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STD 12 - 9. differential equations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 9. differential equations4 Marks
MCQ
A function $y = f(x)$ has a second order derivatives $f''(x) = 6(x - 1)$. If its graph passes through the point $(2, 1)$ and at that point the tangent to the graph is $y = 3x - 5$, then the function is
  • A
    ${(x + 1)^3}$
  • ✓
    ${(x - 1)^3}$
  • C
    ${(x + 1)^2}$
  • D
    ${(x - 1)^2}$
✓

Answer

Correct option: B.
${(x - 1)^3}$
b
(b) Given $f''(x) = 6(x - 1)$

$f'(x) = 3{(x - 1)^2} + {c_1}$…..$(i)$

But at point $(2, 1)$ the line $y = 3x - 5$ is tangent to the graph $y = f(x)$. Hence

${\left. {\frac{{dy}}{{dx}}} \right|_{x = 2}} = 3$ or $f'(2) = 3$.

Then from $(i)$ $f'(2) = 3{(2 - 1)^2} + {c_1}$

$3 = 3 + {c_1}$ ==> ${c_1} = 0$ i.e., $f'(x) = 3{(x - 1)^2}$

Given $f(2) = 1$

$f(x) = {(x - 1)^3} + {c_2}$ ==> $f(2) = 1 + {c_2}$

==> $1 = 1 + {c_2}$ ==> ${c_2} = 0$

Hence $f(x) = {(x - 1)^3}$.

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