Q98 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
If $f(x), g(x)$ be twice differentiable functions, satisfying $f''(x) = g''(x), f'(1) = 2g'(1) = 4$ and $f(2) = 3g(2) = 9$ then $f(x) - g(x)$ at $x = 4$ is equal to
- A$0$
- Correct answerB$10$
- C$8$
- D$2$
Solution
Let $h(x) = f(x) - g(x)$.
Given $f''(x) = g''(x)$, we have $h''(x) = 0$.
Integrating with respect to $x$, we get $h'(x) = c_1$.
We are given $f'(1) = 4$ and $2g'(1) = 4 \Rightarrow g'(1) = 2$.
Thus, $h'(1) = f'(1) - g'(1) = 4 - 2 = 2$.
Since $h'(x)$ is constant, $h'(x) = 2$.
Integrating again with respect to $x$, we get $h(x) = 2x + c_2$.
We are given $f(2) = 9$ and $3g(2) = 9 \Rightarrow g(2) = 3$.
Thus, $h(2) = f(2) - g(2) = 9 - 3 = 6$.
Substituting $x = 2$ into $h(x) = 2x + c_2$, we get $4 + c_2 = 6 \Rightarrow c_2 = 2$.
Therefore, $h(x) = 2x + 2$.
At $x = 4$, $h(4) = 2(4) + 2 = 10$.
Answer: $10$
Given $f''(x) = g''(x)$, we have $h''(x) = 0$.
Integrating with respect to $x$, we get $h'(x) = c_1$.
We are given $f'(1) = 4$ and $2g'(1) = 4 \Rightarrow g'(1) = 2$.
Thus, $h'(1) = f'(1) - g'(1) = 4 - 2 = 2$.
Since $h'(x)$ is constant, $h'(x) = 2$.
Integrating again with respect to $x$, we get $h(x) = 2x + c_2$.
We are given $f(2) = 9$ and $3g(2) = 9 \Rightarrow g(2) = 3$.
Thus, $h(2) = f(2) - g(2) = 9 - 3 = 6$.
Substituting $x = 2$ into $h(x) = 2x + c_2$, we get $4 + c_2 = 6 \Rightarrow c_2 = 2$.
Therefore, $h(x) = 2x + 2$.
At $x = 4$, $h(4) = 2(4) + 2 = 10$.
Answer: $10$
Hence, the correct answer is option: 'B'.