The value of a for which ax2+sin−1(x2−2x+2)+cos−1(x2−2x+2)=0 has a real solution is
A
π2
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B
−π2
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C
2π
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D
−2π
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Solution
The correct option is B−π2 ax2+sin−1((x−1)2+1)+cos−1((x−1)2+1)=0 ax2+π2=0 x2=−π2a x2=−π2a ...(i) And −1≤(x−1)2+1≤1 −2≤(x−1)2≤0 (x−1)2≤0 There is only one possibility (x−1)2=0 x=1 Substituting in 1, we get a=−π2