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Question: If \( f\left( x \right) \) is a twice differentiable function for which \( f\left( 1 \right) = 1 \) ...

If f(x)f\left( x \right) is a twice differentiable function for which f(1)=1f\left( 1 \right) = 1 , f(2)=4f\left( 2 \right) = 4 and then f(3)=9f\left( 3 \right) = 9
(a) f”(x)=2 for all x∈(1,3)\left( a \right){\text{ f''}}\left( x \right) = 2{\text{ for all x}} \in \left( {1,3} \right)
(b) f”(x)=f′(x)=5 for some x∈(2,3)\left( b \right){\text{ f''}}\left( x \right) = f'\left( x \right) = 5{\text{ for some x}} \in \left( {2,3} \right)
(c) f”(x)=3 for all x∈(2,3)\left( c \right){\text{ f''}}\left( x \right) = 3{\text{ for all x}} \in \left( {2,3} \right)
(d) f”(x)=2 for some x∈(1,3)\left( d \right){\text{ f''}}\left( x \right) = 2{\text{ for some x}} \in \left( {1,3} \right)

Explanation

Solution

Here, in this question, we will use the concept of Lagrange’s mean value theorem which is also known as MVT and by using this we will solve this question by substituting the values and checking it. What's more, in this way we can unravel this problem.

Complete step by step solution:
Here we will first apply the MVT in the interval of [1,2]\left[ {1,2} \right] , we get
So now taking out f′(a)=f(2)−f(1)2−1f'\left( a \right) = \dfrac{{f\left( 2 \right) - f\left( 1 \right)}}{{2 - 1}}
And on putting the values, we get
⇒f′(a)=4−12−1\Rightarrow f'\left( a \right) = \dfrac{{4 - 1}}{{2 - 1}}
On solving the above equation, we get
⇒f′(a)=3\Rightarrow f'\left( a \right) = 3 , let’s name it equation 11
So we can say 1<a<21 < a < 2
Now for the interval of [2,3]\left[ {2,3} \right] , we get
So now taking out f′(b)=f(3)−f(2)3−2f'\left( b \right) = \dfrac{{f\left( 3 \right) - f\left( 2 \right)}}{{3 - 2}}
And on putting the values, we get
⇒f′(b)=9−43−2\Rightarrow f'\left( b \right) = \dfrac{{9 - 4}}{{3 - 2}}
On solving the above equation, we get
⇒f′(b)=5\Rightarrow f'\left( b \right) = 5 , let’s name it equation 22
So we can say 2<b<32 < b < 3
Since it is given that f(x)f\left( x \right) is a twice differentiable function, therefore, we can apply LMVT on the first derivative of f(x)f\left( x \right) .
Hence for the interval of [a,b]\left[ {a,b} \right] , we get
⇒f′′(c)=f′(b)−f′(a)b−a\Rightarrow f''\left( c \right) = \dfrac{{f'\left( b \right) - f'\left( a \right)}}{{b - a}}
Now on substituting the values we had obtained in the above equation, we get
⇒f′′(c)=5−3b−a\Rightarrow f''\left( c \right) = \dfrac{{5 - 3}}{{b - a}}
And on solving furthermore, we get
⇒f′′(c)=2b−a\Rightarrow f''\left( c \right) = \dfrac{2}{{b - a}}
On considering the difference between the bb and aa , we get
⇒f′′(c)=2\Rightarrow f''\left( c \right) = 2
And since 1<a<21 < a < 2 and 2<b<32 < b < 3
Therefore, we can say that 1<c<31 < c < 3
Or it can also be written as c∈(1,3)c \in \left( {1,3} \right)

Therefore, the option (d)\left( d \right) is correct.

Note:
There is more way to solve this question, it can be solved by calculating g(x)=f(x)−x2g\left( x \right) = f\left( x \right) - {x^2} at each of the points. And then on taking the derivative of all the terms we will get at least three roots g(x)=f(x)−x2g\left( x \right) = f\left( x \right) - {x^2} . The rest derivative has two and one roots. So through this, we can conclude c∈(1,3)c \in \left( {1,3} \right) .