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Trigonometrical Equations
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निम्न समीकरण में वास्तविक हलों $x$ की संख्या होगी: $\cos ^2(x \sin (2 x))+\frac{1}{1+x^2}=\cos ^2 x+\sec ^2 x$
A
$0$
B
$1$
C
$2$
D
अनंत
(KVPY-2018)
Solution
(b)
We have,
$\cos ^2(x \sin (2 x))+\frac{1}{1+x^2}=\cos ^2 x+\sec ^2 x$
$\text { LHS }=\cos ^2(x \sin 2 x)+\frac{1}{1+x^2}$
$\cos ^2(x \sin 2 x)+\frac{1}{1+x^2} \leq 2$
$\text { RHS }=\cos ^2 x+\sec ^2 x \geq 2$
$\qquad \quad \operatorname{LHS}^2 x+\sec ^2 x=2 \text { at } x=0$
Hence, only one solution.
Standard 11
Mathematics
