Using mathematical induction prove that \(\frac{d}{dx}\)(xn)=nxn-1 for all positive integers n
To prove:P(n):\(\frac{d}{dx}\)(xn)=nxn-1 for all positive integers n
For n=1
P(1)=\(\frac{d}{dx}\)(x)=1=1.x1-1
∴P(n) is true for n=1
Let P(k) be true for some positive integer k.
That is,P(k):\(\frac{d}{dx}\)(xk)=kxk-1
It has to be proved that P(k+1) is also true.
Consider \(\frac{d}{dx}\)(xk+1)=\(\frac{d}{dx}\)(x.xk)
=xk.\(\frac{d}{dx}\)(x)+x.\(\frac{d}{dx}\)(xk)
=xk.1+x.k.xk-1
=xk+kxk
=(k+1).xk
=(k+1).x(k+1)-1
Thus, P(k+1) is true whenever P(k) is true.
Therefore, by the principle of mathematical induction, the statement P(n) is true for every positive integer n.
Hence, it proved.
Match List-I with List-II
List-I | List-II |
---|---|
(A) \( f(x) = |x| \) | (I) Not differentiable at \( x = -2 \) only |
(B) \( f(x) = |x + 2| \) | (II) Not differentiable at \( x = 0 \) only |
(C) \( f(x) = |x^2 - 4| \) | (III) Not differentiable at \( x = 2 \) only |
(D) \( f(x) = |x - 2| \) | (IV) Not differentiable at \( x = 2, -2 \) only |
Choose the correct answer from the options given below:
Match List-I with List-II
List-I | List-II |
---|---|
(A) \( f(x) = |x| \) | (I) Not differentiable at \( x = -2 \) only |
(B) \( f(x) = |x + 2| \) | (II) Not differentiable at \( x = 0 \) only |
(C) \( f(x) = |x^2 - 4| \) | (III) Not differentiable at \( x = 2 \) only |
(D) \( f(x) = |x - 2| \) | (IV) Not differentiable at \( x = 2, -2 \) only |
Choose the correct answer from the options given below:
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.