The solution of ∫2xdtt2 - 1 = π12

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 Multiple Choice QuestionsMultiple Choice Questions

211.

27e9x + e12x13dx is equal to

  • 1427 + e3x13 + C

  • 1427 + e3x23 + C

  • 1327 + e3x43 + C

  • 1427 + e3x43 + C

     


212.

4dxx24 - 9x2 is equal to

  • 4 - 9x2 + C

  • - 234 - 9x2 + C

  • - 4 - 9x2x + C

  • 234 - 9x2x + C


213.

e- xcscx1 + cotxdx is equal to

  • e- xcscx + C

  • - e- xcscx + C

  • e- xcscx + cotx + C

  • e- xcscx - tanx + C


214.

sinx + cosxe- x + sinxdx is equal to

  • log1 - exsinx + C

  • log1 + exsinx + C

  • log1 + exsinx + C

  • log1 - e- xsinx + C


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215.

The solution of 2xdtt2 - 1 = π12 is

  • 1

  • 2

  • 3

  • 4


B.

2

Given,2xdtt2 - 1 = π12  sec-1t2x = π12 sec-1x - sec-12 = π12 sec-1x = π12 + π4 = 4π12 = π3             x = secπ3 = 2


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216.

Let f(x) = x2 - 2. If 36fxdx = 3fc for some c  (3, 6),  then the value of c is equal to

  • 12

  • 21

  • 19

  • 17


217.

0π2sin2x1 + 2cos2xdx is equal to

  • 12log2

  • log2

  • 12log3

  • log3


218.

0π2dx1 +tan3x is equal to

  • 1

  • π

  • π2

  • π4


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219.

If f(x) = 1xsin2t2dt, then the value of limx0fπ +x - fπx is equal to

  • 14

  • 12

  • 34

  • 1


220.

1x2x4 + 134dx is equal to

  • - 1 + x434x + C

  • - 1 + x4142x + C

  • - 1 + x414x + C

  • - 1 + x414x2 + C


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