MCQ
જો $\int\limits_{ - \infty }^\infty {f(x)dx = 1} $ તો $\int\limits_{ - \infty }^\infty {f\left( {x - \frac{1}{x}} \right)dx} $ મેળવો.
- A$0$
- ✓$1$
- C$-1$
- D$2$
$\Rightarrow 1 \Rightarrow \int_{-\infty}^{\infty} f\left(y-\frac{1}{y}\right)\left(1+\frac{1}{y^{2}}\right) d y$
$=\int_{-\infty}^{0} f\left(y-\frac{1}{y}\right) d y+\int_{-\infty}^{0} f\left(y-\frac{1}{y}\right) \frac{d y}{y^{2}}$
Putting $z=-\frac{1}{y}=\int_{-\infty}^{0} f\left(y-\frac{1}{y}\right) d y+\int_{0}^{\infty} f\left(z-\frac{1}{z}\right) d z=1$
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