MCQ
$\sin^2x+\cos^4x,\forall x\in R$ નો વિસ્તાર ................ છે.
- A$\left[\frac{13}{16},1\right]$
- B$\left[1,2\right]$
- C$\left[\frac{3}{4}\frac{13}{16}\right]$
- ✓$\left[\frac{3}{4},1\right]$
$\sin^2x+\cos^4x=\left(\frac{1-\cos2x}{2}\right)+\left(\frac{1+\cos2x}{2}\right)^2$
અહીંથી સાદુરૂપ આપી આગળગણતરી કરતા.
$=\frac{7}{8}+\frac{1}{8}\cos4x$
અહી $\csc4x$ નો વિસ્તાર $[-1,1]$
$\therefore \frac{7}{8}+\frac{1}{8}\cos4x$ નો વિસ્તાર
$\left[\frac{7}{8}-\frac{1}{8},\frac{7}{8}+\frac{1}{8}\right]=\left[\frac{3}{4},1\right]$
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