If the integral I= displaystyle ∫ (2 x2/4 + x2)dx= 2x-f(x)+c , where f(2)=π , then the minimum value of y=f(x)∀ x∈ [- 2,2] is (where, c is the constant of integration)

Question

If the integral I= displaystyle ∫ (2 x2/4 + x2)dx= 2x-f(x)+c , where f(2)=π , then the minimum value of y=f(x)∀ x∈ [- 2,2] is (where, c is the constant of integration) (A) 0 (B) -π  (C) 2π  (D) -4π

Tardigrade Admin · Accepted Answer

I=2 ∫((x2+4-4/x2+4)) d x =2 ∫(1-(4/x2+4)) d x =2(x-(4/2) tan -1((x/2)))+c =2 x-4 tan -1((x/2))+c ∴ f(x)=4 tan -1((x/2)) So, min (f(x))=4 tan -1(-1)=-π

Q. If the integral $I = \int \frac{2 x ^{2}}{4 + x ^{2}} d x =$ $2 x - f (x) + c$ , where $f (2) = π$ , then the minimum value of $y = f (x) \forall x \in [- 2, 2]$ is (where, $c$ is the constant of integration)

$0$

$- π$

$2 π$

$- 4 π$

$I = 2 \int (\frac{x ^{2} + 4 - 4}{x ^{2} + 4}) d x$
$= 2 \int (1 - \frac{4}{x ^{2} + 4}) d x$
$= 2 (x - \frac{4}{2} tan^{- 1} (\frac{x}{2})) + c$
$= 2 x - 4 tan^{- 1} (\frac{x}{2}) + c$
$∴ f (x) = 4 tan^{- 1} (\frac{x}{2})$
So, $min (f (x)) = 4 tan^{- 1} (- 1) = - π$

Q. If the integral I=∫4+x22x2​dx= 2x−f(x)+c , where f(2)=π , then the minimum value of y=f(x)∀x∈[−2,2] is (where, c is the constant of integration)

0

−π

2π

−4π

I=2∫(x2+4x2+4−4​)dx =2∫(1−x2+44​)dx =2(x−24​tan−1(2x​))+c =2x−4tan−1(2x​)+c ∴f(x)=4tan−1(2x​) So, min(f(x))=4tan−1(−1)=−π

Q. If the integral $I = \int \frac{2 x ^{2}}{4 + x ^{2}} d x =$ $2 x - f (x) + c$ , where $f (2) = π$ , then the minimum value of $y = f (x) \forall x \in [- 2, 2]$ is (where, $c$ is the constant of integration)

$0$

$- π$

$2 π$

$- 4 π$

$I = 2 \int (\frac{x ^{2} + 4 - 4}{x ^{2} + 4}) d x$
$= 2 \int (1 - \frac{4}{x ^{2} + 4}) d x$
$= 2 (x - \frac{4}{2} tan^{- 1} (\frac{x}{2})) + c$
$= 2 x - 4 tan^{- 1} (\frac{x}{2}) + c$
$∴ f (x) = 4 tan^{- 1} (\frac{x}{2})$
So, $min (f (x)) = 4 tan^{- 1} (- 1) = - π$