For non-zero x- let 3fx-4f 1x -1x-10- If - 23fxdx-abln ca- where a- b- c are co-prime- then the value of a - b - c is

3f(x)+4f ( 1x )=1x 10 ⋯(1) Replace x by 1x, we get 3f ( 1x )+4f(x)=x 10 ⋯(2) Solving (1) and (2), we get f(x)=17(4x 3x 10) Now, nbsp;∫ _2^317(4x 3x 10)dx ...

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Byju's Answer

Standard XII

Mathematics

Multiplication of Matrices

For non-zero ...

Question

For non-zero $x,$ let $3 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} - 10.$ If $3 \int 2 f (x) d x = \frac{a}{b} ln (\frac{c}{a}),$ where $a, b, c$ are co-prime, then the value of $(a + b + c)$ is

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Solution

$3 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} - 10 \dots (1)$
Replace $x$ by $\frac{1}{x},$ we get
$3 f (\frac{1}{x}) + 4 f (x) = x - 10 \dots (2)$
Solving $(1)$ and $(2),$ we get
$f (x) = \frac{1}{7} (4 x - \frac{3}{x} - 10)$

Now, $3 \int 2 \frac{1}{7} (4 x - \frac{3}{x} - 10) d x$
$= \frac{1}{7} {[2 x^{2} - 3 ln x - 10 x]}_{2}^{2}$
$= \frac{3}{7} ln (\frac{2}{3})$
$∴ a = 3, b = 7, c = 2$
Hence, $a + b + c = 12$

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For non-zero x, let 3f(x)+4f(1x)=1x−10. If 3∫2f(x)dx=abln(ca), where a,b,c are co-prime, then the value of (a+b+c) is

3f(x)+4f(1x)=1x−10 ⋯(1) Replace x by 1x, we get 3f(1x)+4f(x)=x−10 ⋯(2) Solving (1) and (2), we get f(x)=17(4x−3x−10) Now, 3∫217(4x−3x−10)dx =17[2x2−3lnx−10x]32 =37ln(23) ∴a=3,b=7,c=2 Hence, a+b+c=12

For non-zero $x,$ let $3 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} - 10.$ If $3 \int 2 f (x) d x = \frac{a}{b} ln (\frac{c}{a}),$ where $a, b, c$ are co-prime, then the value of $(a + b + c)$ is