If `cosec^(-1) (cosec x) and cosec (cosec^(-1) x)` are equal function then the maximum range of value of x is
A. `[-(pi)/(2), -1] uu [1, (pi)/(2)]`
B. `[-(pi)/(2), 0)uu[0, (pi)/(2)]`
C. `(-oo, -1] uu [1, oo)`
D. `[-1, 0) uu [0, 1)`
Correct Answer - A
`cosec (cosec^(-1) x) = x AA x in R -(-1, 1)`
Also range of `cosec^(-1) (cosec x) in [-(pi)/(2), 0) uu (0, (pi)/(2]`
So combing these two, we get
`x in [-(pi)/(2), -1] uu [1, (pi)/(2)]`
Correct Answer - C
Maximum range for a given velocity is attained at `theta=45^(@)`
`R_("max")=(u^(2)sin2theta)/(g)rArrR_("max")=(u^(2)sin90^(@))/(g)=(u^(2))/(g)`
`therefore R_("max")=4.H_("max")`
If `T_(r+1)` is the greatest term in `(a+bx)^(n)`,
`(r/(n-r+1)) a/b le |x|le ((r+1)/(n-r)) a/b`
Given that the fourth terms is in the expansion of `(2+3/(8)x)^(10)`.
`rArr (3/(10-3+1)) (2)/((3//8)) le |x|le...
Correct Answer - C
`(c )` Consider positive numbers
`(2a)/(4)`, `(2a)/(4)`,`(2a)/(4)`,`(2a)/(4)`,`(b)/(2)`,`(b)/(2)`,`(3c)/(2)`,`(3c)/(2)`
Using `A.M. ge G.M.`
`((2a)/(4)+(2a)/(4)+(2a)/(4)+(2a)/(4)+(b)/(2)+(b)/(2)+(3c)/(2)+(3c)/(2))/(8)`
`ge (((2a)/(4)*(2a)/(4)*(2a)/(4)*(2a)/(4)*(b)/(2)*(b)/(2)*(3c)/(2)*(3c)/(2))/(8))^((1)/(8)`
`implies(2a+b+3c)/(8) ge ((3^(2))/(2^(8))*a^(4)b^(2)c^(2))^((1)/(8))`
`(1)/(8) ge ((3^(2)a^(4)b^(2)c^(2))^((1)/(8)))/(2)`
`impliesa^(4)b^(2)c^(2) le (1)/(4^(8)*3^(2))`