Aspire Faculty ID #17152 · Topic: AMU MCA 2020 · Just now
AMU MCA 2020

Let $f(x,y,z) = e^{\sqrt{1 - x^2 - y^2}}$. Then the range of $f$ is

Solution

For real values:

$1 - x^2 - y^2 \ge 0$

$x^2 + y^2 \le 1$

Minimum value of $\sqrt{1 - x^2 - y^2}$ is $0$ (when $x^2+y^2=1$)

Maximum value is $1$ (when $x=y=0$)

So exponent varies from $0$ to $1$

$e^0 = 1$

$e^1 = e$

Hence range is $[1, e]$

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