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If $f(x)$ is a real valued function such that $2f(x)+3f(-x)=15-4x$, for every $x \in \mathbb{R}$, then $f(2)$ is

  1. $-15$
  2. $22$
  3. $11$
  4. $0$
in Calculus by Veteran (431k points)
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1 Answer

+6 votes

Answer: $\mathbf C$

Solution:

Given: $\mathrm {2f(x) + 3f(-x) = 15 - 4x }\tag{1}$

Substitute $\color {red} {\mathrm x}$ with $\color {red} {\mathrm {-x}}$, in the above question, we get:

$\mathrm {2f(-x) + 3 f(x) = 15 +4 x} \tag {2}$

On multiplying, Equation $(1)$ $\times$ $3$ and Equation $(2)$ $\times$ $2$, we get:

$\mathrm {6f(x) + 9f(-x) = 45-12x} \tag {3}$

and,

$\mathrm {4f(-x) + 6f(x) = 30 + 8x} \Rightarrow \mathrm {6f(x) + 4f(-x) = 30+8x} \tag{4}$



Subtracting equation $(3)$ and $(4)$,we get:



$\mathrm {5f(-x) = 15-20x} $



Now, substitute, $\mathrm x = -2$, we get:



$\mathrm {5f(2) = [15 - 20\times (-2)]}$

 

$ \Rightarrow \mathrm{5f(2)} = 55$

 

$\Rightarrow \mathrm{f(2) = \bf{11}}$



$\therefore \mathbf C$ is the correct option.

by Boss (18.9k points)
edited by

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