If y-sin nx - cos nx x- n real- then -2- y -2

We 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 + π/4 ) Since 1 ≤ sin (nx + π ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Inequality

If y=sin nx +...

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

False

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $- \sqrt{2} \leq y \leq \sqrt{2}$

Suggest Corrections

Similar questions

y = A cos n x + B sin n x

Prove that

\frac{d^{2} y}{{d x}^{2}} + n^{2} y = 0

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 c o s (n x) & c o s (n + 1) x & c o s (n + 2) x s i n (n x) & s i n (n + 1) x & s i n (n + 2) x \end{matrix} ∣ ∣ ∣ ∣ ∣

does not depend

Q. The determinant

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & a & 1 cos (n x) & cos (n + 1) x & cos (n + 2) x sin (n x) & sin (n + 1) x & sin (n + 2) x \end{matrix} ∣ ∣ ∣ ∣ ∣

is independent of

Q. If

n (X) = m, n (Y) = n

and

n (X \cap Y) = p

then

n (X \cup Y) =

Q. If

cos x + sec x = - 2

for a positive odd integer

n

then

{cos}^{n} x + {sec}^{n} x

Join BYJU'S Learning Program

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 $- \sqrt{2} \leq y \leq \sqrt{2}$