The value of the integral Ι=∫ ex(sin x + cos ⁡ x)dx is equal to ex⋅ f(x)+C, C being the constant of integration. Then the maximum value of y=f(x2),∀ x∈ R is equal to

Question

The value of the integral Ι=∫ ex(sin x + cos ⁡ x)dx is equal to ex⋅ f(x)+C, C being the constant of integration. Then the maximum value of y=f(x2),∀ x∈ R is equal to (A) 0 (B) -1 (C) 1 (D) (1/2)

Tardigrade Admin · Accepted Answer

As we know, ∫ ex(f(x)+f prime(x)) d x=ex ⋅ f(x)+C Thus, ∫ ex (sin x + cos ⁡ x) d x = ex ⋅ sin ⁡ x + C i.e. f(x)=sin x Hence, f(x2)=sin(x2): which has the maximum value of ‘1’

Q. The value of the integral $I = \int e^{x} (s in x + cos x) d x$ is equal to $e^{x} \cdot f (x) + C,$ $C$ being the constant of integration. Then the maximum value of $y = f (x^{2}), \forall x \in R$ is equal to

$0$

$- 1$

$1$

$\frac{1}{2}$

As we know,
$\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} \cdot f (x) + C$
Thus, $\int e^{x} (s in x + cos x) d x = e^{x} \cdot s in x + C$
i.e. $f (x) = s in x$
Hence, $f (x^{2}) = s in (x^{2}) :$ which has the maximum value of $‘1’$

Q. The value of the integral I=∫ex(sinx+cos⁡x)dx is equal to ex⋅f(x)+C, C being the constant of integration. Then the maximum value of y=f(x2),∀x∈R is equal to

0

−1

1

21​

As we know, ∫ex(f(x)+f′(x))dx=ex⋅f(x)+C Thus, ∫ex(sinx+cos⁡x)dx=ex⋅sin⁡x+C i.e. f(x)=sinx Hence, f(x2)=sin(x2): which has the maximum value of ‘1’

Q. The value of the integral $I = \int e^{x} (s in x + cos x) d x$ is equal to $e^{x} \cdot f (x) + C,$ $C$ being the constant of integration. Then the maximum value of $y = f (x^{2}), \forall x \in R$ is equal to

$0$

$- 1$

$1$

$\frac{1}{2}$

As we know,
$\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} \cdot f (x) + C$
Thus, $\int e^{x} (s in x + cos x) d x = e^{x} \cdot s in x + C$
i.e. $f (x) = s in x$
Hence, $f (x^{2}) = s in (x^{2}) :$ which has the maximum value of $‘1’$