Let fx is a derivable function satisfying fx-x0etsinx-tdt and gx-f-x-fx- then the number of possible integers in the range of gx is

F(x)=∫^x_0e^tsin(x t)dt Applying integration by parts succesively : taking u=sin(x t),v=e^t ⇒ f(x)=[sin (x t)e^t ]_0^x +∫_0^xe^tcos(x t)dt = sin x+ [[cos (x t ...

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

Let fx is a d...

Question

Let $f (x)$ is a derivable function satisfying $f (x) = x \int 0 e^{t} sin (x - t) d t$ and $g (x) = f^{''} (x) - f (x),$ then the number of possible integers in the range of $g (x)$ is

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Solution

$f (x) = x \int 0 e^{t} sin (x - t) d t$
Applying integration by parts succesively : taking $u = sin (x - t), v = e^{t}$
$\Rightarrow f (x) = {[sin (x - t) e^{t}]}_{0}^{t} + x \int 0 e^{t} cos (x - t) d t = - sin x + ⎡ ⎢ ⎣ {[cos (x - t) e^{t}]}_{0}^{t} - x \int 0 e^{t} sin (x - t) d t ⎤ ⎥ ⎦ = - sin x + e^{x} - cos x - f (x) \Rightarrow f (x) = \frac{1}{2} (- cos x - sin x + e^{x}) f^{'} (x) = \frac{1}{2} [sin x - cos x + e^{x}] f^{''} (x) = \frac{1}{2} [cos x + sin x + e^{x}] g (x) = f^{''} (x) - f (x) = \frac{1}{2} [cos x + sin x + e^{x} + cos x + sin x - e^{x}] = \frac{1}{2} [2 (cos x + sin x)] g (x) = cos x + sin x$
$\Rightarrow$ Range of $g (x)$ is $[- \sqrt{2}, \sqrt{2}]$
Thus, $3$ integral points in range of $g (x)$

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Let f(x) is a derivable function satisfying f(x)=x∫0etsin(x−t)dt and g(x)=f′′(x)−f(x), then the number of possible integers in the range of g(x) is

Let $f (x)$ is a derivable function satisfying $f (x) = x \int 0 e^{t} sin (x - t) d t$ and $g (x) = f^{''} (x) - f (x),$ then the number of possible integers in the range of $g (x)$ is