If exactly one root of equation x2-(p+1) x-p2=0 lie between 1 and 4 then number of integral values of p is -

Question

If exactly one root of equation x2-(p+1) x-p2=0 lie between 1 and 4 then number of integral values of p is - (A) 4 (B) 5 (C) 7 (D) 9

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/8062de9f709a878050058367d34a8855-.png" /> f(1) f(4)< 0 (-p-p2)(12-4 p-p2)< 0 p(p+1)(p+6)(p-2)< 0 so p ∈(-6,-1) ∪(0,2) number of integral values of p is 5

Q. If exactly one root of equation $x^{2} - (p + 1) x - p^{2} = 0$ lie between 1 and 4 then number of integral values of $p$ is -

4

5

7

9

$f (1) f (4) < 0$
$(- p - p^{2}) (12 - 4 p - p^{2}) < 0$
$p (p + 1) (p + 6) (p - 2) < 0$
so $p \in (- 6, - 1) \cup (0, 2)$
number of integral values of $p$ is $5$

Q. If exactly one root of equation x2−(p+1)x−p2=0 lie between 1 and 4 then number of integral values of p is -

4

5

7

9

f(1)f(4)<0 (−p−p2)(12−4p−p2)<0 p(p+1)(p+6)(p−2)<0 so p∈(−6,−1)∪(0,2) number of integral values of p is 5

Q. If exactly one root of equation $x^{2} - (p + 1) x - p^{2} = 0$ lie between 1 and 4 then number of integral values of $p$ is -

$f (1) f (4) < 0$
$(- p - p^{2}) (12 - 4 p - p^{2}) < 0$
$p (p + 1) (p + 6) (p - 2) < 0$
so $p \in (- 6, - 1) \cup (0, 2)$
number of integral values of $p$ is $5$