AP EAMCET202420 May 2024Morning ShiftMathematicsLimitsActual

_ x 0 1+ 1+x^4 - 2+x^5+x^6 x^4 =

Options

A1 4 2
B1 2 2
C1 2
D1 3 2

Correct answer

A. 1 4 2

Step-by-step solution

_ x 0 1+ 1+x^4 - 2+x^5+x^6 x^4 = _ x 0 ( 1+x^4 -x^5-x^6-1 ) x^4 ( 1+ 1+x^4 + 2+x^5+x^6 ) = _ x 0 (1+ x^4 2 -x^5-x^6-1 ) x^4 ( 1+ 1+x^4 + 2+x^5+x^6 ) = _ x 0 1 2 ( 2 + 2 ) = 1 4 2

Practice Limits on Quantrex Academy →

More from Limits

If _ x 0 ( p 2x + 1 - 2x x + x ) = 1 then the value of ' p ' is 2026 _ x 2 ( 1 - x x ) is equal to 2026 The value of _ x 3 [ 1 x-3 + 9x 27-x^3 ] is: 2026 _ x 0 (1 - 2x)(3 + x) x 4x is equal to: 2026 If _ x 3 ( x^2 - ax - 3b x - 3 ) = 5 , then a + b = 2026 If f(x) = cases x^2 - 1 & if x 2 x + 1 & if x < 2 cases , then _ x 1 f(x) + _ x 2 f(x) = 2026 _ x 4 2 2 -( x+ x)^3 1- 2 x = 2025 Let [x] denote the greatest integer less than or equal to x . Then _ x 2⁺ ( [x]^3 3 - [ x 3 ]^3 )= 2025 Full Limits list All AP EAMCET PYQs