The number of elements in the set {n in mathb | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given inequality is $|n^2 - 10n + 19| < 6$.This is equivalent to $-6 < n^2 - 10n + 19 < 6$.Case $1$: $n^2 - 10n + 19 < 6 \Rightarrow n^2 - 10n

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) The given inequality is $|n^2 - 10n + 19| This is equivalent to $-6 Case $1$: $n^2 - 10n + 19 The roots of $n^2 - 10n + 13 = 0$ are $n = \frac{10 \pm \sqrt{100 - 52}}{2} = 5 \pm \sqrt{12} = 5 \pm 2\sqrt{3}$.Since $2\sqrt{3} \approx 3.46$,the range is $n \in (5 - 3.46, 5 + 3.46) = (1.54, 8.46)$.Case $2$: $n^2 - 10n + 19 > -6$ $\Rightarrow n^2 - 10n + 25 > 0$ $\Rightarrow (n - 5)^2 > 0$.This is true for all $n \in \mathbb{Z}$ except $n = 5$.Combining these,$n \in \{2, 3, 4, 6, 7, 8\}$.The number of such elements is $6$.

The number of elements in the set $\{n \in \mathbb{Z} : |n^2 - 10n + 19| < 6\}$ is $...........$

Similar Questions

Solve the following inequality: $\frac{|x-1|}{x+2} < 1$.

If $x$ is real and satisfies $x + 2 > \sqrt{x + 4}$,then

The feasible region represented by the inequations $2x + 3y \leq 18$,$x + y \geq 10$,$x \geq 0$,$y \geq 0$ is

Find all pairs of consecutive odd positive integers,both of which are smaller than $10$,such that their sum is more than $11$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module