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Inverse Trigonometric Functions · Question 11

JEE Main 2026 (Online) 21st January Evening ShiftNumericalMedium📈 Topic in 19% papers+4
1

Let the maximum value of ( -1x )2 + ( -1x )2 for x [- 32, 1 2 ] be mn 2, where (m, n) = 1. Then m + n is equal to .

Answer

65

Solution

Let t = -1x; then -1x = 2 - t

For x [- 32, 1 2 ]: t [- 3, 4 ]

E = t2 + ( 2 - t )2 = 2t2 - t + 24

= 2 (t - 4 )2 + 28

E decreases on [- 3, 4 ], so it is maximum at t = - 3:

E = 29 + ( 5 6 )2 = 4 2 + 25 236 = 2936 2

m = 29, n = 36

m + n = 65

Answer: 65

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2

The value of -1 ( 1+ 2(2)-1 (2) ) - -1 ( 1+ 2 ( 12 )+1 ( 12 ) ) is equal to

  1. - 32
  2. + 52
  3. - 54
  4. + 32
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3

Considering the principal values of the inverse trigonometric functions, -1 ( 32x + 12 1 - x2 ), - 12 < x < 1 2, is equal to

  1. -5 6 - -1x
  2. 5 6 - -1x
  3. 6 + -1x
  4. 4 + -1x
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4

The sum of the infinite series -1 ( 74 )+ -1 ( 194 )+ -1 ( 394 )+ -1 ( 674 )+ is :

  1. 2+ -1 ( 12 )
  2. 2- -1 ( 12 )
  3. 2- -1 ( 12 )
  4. 2+ -1 ( 12 )
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5

If y = ( 3 + -1 x2 ), then (x - y)2 + 3y2 is equal to .

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6

Let S = \x : -1x = + -1x + -1(2x + 1) \. Then x S(2x - 1)2 is equal to .

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7

( -1 35 + -1 513 + -1 3365 ) is equal to:

  1. 3365
  2. 1
  3. 3265
  4. 0
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8

Let [x] denote the greatest integer less than or equal to x. Then the domain of f(x) = -1(2[x] + 1) is:

  1. (- , )
  2. (- , ) - \0\
  3. (- , -1] [0, )
  4. (- , -1] [1, )
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