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Ishan wants to donate a rectangular plot of land for a school in his village. When he was asked to give dimensions of the plot ,he told that if its lenght is decreased by 50m and breadth increased by 50m,then area will remain same but if length is decreased by 10m and breadth decreased by 20m then area will decreased by 5300 metre square. using Matrix method, find dimensions of the plot

Posted by Prashant Jha (May 11, 2017 9:51 p.m.) (Question ID: 5245)

  • Thanks naveen sharma

     

    Posted by Prashant Jha (May 12, 2017 10:53 a.m.)
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  • Answers:
  • Ans. Let the length of Plot = x m

    Breadth of plot = y m

    Area of Plot = xy

    According to First condition, 

    (x-50) (y+50) = xy 

    => xy + 50x -50y - 2500 = xy 

    => 50x - 50y = 2500

    => x- y = 50 ........ (1)

    According to 2nd condition, 

    (x-10) (y-20) = xy - 5300

    => xy - 20x - 10y + 200 = xy -  5300

    => -20x - 10y = -5500

    => 20x + 10y = 5500

    => 2x + y = 550 ........(2) 

    We can represent these two equation using Matrix 

    A = {tex}\begin{bmatrix} 1 & -1 \\ 2& 1 \end{bmatrix}{/tex},  |A| = 1+2 = 3

    X = {tex}\begin{bmatrix} x \\ y \end{bmatrix}{/tex}

    B = {tex}\begin{bmatrix} 50 \\ 550 \end{bmatrix}{/tex}

    Such that AX = B

    X = A-1B
    A-1 {tex}Adjoint \space of \space A\over Determinant \space of \space A {/tex}

    => X= {tex}{1\over 3} \begin{bmatrix} 1 & 1 \\ -2& 1 \end{bmatrix}\begin{bmatrix} 50 \\ 550 \end{bmatrix}{/tex}

    => X = {tex}{1\over 3}\begin{bmatrix} 600 \\ 450 \end{bmatrix}{/tex}

    =>X =  {tex}\begin{bmatrix} 200 \\ 150 \end{bmatrix}{/tex}

    x = 200

    y = 150 

    So Length of Plot = 200 m

    Breadth of Plot = 150 m

    Answered by Naveen Sharma (May 12, 2017 9:36 a.m.)
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Differtiate w.r to x    Sin2(2x+1)

Posted by Ajhar Sultan (May 05, 2017 11:25 p.m.) (Question ID: 5131)

  • Answers:
  • Ans. {tex}{d\over dx}(sin^2(2x+1)){/tex}

    Using chain rule, we get

    ={tex}cos^2(2x+1).{d\over dx}(2x+1){/tex}

    {tex}= cos^2(2x+1).2{/tex}

    {tex}=2. cos^2(2x+1){/tex}

     

     

    Answered by Payal Singh (May 06, 2017 7:27 a.m.)
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If X≠Y≠Z AND |x x3 x4-1

y y3 y4-1

z z3 z4-1|then prove that XYZ(XYZ(XY+YZ+ ZX)=X+Y+Z

Posted by Gaurav Gogoi (May 04, 2017 8:58 p.m.) (Question ID: 5119)

 What is the value of the arcsin(sin3)

 

Posted by Harshit Mishra (Apr 27, 2017 10:01 p.m.) (Question ID: 5047)

  • Answers:
  • Ans. 

    {tex}arcsin(sin3) =  \pi - 3{/tex}

    Explanation:

     arcsin maps [-1,1] to {tex}\left [-{\pi\over 2}, {\pi\over 2}\right ]{/tex}.

    We need to get an angle {tex}x \in \left [-{\pi\over 2}, {\pi\over 2}\right ]{/tex}where sin(x) = sin(3).

    Since sin x is symmetric about {tex}x = {\pi\over 2}{/tex}, we have {tex}sin (3) = sin (\pi -3){/tex}{tex}\pi - 3{/tex} is in {tex}\left [-{\pi\over 2}, {\pi\over 2}\right ]{/tex}

    {tex}\therefore we \space have {/tex} {tex}arcsin(sin3) =  \pi - 3{/tex}

    Answered by Naveen Sharma (Apr 28, 2017 11:36 a.m.)
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ltx-0 [cosx]

Posted by ASHUTOSH DAS (Apr 20, 2017 9:36 a.m.) (Question ID: 4962)

ltx-0 (tanx-x)/x2tanx

Posted by ASHUTOSH DAS (Apr 20, 2017 9:36 a.m.) (Question ID: 4961)

What is the inverse of x-[x] ?

[x] is greatest integer function. 

Posted by Akanksha Tomar (Apr 19, 2017 1:02 a.m.) (Question ID: 4945)

  • Answers:
  • Ans. For inverse of a function to be defined it should be bijective function.

     

    f(x) = x−n when x∈(n,n+1) and f(x)=0 when x = n

    => f(x) is not one-one function but is many to one function.

    => f(x) is not bijective function

    So f-1(x) is not defined.

    Answered by Naveen Sharma (Apr 19, 2017 10:45 a.m.)
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Can phi be included in any relation R from A to B...??

Posted by Achhaya Pathak (Apr 01, 2017 4:09 p.m.) (Question ID: 4651)

Sin y =x sin (a+y)

Prove that :- dy/dx= sin x/1-2xcosx+ x^2

 

Posted by Mohana Paul (Apr 01, 2017 8:02 a.m.) (Question ID: 4642)

Sin-1(tan5π/4)

 

Posted by Mohana Paul (Mar 30, 2017 7:05 a.m.) (Question ID: 4606)

  • Answers:
  • {tex}\tan{5\pi\over4}{/tex} = {tex}\tan({\pi + {\pi\over4}}){/tex} = {tex}\tan{\pi\over4}{/tex}

    {tex}\sin^{-1}({\tan{\pi\over4}}){/tex} = {tex}\sin^{-1}({1}){/tex} = {tex}\pi\over2{/tex}

     

    Answered by Lokendra Soni (Apr 08, 2017 8:27 a.m.)
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If f (x) = 3x^2 -9x+7 for any square matrix A then write f (A)

 

Posted by Madhu mitha (Mar 29, 2017 11:27 a.m.) (Question ID: 4582)

  • Answers:
  • {tex}f\left( A \right) = 3{A^2} - 9A + 7I{/tex}

    Answered by Neeraj Sharma (Apr 10, 2017 10:18 p.m.)
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Find the principal value of sin-1(√3-1/2√2)

Posted by Kadam cool (Mar 26, 2017 12:51 p.m.) (Question ID: 4520)

tan-1 x-3/x-4.  +.   tan-1 x+3/x+4. =.  π/4

Posted by Vishaal Velu (Mar 21, 2017 1:02 p.m.) (Question ID: 4324)

  • Answers:
  • I think it is x-2 and x +2 instead of x-3 and x +3 respectively

    Ans.

    {tex}tan^{-1} ({x-2\over x-4})+ tan^{-1}( {x+2\over x+4}) = {\pi \over 4}{/tex}

    Using{tex}tan^{-1} x + tan^{-1}y = tan^{-1} ({x+y\over 1-xy}){/tex}

    => {tex}tan^{-1} ({{x-2\over x-4} + {x+2\over x+4}\over 1-( {x-2\over x-4} \times {x+2\over x+4})}) = {\pi \over 4}{/tex} 

    => {tex}tan^{-1}({2x^2-16\over -12}) = {\pi \over 4}{/tex}

    {tex}=> {2x^2-16\over -12}= tan^{-1} {\pi \over 4}{/tex}

    {tex}=> {2x^2-16\over -12}= 1{/tex}

    {tex}=> 2x^2-16= -12{/tex}

    {tex}=> 2x^2= 4 => x^2 = 2{/tex}

    {tex}x = \sqrt 2{/tex}

    if you use 3 answer ll be {tex}x = \sqrt {17\over 2}{/tex}

    Answered by Naveen Sharma (Mar 21, 2017 7:07 p.m.)
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Find the matrix A such that    

 Matrix of ( 2. -1.                  (-1.   -8

                    1.  0.   * A=.         1.    -2

                     -3. 4)                    9.    22)

                                       

Posted by Vishaal Velu (Mar 21, 2017 12:56 p.m.) (Question ID: 4323)

There are 4 cards numbered 1,3,5and 7 one number on one card . Two cards are drawn at random! Without replacement . Let x denote the sum of the numbers on the two cards .Find the mean and variance of x

Posted by Vishaal Velu (Mar 21, 2017 12:49 p.m.) (Question ID: 4322)

Find the value of x such that the points are coplanar A(3,2,1), B(4,x,5),. C(4,2,-2) and D(6,5,-1)

Posted by Vishaal Velu (Mar 21, 2017 12:47 p.m.) (Question ID: 4321)

  • Answers:
  • Ans.

    Points are A(3,2,1), B(4,x,5), C(4,2,-2) And D(6,5,-1).

    Position Vector of A = {tex}3\hat i + 2\hat j + \hat k{/tex}

    Position Vector of B = {tex}4\hat i + x\hat j + 5\hat k{/tex}

    Position Vector of C = {tex}4\hat i + 2\hat j -2\hat k{/tex}

    Position Vector of D = {tex}6\hat i + 5\hat j -\hat k{/tex}

    {tex}=> \overrightarrow {AB} = \hat i + (2-x)\hat j + 4\hat k{/tex}

    {tex}=> \overrightarrow {AC} = \hat i + 0\hat j -3\hat k{/tex} 

    {tex}=> \overrightarrow {AD} = 3\hat i + 3\hat j -2\hat k{/tex}

    As points are coplaner, then triple scaler product is zero,

    So,

    {tex}\begin{bmatrix} 1& (x-2) & 4 \\[0.3em] 1 & 0 & -3 \\[0.3em] 3 & 3 & -2 \end{bmatrix} = 0{/tex}

    Expanding along R2

    => -1[-2(x-2) -12] -(-3)[3-3(x-2)] = 0

    => 2x -4+12 +9-9x +18 = 0

    => -7x +35 = 0

    => 7x = 35

    => x = 5

     

    Answered by Naveen Sharma (Mar 21, 2017 7:40 p.m.)
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Using the method of integration find the area of the triangle ABC , coordinates of whose vertices are A(4,1). B(6,6),.  C(8,4)

 

 

Posted by Vishaal Velu (Mar 21, 2017 12:45 p.m.) (Question ID: 4320)

Find the coordinates of the point where the line through the points (3,-4,-5)and (2,-3,1),crosses the plane determined by the points (1,2,3),(4,2,-3)and (0,4,3).

Posted by Vishaal Velu (Mar 21, 2017 12:44 p.m.) (Question ID: 4319)

  • Answers:
  • (1,-2,7)

    Answered by Pratit Luthra (Mar 22, 2017 1:08 p.m.)
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Will teachers will do linent marking or strict marking in mathematics

Posted by Singh Gajendra (Mar 21, 2017 9:41 a.m.) (Question ID: 4308)

What is the minimum passing marks required for mathematics in class12?

 

Posted by Vinayak Parihar (Mar 21, 2017 12:12 a.m.) (Question ID: 4294)

  • Answers:
  • 33% marks are required to pass any subject.

    Answered by Naveen Sharma (Mar 21, 2017 7:56 a.m.)
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find the area enclosed between the parabola 4y= 3x2 and the straight line 3x -2y+12=0

Posted by mayank pradhan (Mar 20, 2017 6:32 p.m.) (Question ID: 4222)

how will be the evalution of maths examination conducted today be?

Posted by sandhra francis (Mar 20, 2017 5:36 p.m.) (Question ID: 4218)

integration of sin inverse x divide by x

 

Posted by Shikhar Gupta (Mar 20, 2017 9:38 a.m.) (Question ID: 4180)

For what value of k ,the system of linear equatoin 

            X+y+z=2

           2x+y-z=3

         3x+2y+kz=4

Posted by rabi heishnam (Mar 20, 2017 4:18 a.m.) (Question ID: 4161)

Important. Question given by u r from ncert only?

 

 

Posted by Gurjot Singh (Mar 20, 2017 12:01 a.m.) (Question ID: 4158)

  • Are most of the questions given tomorrow in the exam are from ncert examples?

    Posted by Sreeja Samanta (Mar 20, 2017 12:13 a.m.)
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  • Answers:
  • Yeah......it is for sure.but dont depend on it fullt.practice other problems also.

    Answered by anjana krishnan (Mar 20, 2017 7:37 a.m.)
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Evaluate: Sin a/ sin( x+a) 

Posted by Deepali Kar (Mar 19, 2017 8:54 p.m.) (Question ID: 4151)

Find the shortest distance between the line x-y+1=0 and thr curve y2= x

Posted by Gaurav Pandey (Mar 19, 2017 6:49 p.m.) (Question ID: 4142)

  • pls teachers answer it quickly i really need ur help 

    Posted by Gaurav Pandey (Mar 19, 2017 7:15 p.m.)
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range of |sinx|

Posted by vishal singla (Mar 19, 2017 5:59 p.m.) (Question ID: 4139)

  • Answers:
  • [0,1]

    Answered by Pratit Luthra (Mar 19, 2017 8:05 p.m.)
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A signal which can be green or red with probability 4/5 and 1/5 respectivEly, is received by station A and then transmitted to station B. The probability of each receiving the signal correctly is 3/4. If the signal received at station B is green, then find the probability that original signal is green.

 

Posted by Shivang Puri (Mar 19, 2017 12:54 p.m.) (Question ID: 4110)

3 no.s are selected at random from first 6 +ve integers. x denotes the smallest of 3 no.s selected . find the probabiloty distribution of x?

Posted by Shivam .G (Mar 19, 2017 12:26 p.m.) (Question ID: 4101)

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