JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 2)
If the equation cos4 $$\theta $$ + sin4 $$\theta $$ +
$$\lambda $$ = 0 has real
solutions for
$$\theta $$, then
$$\lambda $$ lies in the interval :
$$\left[ { - {3 \over 2}, - {5 \over 4}} \right]$$
$$\left( { - {1 \over 2}, - {1 \over 4}} \right]$$
$$\left( { - {5 \over 4}, - 1} \right]$$
$$\left[ { - 1, - {1 \over 2}} \right]$$
Explanation
cos4 $$\theta $$ + sin4 $$\theta $$ +
$$\lambda $$ = 0
$$ \Rightarrow $$ 1 – 2sin2 $$\theta $$ cos2 $$\theta $$ = -$$\lambda $$
$$ \Rightarrow $$ 1 - $$\frac{1}{2} \times 4$$sin2 $$\theta $$ cos2 $$\theta $$ = -$$\lambda $$
$$ \Rightarrow $$ 1 - $$\frac{\sin^{2} 2\theta }{2} $$ = -$$\lambda $$
$$ \Rightarrow $$2($$\lambda $$ + 1) = sin2 2$$\theta $$
0 $$ \le $$ 2 ($$\lambda $$ + 1) $$ \le $$ 1
0 $$ \le $$ ($$\lambda $$ + 1) $$ \le $$ $$\frac{1}{2} $$
-1 $$ \le $$ $$\lambda $$ $$ \le $$ -$$\frac{1}{2} $$
$$ \Rightarrow $$ 1 – 2sin2 $$\theta $$ cos2 $$\theta $$ = -$$\lambda $$
$$ \Rightarrow $$ 1 - $$\frac{1}{2} \times 4$$sin2 $$\theta $$ cos2 $$\theta $$ = -$$\lambda $$
$$ \Rightarrow $$ 1 - $$\frac{\sin^{2} 2\theta }{2} $$ = -$$\lambda $$
$$ \Rightarrow $$2($$\lambda $$ + 1) = sin2 2$$\theta $$
0 $$ \le $$ 2 ($$\lambda $$ + 1) $$ \le $$ 1
0 $$ \le $$ ($$\lambda $$ + 1) $$ \le $$ $$\frac{1}{2} $$
-1 $$ \le $$ $$\lambda $$ $$ \le $$ -$$\frac{1}{2} $$
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