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.
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?
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
Use product to solve the system of equations x + 3z = 9,
–x + 2y – 2z = 4, 2x – 3y + 4z = –3
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.
Consider f : R+ → [−5, ∞), given by f(x) = 9x2 + 6x − 5. Show that f is invertible with f−1(y).
Hence Find
(i) f−1(10)
(ii) y if f−1(y)=43,
where R+ is the set of all non-negative real numbers.
f : R+ → [−5, ∞) given by f(x) = 9x2+ 6x − 5
To show: f is one-one and onto.
Let us assume that f is not one-one.
Therefore there exist two or more numbers for which images are same.
For x1, x2∈ R+ and x1≠ x2
Let f(x1)=f(x2)
⇒9x12+6x1−5=9x22+6x2−5
⇒9x12+6x1=9x22+6x2
⇒9x12−9x22+6x1−6x2=0
⇒9(x12−x22)+6(x1−x2)=0
⇒(x1−x2)[9(x1+x2)+6]=0
Since x1 and x2 are positive,
9(x1+x2)+6>0
∴x1−x2=0
⇒x1=x2
Therefore, it contradicts our assumption.
Hence the function f is one-one.
Now, let is prove that f is onto.
A function f : X → Y is onto if for every y ∈ Y, there exist a pre-image in X.
f(x)=9x2+6x−5
=9x2+6x+1−6
=(3x+1)2−6
Now, for all x∈R+ or [0,∞), f(x)∈[−5, ∞).
∴ Range = co-domain.
Hence, f is onto.
Therefore, function f is invertible.
Now, let y = 9x2 + 6x − 5
9x2+6x−5−y=0
or
9x2+6x−(5+y)=0 where x∈R+
As x∈R+ i.e., is a positive real number
x cannot be equal to
Since f: R+ →[-5,∞)
so y ∈ [-5,∞)
i.e y is greater than or equal to -5
i.e. y ≥-5
y+5 ≥0
⇒ Hence the value inside root is positive
Hence √y +6≥0
⇒ x≥0
Hence x is a real number which is greater than or equal to 0.
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.
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.
If are mutually perpendicular vectors of equal magnitudes, show that the vector is equally inclined to Also, find the angle which with .