Question

Let \[f,g:[ - 1,2] \to R\] be continuous functions which are twice differentiable on the interval \[( - 1,2)\]. Let the values of f and g at the points \[ - 1,0\] and \[2\] be as given in the following table:
\[f(x) = \left\{
  3;x = - 1 \\
  6;x = 0 \\
  0;x = 2 \\
  \right.\]and \[g(x) = \left\{
  0;x = - 1 \\
  1;x = 0 \\
  0;x = - 1 \\
  \right.\]
In each of the intervals \[( - 1,0)\] and \[(0,2)\], the function \[(f - 3g)''\]never vanishes. Then the correct statement(s) is (are) THIS QUESTION HAS MULTIPLE CORRECT OPTIONS
A.\[f'(x) - 3g'(x) = 0\] has exactly three solutions in \[( - 1,0) \cup (0,2)\]
B.\[f'(x) - 3g'(x) = 0\] has exactly one solution in \[( - 1,0)\]
C.\[f'(x) - 3g'(x) = 0\] has exactly one solution in\[(0,2)\]
D.\[f'(x) - 3g'(x) = 0\] has exactly two solutions in \[( - 1,0)\] and exactly two solutions in \[(0,2)\]

Answer 1

Hint: First form the equation as per the given information in the question. And then apply Rolle’s theorem to calculate their differentiation, as Rolle's theorem states that if a function f is continuous on the closed interval \[[a,b]\] and differentiable on the open interval \[(a,b)\] such that \[f(a) = f(b)\], then \[f'(x) = 0\] for some x with \[a \leqslant x \leqslant b\]. Hence, after differentiation of the above formed term we can reach the solution required.

Complete step-by-step answer:
As the given conditions are of \[f,g:[ - 1,2] \to R\] be continuous functions which are twice differentiable on the interval \[( - 1,2)\].
And the values of f and g at the points \[ - 1,0\] and \[2\] be as given in the following table:
\[f(x) = \left\{
  3;x = - 1 \\
  6;x = 0 \\
  0;x = 2 \\
  \right.\]and \[g(x) = \left\{
  0;x = - 1 \\
  1;x = 0 \\
  0;x = - 1 \\
  \right.\]
Now, let the new equation be
\[h(x) = f(x) - 3g(x)\]
Now calculate the value of \[h(x)\]at various points \[ - 1,0\] and \[2\]
\[
  h( - 1) = f( - 1) - 3g( - 1) \\
   = 3 - 0 \\
   = 3 \\
 \]
Similarly calculate for remaining point as,
\[
  h(0) = f(0) - 3g(0) \\
   = 6 - 3(1) \\
   = 3 \\
 \]
And,
\[
  h(2) = f(2) - 3g(2) \\
   = 0 - 3( - 1) \\
   = 3 \\
 \]
Hence, now calculate the differentiation of \[h'(x)\], so it can be given as
Let us calculate the value of \[h'(x)\]applying Rolle’s for interval \[( - 1,0)\].
As the function satisfies all the Rolle’s condition so it can be given as,
\[h'(x) = 0\]
Thus, has exactly one root in the interval \[( - 1,0)\]
So, \[f'(x) - 3g'(x) = 0\] has exactly one solution in \[( - 1,0)\]
Let us calculate the value of \[h'(x)\]applying Rolle’s for interval\[(0,2)\].
As the function satisfies all the Rolle’s condition so it can be given as,
\[h'(x) = 0\]
Thus, has exactly one root in the interval \[(0,2)\]
So, \[f'(x) - 3g'(x) = 0\] has exactly one solution in \[(0,2)\]
Hence, option (B) and option (C) require the correct answer.

Note: Rolle's theorem, in analysis, special case of the mean-value theorem of differential calculus. Rolle's theorem states that if a function f is continuous on the closed interval \[[a,b]\] and differentiable on the open interval \[(a,b)\] such that \[f(a) = f(b)\], then \[f'(x) = 0\] for some x with \[a \leqslant x \leqslant b\].
Calculate the function \[h\left( x \right)\]properly and make use of it’s derivative without any mistake, also use the value of given functions at stated points properly.