equals space fraction numerator vertical line Adj space straight A vertical line over denominator vertical line straight A vertical line end fraction
equals space fraction numerator vertical line straight A vertical line to the power of 3 minus 1 end exponent over denominator vertical line straight A vertical line end fraction
left square bracket because space If space straight A space is space straight a space non space singular space matrix space of space order space straight n comma space then space vertical line adj space left parenthesis straight A right parenthesis space equals space vertical line straight A vertical line to the power of straight n minus 1 end exponent right square bracket

equals space fraction numerator vertical line straight A vertical line squared over denominator vertical line straight A vertical line end fraction
As space we space are space given space that space vertical line straight A to the power of negative 1 end exponent vertical line space equals space vertical line straight A vertical line to the power of straight k
therefore space straight k space equals negative 1
3628 Views

6. Find space dy over dx space at space straight x space equals 1 comma space straight y space equals space straight pi over 4 space if space sin squared space straight y space plus space cos space xy space equals space straight K

From the given equation

2 space sin space straight y space cos space straight y. space dy over dx minus space sin space xy. space open square brackets straight x. dy over dx plus straight y.1 close square brackets space equals 0
rightwards double arrow space dy over dx space equals space fraction numerator straight y space sin space xy over denominator sin space 2 straight y space minus space straight x space sin left parenthesis space xy right parenthesis end fraction

therefore space right enclose dy over dx end enclose subscript straight x space equals 1 comma space straight y space equals straight pi over 4 end subscript space equals space fraction numerator straight pi over denominator 4 left parenthesis square root of 2 minus 1 right parenthesis end fraction

695 Views

7.

Determine the value of the constant ‘k’ so that the functionstraight f left parenthesis straight x right parenthesis space equals space open curly brackets table attributes columnalign left end attributes row cell fraction numerator Kx over denominator vertical line straight x vertical line end fraction comma space If space straight x space less than 0 end cell row cell 3 comma space if space straight x greater or equal than space 0 end cell end table close  is continuous at x = 0.


straight f left parenthesis straight x right parenthesis space equals space open curly brackets table attributes columnalign left end attributes row cell fraction numerator Kx over denominator vertical line straight x vertical line end fraction comma space If space straight x space less than 0 end cell row cell 3 comma space if space straight x greater or equal than space 0 end cell end table close
stack lim space straight f left parenthesis straight x right parenthesis with straight x space rightwards arrow 0 below space equals stack space lim with straight x rightwards arrow 0 below space straight f left parenthesis space straight x right parenthesis space equals space straight f left parenthesis 0 right parenthesis
rightwards double arrow space limit as straight x space rightwards arrow 0 of space fraction numerator negative kx over denominator straight x end fraction space equals space stack lim space 3 with straight x rightwards arrow 0 below space equals space 3
rightwards double arrow negative straight k space equals 3
rightwards double arrow straight k equals negative 3
837 Views

8.

Let straight A space equals space open parentheses table row 2 cell negative 1 end cell row 3 4 end table close parentheses comma space straight B space equals space open parentheses table row 5 2 row 7 4 end table close parentheses comma space straight C equals space open parentheses table row 2 5 row 3 8 end table close parenthesesfind a matrix D such that CD – AB = O.


Let space straight D space equals space open square brackets table row straight p straight q row straight r straight s end table close square brackets
Then CD-AB = O

open square brackets table row 2 5 row 3 8 end table close square brackets open square brackets table row straight p straight q row straight r straight s end table close square brackets space minus space open square brackets table row 2 cell negative 1 end cell row 3 cell space space 4 end cell end table close square brackets open square brackets table row 5 2 row 7 4 end table close square brackets space equals straight O

rightwards double arrow space open square brackets table row cell 2 straight p space plus space 5 straight r end cell cell 2 straight q space plus space 5 straight s end cell row cell 3 straight p space plus space 8 straight r end cell cell 3 straight q space plus space 8 straight s end cell end table close square brackets minus open square brackets table row 3 0 row 43 22 end table close square brackets space equals space open square brackets table row 0 0 row 0 0 end table close square brackets

rightwards double arrow space open square brackets table row cell 2 straight p plus space 5 straight r minus 3 end cell cell 2 straight q space plus 5 straight s end cell row cell 3 straight p plus 8 straight r minus 43 end cell cell 3 straight q plus 8 straight s minus 22 end cell end table close square brackets space equals space open square brackets table row 0 0 row 0 0 end table close square brackets

By equality of matrices we get,
2p +5r-3 = 0 ....(1)
3p +8r-43 = 0 ..(2)
2q +5s = 0 ......(3)
3q +8s-22 = 0 ..(4)
By solving (1) and (2) we get p = -191 and r = 77
Similarly, on solving (3) and (4) we get q = - 110 and s = 44

Let space straight D space equals space open square brackets table row straight p straight q row straight r straight s end table close square brackets space equals space open square brackets table row cell negative 191 end cell cell negative 110 end cell row 77 44 end table close square brackets

530 Views

9.

If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of the z-axis.


suppose the direction cosines of the line be l,m,and n.
we know that l2 + m2+n2 = 1
Let the line make angle θ with the positive direction of the z-axis.
α = 90°, β = 60° γ = θ
Thus,
cos2 90 + cos260 + cos2θ =1
rightwards double arrow space 0 space plus space open parentheses 1 half close parentheses squared space plus space cos squared straight theta space equals space 1
rightwards double arrow space cos squared space straight theta space equals space 1 minus 1 fourth
rightwards double arrow space cos squared space straight theta space equals 3 over 4
rightwards double arrow space cos space straight theta space equals space plus-or-minus fraction numerator square root of 3 over denominator 2 end fraction
rightwards double arrow space straight theta space equals space 30 to the power of straight o

1682 Views

10.

Show that the function f(x) = 4x3 – 18x2 + 27x – 7 is always increasing on R.


The given function is f(x) =4x3 – 18x2 + 27x – 7 

On differentiating both sides with respect to x, we get
f'(x) = 12x2-36x +27
⇒ f'(x) = 3(4x2-12x+9)
⇒ f'(x) = 3(2x-3)2
Which is always positive for all x ε R.
Since, f'(x) ≥ 0 ∀ x ε R,
Therefore, f(x) is always increasing on R.

1063 Views