The all possible values of ‘a’ such that sin2(2π5x)+asin(2π5x)+sin−1(x2−10x+26)+cos−1(x2−10x+26)=0has atleast one solution is/are

−1≤X2−10X+26≤1⇒X=5 Only and at x=5 π2=0 which not possible. Hence a ∈ϕ.

The all possible values of ‘a’ such that sin2(2π5x)+asin(2π5x)+sin−1(x2−10x+26)+cos−1(x2−10x+26)=0has atleast one solution is/are

−1≤X2−10X+26≤1⇒X=5 Only and at x=5 π2=0 which not possible. Hence a ∈ϕ.

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The all possible values of ‘a’ such that $\sin^{2} (\frac{2 π}{5} x) + a \sin (\frac{2 π}{5} x) + \sin^{- 1} (x^{2} - 10 x + 26) + \cos^{- 1} (x^{2} - 10 x + 26) = 0$
has atleast one solution is/are

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$(- \sqrt{2 π,} \sqrt{2 π})$

$(- \infty, - \sqrt{2 π}] \cup (\sqrt{π}, \infty)$

$(- \infty, - \sqrt{2 π}] \cup [\sqrt{2 π}, \infty)$

$ϕ$

answer is D.

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$- 1 \leq X^{2} - 10 X + 26 \leq 1 \Rightarrow X = 5$ Only and at x=5 $\frac{π}{2} = 0$ which not possible. Hence a $\in ϕ .$

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The all possible values of ‘a’ such that sin2(2π5x)+asin(2π5x)+sin−1(x2−10x+26)+cos−1(x2−10x+26)=0has atleast one solution is/are

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The all possible values of ‘a’ such that $\sin^{2} (\frac{2 π}{5} x) + a \sin (\frac{2 π}{5} x) + \sin^{- 1} (x^{2} - 10 x + 26) + \cos^{- 1} (x^{2} - 10 x + 26) = 0$
has atleast one solution is/are