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3.Trigonometrical Ratios, Functions and Identities
easy
જો ${\sin ^2}\theta = \frac{{{x^2} + {y^2} + 1}}{{2x}}$, તો $x$ એ ફરજિયાત . . . હોવો જોઈએ.
A
$-3$
B
$-2$
C
$1$
D
એકપણ નહી
Solution
(d) ${\sin ^2}\theta \le 1$
$\therefore$ $\frac{{{x^2} + {y^2} + 1}}{{2x}} \le 1$
$\Rightarrow$ ${x^2} + {y^2} – 2x + 1 \le 0$.
$\Rightarrow$ ${(x – 1)^2} + {y^2} \le 0$
It is possible, if $x = 1$ and $y = 0$,
$i.e.$, It also depends on value of $y$.
Hence, option $(d)$ is correct.
Standard 11
Mathematics
