The number of values of b for which there is | vedclass

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

Question

(C) For a triangle to exist,the sum of any two sides must be greater than the third side,and all side lengths must be positive.Let the sides be $a = b

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) For a triangle to exist,the sum of any two sides must be greater than the third side,and all side lengths must be positive.Let the sides be $a = b+5$,$c = 3b-2$,and $d = 6-b$.For sides to be positive: $b+5 > 0 \Rightarrow b > -5$,$3b-2 > 0 \Rightarrow b > 2/3$,and $6-b > 0 \Rightarrow b Case $I$: $b+5 = 3b-2$$2b = 7 \Rightarrow b = 3.5$. Sides are $8.5, 8.5, 2.5$. Since $8.5+8.5 > 2.5$,$8.5+2.5 > 8.5$,this is a valid triangle.Case $II$: $3b-2 = 6-b$$4b = 8 \Rightarrow b = 2$.Sides are $7, 4, 4$. Since $4+4 > 7$,this is a valid triangle.Case $III$: $b+5 = 6-b$$2b = 1 \Rightarrow b = 0.5$.This contradicts the condition $b > 2/3$. Also,the side $3b-2$ would be $3(0.5)-2 = -0.5$,which is not possible.Thus,there are $2$ such values of $b$.

The number of values of $b$ for which there is an isosceles triangle with sides of lengths $b+5$,$3b-2$,and $6-b$ is

Similar Questions

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-Px^2+Qx-R=0$ and $(\alpha-2)^2, (\beta-2)^2, (\gamma-2)^2$ are the roots of the equation $x^3-5x^2+4x=0$,then the possible least value of $P+Q+R$ is

Let $\alpha \neq 1$ be a real root of the equation $x^3-a x^2+a x-1=0$,where $a \neq -1$ is a real number. Then,a root of this equation,among the following,is

The value of $a$ for which the quadratic equation $3x^2 + 2(a^2 + 1)x + (a^2 - 3a + 2) = 0$ possesses roots with opposite signs,lies in

If $a > 0,$ then $\sqrt {a + \sqrt {a + \sqrt {a + ....\infty } } } $ is

The product of the roots of the equation $9x^{2}-18|x|+5=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module