JEE MAIN - Mathematics (2023 - 24th January Evening Shift - No. 13)
If $$f(x) = {x^3} - {x^2}f'(1) + xf''(2) - f'''(3),x \in \mathbb{R}$$, then
$$2f(0) - f(1) + f(3) = f(2)$$
$$f(1) + f(2) + f(3) = f(0)$$
$$f(3) - f(2) = f(1)$$
$$3f(1) + f(2) = f(3)$$
Explanation
$$
f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R
$$
Let $\mathrm{f}^{\prime}(1)=\mathrm{a}, \mathrm{f}^{\prime \prime}(2)=\mathrm{b}, \mathrm{f}^{\prime \prime \prime}(3)=\mathrm{c}$
$$ \begin{aligned} & f(x)=x^3-a x^2+b x-c \\\\ & f^{\prime}(x)=3 x^2-2 a x+b \\\\ & f^{\prime \prime}(x)=6 x-2 a \\\\ & f^{\prime \prime \prime}(x)=6 \\\\ & c=6, a=3, b=6 \\\\ & f(x)=x^3-3 x^2+6 x-6 \\\\ & f(1)=-2, f(2)=2, f(3)=12, f(0)=-6 \\\\ & 2 f(0)-f(1)+f(3)=2=f(2) \end{aligned} $$
Let $\mathrm{f}^{\prime}(1)=\mathrm{a}, \mathrm{f}^{\prime \prime}(2)=\mathrm{b}, \mathrm{f}^{\prime \prime \prime}(3)=\mathrm{c}$
$$ \begin{aligned} & f(x)=x^3-a x^2+b x-c \\\\ & f^{\prime}(x)=3 x^2-2 a x+b \\\\ & f^{\prime \prime}(x)=6 x-2 a \\\\ & f^{\prime \prime \prime}(x)=6 \\\\ & c=6, a=3, b=6 \\\\ & f(x)=x^3-3 x^2+6 x-6 \\\\ & f(1)=-2, f(2)=2, f(3)=12, f(0)=-6 \\\\ & 2 f(0)-f(1)+f(3)=2=f(2) \end{aligned} $$
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