JEE Mathematics 2024 Question with Solution

Given: $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3)$

Find: $f'(10)$

Differentiate the given function:

Differentiate again:

And the third derivative is:

Now use the relations shown in the solution:

Substitute these values into the expression for $f(x)$:

Differentiate this simplified form:

Now evaluate at $x = 10$:

Therefore, the correct option is A and $f'(10) = 202$.

Given: $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3)$

Find: $f'(10)$

From the given form,

so at $x=1$,

Using $f''(2)=2$ from the solution,

which gives

Also,

so at $x=2$,

Using $f'(1)=-5$,

This is consistent with the value used above.

Now compute

Hence, $f'(10) = 202$, so the correct option is A.