If $ x^2 + y^2 = 1$, then
- $ yy" - 2 (y' )^2 + 1 = 0 $
- $ yy" + (y' )^2 + 1 = 0 $
- $ yy" + (y' )^2 - 1 = 0 $
- $ yy" + 2 (y' )^2 + 1 = 0 $
The Correct Option is B
Solution and Explanation
Given that;
\(x^2+y^2=1\)
Now differentiate w.r.t ‘x’
\(\therefore 2x+2y\frac{dy}{dx}=0\)
again differentiate w.r.t ‘x’
\(\therefore 2+2(y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2)=0\)
⇒\(1+yf''(x)+(y')^2=0\)
⇒\((y')^2+yy''+1=0\)
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.