Let f and g be twice differentiable functions on R such that f''(x) = g''(x) + 6x, f'(1) = 4g'(1) - 3 = 9, and f(2) = 3g(2) = 12. Which of the following is NOT true?

Let f and g be twice differentiable functions on R such that f''(… — Solution | JEE Mathematics

MCQMediumJEE 2023Linear Differential Equations

Let $f$ and $g$ be twice differentiable functions on $\mathbb{R}$ such that $f''(x) = g''(x) + 6x$ , $f'(1) = 4g'(1) - 3 = 9$ , and $f(2) = 3g(2) = 12$ . Which of the following is NOT true?

A
$g(-2) - f(-2) = 20$
B
If $-1 < x < 2$ , then $|f(x) - g(x)| < 8$
C
$|f'(x) - g'(x)| < 6, -1 < x < 1$
D
There exists $x \in [1, 3/2]$ such that $f(x_1) = g(x_1)$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $f''(x) = g''(x) + 6x$ , $f'(1) = 4g'(1) - 3 = 9$ , and $f(2) = 3g(2) = 12$ .

Find: Which statement is not true.

From the given second derivative relation,

f''(x) = g''(x) + 6x

integration gives

f'(x) = g'(x) + 3x^2 + C_1

Using the given conditions

Now use $f'(1) = 9$ . From $f'(1) = 4g'(1) - 3 = 9$ , we get

4g'(1) = 12

g'(1) = 3

Substituting $x=1$ in

f'(x) = g'(x) + 3x^2 + C_1

gives

9 = 3 + 3 + C_1

Hence,

C_1 = 3

Therefore,

f'(x) - g'(x) = 3x^2 + 3

Compare through the difference function

Let

h(x) = f(x) - g(x)

Then

h''(x) = 6x

Integrating twice,

h'(x) = 3x^2 + C_1

and

h(x) = x^3 + C_1x + C_2

Since $f'(1)=9$ and $4g'(1)-3=9$ , we get $g'(1)=3$ , so

h'(1) = f'(1)-g'(1) = 9-3 = 6

Thus

3 + C_1 = 6

which gives

C_1 = 3

Also, from $f(2)=12$ and $3g(2)=12$ , we get $g(2)=4$ , hence

h(2) = f(2)-g(2) = 12-4 = 8

8 + 6 + C_2 = 8

which gives

C_2 = -6

Therefore,

h(x) = x^3 + 3x - 6

Now check the options using this expression. For option A,

g(-2) - f(-2) = -h(-2) = -(-8-6-6) = 20

So option A is true. Hence the solution's stated answer conflicts with the working, and the listed answer is discrepant.

Common mistakes

Using the incorrect condition $f'(1)=9$ alone and ignoring that $4g'(1)-3=9$ also determines $g'(1)$ . This gives a wrong constant of integration. First compute $g'(1)=3$ , then use $f'(1)-g'(1)$ .
Integrating $f''(x)-g''(x)=6x$ incorrectly. The integral of $6x$ is $3x^2$ , and after the second integration it becomes $x^3$ . Missing these constants changes every option check.
Misreading $f(2)=3g(2)=12$ as $f(2)=g(2)=12$ . The correct interpretation is $f(2)=12$ and $g(2)=4$ . This is essential for finding the second constant.

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Step-by-step solution

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