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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
$f(x)=$
, $g(x)=$
If $h(x)=f[g(x)]$, then an interval in which $h(x)=1$ for all values of $x$ in that interval is:
| { | $1$ | if | $|x|\le 1$ |
| $0$ | if | $|x|>1$ |
| { | $2-x^2$ | if | $|x|\le 2$ |
| $2$ | if | $|x|>2$ |
If $h(x)=f[g(x)]$, then an interval in which $h(x)=1$ for all values of $x$ in that interval is:
Solution
Given,
$h(x)=f[g(x)]$
Now, $f(t)=1$ when
$|t|\le 1$
So, $h(x)=1$ when
$|g(x)|\le 1$
For $|x|\le 2$,
$g(x)=2-x^2$
So,
$|2-x^2|\le 1$
This gives
$-1\le 2-x^2\le 1$
Subtract $2$ from all sides:
$-3\le -x^2\le -1$
Multiplying by $-1$ reverses the inequalities:
$1\le x^2\le 3$
So,
$1\le |x|\le \sqrt{3}$
For $|x|>2$,
$g(x)=2$
So,
$|g(x)|=2>1$
Hence, this case is not valid.
Therefore, the required interval is
$1\le |x|\le \sqrt{3}$
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