MCQ
$\mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{r = 0}^n {\frac{n}{{{{\left( {2r + n} \right)}^2}}}} $ is equal to
- A$1$
- B$-1$
- C$2$
- ✓$\frac {1}{3}$
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$\int_0^1 {(1 + {{\cos }^8}x)(a{x^2} + bx + c)\,dx} = \int_0^2 {(1 + {{\cos }^8}x)(a{x^2} + bx + c)\,dx} $
Then the quadratic equation $a{x^2} + bx + c = 0$ has