If y-sin n x-cos n xx- n real - then -2- y -2A- TrueB- False

The correct option is A TrueWe have y = sin nx + cos nx =√(2) [ 1/√((2)) sin nx+1/√((2))cos nx ] =√(2) [ cos π/4 sin nx + sin π/4 cos nx ] =√(2) sin ( nx + ...

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Byju's Answer

Standard XI

Mathematics

Inequality

If y=sin n x+...

Question

If $y = s i n n x + c o s n x$ (x, n real), then $- \sqrt{2} \leq y \leq \sqrt{2}$

True

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False

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Solution

The correct option is A True
We have
$y = s i n n x + c o s n x$
$= \sqrt{2} [\frac{1}{\sqrt{(2)}} s i n n x + \frac{1}{\sqrt{(2)}} c o s n x]$
$= \sqrt{2} [c o s \frac{π}{4} s i n n x + s i n \frac{π}{4} c o s n x]$
$= \sqrt{2} s i n (n x + \frac{π}{4})$
Since $- 1 \leq s i n (n x + \frac{π}{4}) \leq 1$
we have $- \sqrt{2} \leq \sqrt{2} s i n (n x + \frac{π}{4}) \leq \sqrt{2}$
that is $- \sqrt{2} \leq y \leq \sqrt{2}$

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If y=sin nx+cos nx (x, n real), then −√2≤y≤√2

The correct option is A TrueWe have y=sin nx+cos nx =√2[1√(2)sin nx+1√(2)cos nx] =√2[cos π4sin nx+sin π4cos nx] =√2sin(nx+π4) Since −1≤sin(nx+π4)≤1 we have −√2≤√2sin(nx+π4)≤√2 that is −√2≤y≤√2

If $y = s i n n x + c o s n x$ (x, n real), then $- \sqrt{2} \leq y \leq \sqrt{2}$