Find the solution set of [ x ^ 2 ]+( x +1)^ 2 =25 , where [x] is the least integer…
Mathematics · JEE Main · NTA Exams — Sets, Relations and Functions
Find the solution set of \(\left[\mathbf{x}^{2}\right]+(\mathbf{x}+1)^{2}=25\), where \([x]\) is the least integer greater than or equal to\(\mathbf{x}\)
- \((-5,-4)\)
- \(\left[\begin{array}{ll} -5 & -4 \end{array}\right] \cup\left[\begin{array}{l} 2 \\ 3 \end{array}\right]\)
- \((2,3)\)
- None of these
Answer
(D) None of these
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- The range of the function y = l og 3 (5 + 4x – x 2 ) is
- The set of values of x which satisfy the inequations 5x + 2 < 3x + 8 and x+2 x-1 <4 is
- If x 2 + 6x – 27 > 0 and x 2 – 3x – 4 < 0, then
- If A= x R:|x|<2 and B= x R:|x-2| 3 then :
- If cos –1 x – cos –1 y 2 = , then 4x 2 – 4xy cos α + y 2 is equal to
- The function f (x) = cos ( (x+ x^ 2 +1 ) ) is :
- If the domain of the function f ( x ) = cos –1 ( 2-|x| 4 )+ _ e (3-x) ^ -1 is [–a, b) – γ , then + + is equal…
- Let _ k=1 ^ 10 f(a+k)=16 (z^ 1 a -1 ) , where the function f satisfies f(x+y)=f(x) f(y) for all natural…
More Sets, Relations and Functions questions · Browse all practice questions