The maximum value of sec−1⁡7−5x2+32x2+2 is :

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5π6

5π12

7π12

2π3

answer is D.

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Detailed Solution

sec−1⁡−52+22x2+2 x2+2≥2=sec−1⁡−52+1x2+2 1x2+2≤12≤sec−1⁡(−2)=π−sec−1⁡(2) −52+1x2+2≤−52+12=−2=2π3

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