Prove that x2 – y2 = c(x2 + y2)2 is the general solution of th

Subject

Mathematics

Class

CBSE Class 12

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 Multiple Choice QuestionsShort Answer Type

21.

The random variable X can take only the values 0, 1, 2, 3. Given that P(X =0) = P(X = 1) = p and P(X = 2) = P(X = 3) such that Σpixi2 = 2Σpixi, find the value of p.

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22.

Often it is taken that a truthful person commands, more respect in the society. A man is known to speak the truth 4 out of 5 times. He throws a die and reports that it is a six.
Find the probability that it is actually a six.
Do you also agree that the value of truthfulness leads to more respect in the society?

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23.

Solve the following L.P.P. graphically :
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0
and x, y ≥ 0

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 Multiple Choice QuestionsLong Answer Type

24.

Use product open square brackets table row 1 cell negative 1 end cell 2 row 0 2 cell negative 3 end cell row 3 cell negative 2 end cell 4 end table close square brackets space space open square brackets table row cell negative 2 end cell 0 1 row 9 2 cell negative 3 end cell row 6 1 cell negative 2 end cell end table close square bracketsto solve the system of equations x + 3z = 9,
–x + 2y – 2z = 4, 2x – 3y + 4z = –3

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25.

Discuss the commutativity and associativity of binary operation '*' defined on A = Q − {1} by the rule a * b = a − b + ab for all, a, b ∊ A. Also find the identity element of * in A and hence find the invertible elements of A. 

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26.

Consider f : R+ → [−5, ∞), given by f(x) = 9x2 + 6x − 5. Show that f is invertible with f−1(y)open parentheses fraction numerator square root of straight y plus 6 end root minus 1 over denominator 3 end fraction close parentheses.

Hence Find
(i) f−1(10)
(ii) y if f−1(y)=43,

where R+ is the set of all non-negative real numbers.

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27.

If the sum of lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum, when the angle between them is π/3.

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28.

Solve the differential equation straight x dy over dx space plus straight y space equals space straight x space cos space straight x space plus space sin space straight x comma space given space that space straight y space equals space 1 space when space straight x space equals space straight x over 2

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29.

Prove that x2 – y2 = c(x2 + y2)2 is the general solution of the differential equation (x3 – 3xy2)dx = (y3 – 3x2y) dy, where C is a parameter.


Given equation is
(x3 – 3xy2)dx = (y3 – 3x2y) dy
rightwards double arrow dy over dx space equals space fraction numerator straight x cubed minus 3 xy squared over denominator straight y cubed minus 3 straight x squared straight y end fraction
which is a homogeneous equation.
Therefore substituting y = vx
We have,
dy over dx space equals space straight v space plus straight x. dv over dx
straight v plus straight x. dv over dx space equals space fraction numerator straight x cubed minus 3 straight x. straight v squared straight x squared over denominator straight v cubed straight x cubed minus 3 straight x squared. vx end fraction
straight v plus straight x. dv over dx space equals space fraction numerator 1 minus 3 straight v squared over denominator straight v cubed minus 3 straight v end fraction
straight x. dv over dx space equals space fraction numerator begin display style 1 minus 3 straight v squared end style over denominator begin display style straight v cubed minus 3 straight v end style end fraction space minus straight V
straight x. dv over dx space equals space fraction numerator begin display style 1 minus 3 straight v squared space minus straight v to the power of 4 space plus 3 straight v squared end style over denominator straight v cubed minus 3 straight v end fraction
fraction numerator straight v cubed minus 3 straight v over denominator 1 minus straight v to the power of 4 end fraction. dv space equals 1 over straight x. dx
Integrating the equation both sides, 
log space straight x space equals space integral fraction numerator straight v cubed minus 3 straight v over denominator 1 minus straight v to the power of 4 end fraction dv
rightwards double arrow space log space straight C apostrophe space plus log space straight x space equals space integral fraction numerator straight v cubed over denominator 1 minus straight v to the power of 4 end fraction. dv minus 3. integral fraction numerator straight v over denominator 1 minus straight v to the power of 4 end fraction. dv
log space straight C apostrophe straight x space equals space straight I subscript 1 minus 3. straight I subscript 2.......... space left parenthesis 1 right parenthesis space where space straight C space is space an space integration space constant
straight I subscript 1 space equals space integral fraction numerator straight v cubed over denominator 1 minus straight v to the power of 4 end fraction. dv
let space 1 minus straight v to the power of 4 space equals space straight t
rightwards double arrow space minus 4 straight v cubed. dv space equals space dt
straight I subscript 1 space equals space integral 1 over straight t. open parentheses fraction numerator dt over denominator negative 4 end fraction close parentheses space equals space minus 1 fourth. log space straight t
straight I subscript 1 space equals space minus 1 fourth space lof space left parenthesis 1 minus straight v to the power of 4 right parenthesis space.... left parenthesis 2 right parenthesis
straight I subscript 2 space equals space integral fraction numerator straight v over denominator 1 minus straight v to the power of 4 end fraction. dv
let space straight v squared space equals space straight t
rightwards double arrow space 2 straight v. dv space equals space dt
straight I subscript 2 space equals space integral fraction numerator 1 over denominator 1 minus straight t squared end fraction. open parentheses dt over 2 close parentheses space equals space 1 fourth. log fraction numerator 1 plus straight t over denominator 1 minus straight t end fraction
straight I subscript 2 space equals space 1 fourth. log space open parentheses fraction numerator 1 plus straight v squared over denominator 1 minus straight v squared end fraction close parentheses
thus substituting the values of I1 and I2, 
log space straight C apostrophe straight x space equals space minus 1 fourth. log space left parenthesis 1 minus straight v to the power of 4 right parenthesis minus 3 over 4. log space open parentheses fraction numerator 1 plus straight v squared over denominator 1 minus straight v squared end fraction close parentheses
log space straight C apostrophe straight x space equals space minus space 1 fourth. space open square brackets log space open parentheses 1 minus straight v to the power of 4 close parentheses minus 3 over 4. log open parentheses fraction numerator 1 plus straight v squared over denominator 1 minus straight v squared end fraction close parentheses cubed close square brackets
log space straight C apostrophe straight x space equals space minus 1 fourth. open square brackets log space open parentheses 1 minus straight v to the power of 4 close parentheses. open parentheses fraction numerator 1 plus straight v squared over denominator 1 minus straight v squared end fraction close parentheses cubed close square brackets
log space straight C apostrophe straight x space equals space minus 1 fourth. space log space open square brackets left parenthesis 1 minus straight v squared right parenthesis left parenthesis 1 plus straight v squared right parenthesis fraction numerator left parenthesis 1 plus straight v squared right parenthesis cubed over denominator left parenthesis 1 minus straight v squared right parenthesis cubed end fraction close square brackets
log space straight C apostrophe straight x space equals space 1 fourth. log space open square brackets fraction numerator left parenthesis 1 plus straight v squared right parenthesis to the power of 4 over denominator left parenthesis negative straight v squared right parenthesis squared end fraction close square brackets
space equals space log space open square brackets fraction numerator left parenthesis 1 plus straight v squared right parenthesis to the power of 4 over denominator left parenthesis 1 minus straight v squared right parenthesis squared end fraction close square brackets to the power of 1 divided by 4 end exponent
space equals space log space open square brackets fraction numerator left parenthesis 1 minus straight v squared right parenthesis to the power of 1 divided by 2 end exponent over denominator left parenthesis 1 plus straight v squared right parenthesis end fraction close square brackets
rightwards double arrow straight C apostrophe straight x space equals space fraction numerator left parenthesis 1 minus straight v squared right parenthesis to the power of 1 divided by 2 end exponent over denominator left parenthesis 1 plus straight v squared right parenthesis end fraction
straight C apostrophe straight x space equals space fraction numerator open parentheses 1 minus begin display style straight y squared over straight x squared end style close parentheses to the power of 1 divided by 2 end exponent over denominator 1 plus begin display style straight y squared over straight x squared end style end fraction
straight C apostrophe straight x space equals space fraction numerator begin display style fraction numerator left parenthesis straight x squared minus straight y squared right parenthesis to the power of 1 divided by 2 end exponent over denominator straight x end fraction end style over denominator left parenthesis straight x squared plus straight y squared right parenthesis divided by straight x squared end fraction space equals space fraction numerator left parenthesis straight x squared minus straight y squared right parenthesis to the power of 1 divided by 2 end exponent over denominator left parenthesis straight x squared plus straight y squared right parenthesis end fraction
straight C apostrophe left parenthesis straight x squared plus straight y squared right parenthesis space equals space left parenthesis straight x squared minus straight y squared right parenthesis
squaring space both space sides space
straight C apostrophe left parenthesis straight x squared plus straight y squared right parenthesis squared space equals space left parenthesis straight x squared minus straight y squared right parenthesis

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30.

If straight a with rightwards arrow on top comma straight b with rightwards arrow on top comma space and space straight c with rightwards arrow on top are mutually perpendicular vectors of equal magnitudes, show that the vector stack straight a space with rightwards arrow on top space plus straight b with rightwards arrow on top space plus straight c with rightwards arrow on top is equally straight a with rightwards arrow on top comma straight b with rightwards arrow on top comma space and space straight c with rightwards arrow on top inclined to Also, find the angle which stack straight a space with rightwards arrow on top space plus straight b with rightwards arrow on top space plus straight c with rightwards arrow on top with straight a with rightwards arrow on top comma space straight b with rightwards arrow on top space or space straight c with rightwards arrow on top.

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