Let alpha(a) and beta(a) be the roots of the | vedclass

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

Question

(B) The given equation is $((1+a)^{1/3}-1)x^2 + ((1+a)^{1/2}-1)x + ((1+a)^{1/6}-1) = 0$.Let $1+a = t^6$. As $a \rightarrow 0^{+}$,$t \rightarrow 1^{+}

Vedclass · Accepted Answer

$-\frac{1}{2}$ and $-1$

Vedclass · Answer

(B) The given equation is $((1+a)^{1/3}-1)x^2 + ((1+a)^{1/2}-1)x + ((1+a)^{1/6}-1) = 0$.Let $1+a = t^6$. As $a ightarrow 0^{+}$,$t ightarrow 1^{+}$.The equation becomes $(t^2-1)x^2 + (t^3-1)x + (t-1) = 0$.Dividing by $(t-1)$ (since $t 
eq 1$ as $a > 0$):$(t+1)x^2 + (t^2+t+1)x + 1 = 0$.Taking the limit as $t ightarrow 1$:$(1+1)x^2 + (1^2+1+1)x + 1 = 0$.$2x^2 + 3x + 1 = 0$.Factoring the quadratic equation:$(2x+1)(x+1) = 0$.Thus,the roots are $x = -\frac{1}{2}$ and $x = -1$.

Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $(\sqrt[3]{1+a}-1) x^2+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a > -1$. Then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{a \rightarrow 0^{+}} \beta(a)$ are

Similar Questions

If $\lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8(P x^4-16)}{(x+\sqrt{x^2-2})^8+(x-\sqrt{x^2-2})^8} = 1$,then $P=$

If $\lim_{x \to 1} \frac{x^3 + ax^2 + bx + c}{x^2 - 2x + 1} = 2026$, then the value of $a - c$ is...

If $\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - ax + b}}{{x - 1}} = 3$,then $a + b$ is equal to

If $\lim_{x \rightarrow 0} \frac{e^{(a-1)x} + 2 \cos(bx) + (c-2)e^{-x}}{x \cos x - \log_{e}(1+x)} = 2$, then $a^{2} + b^{2} + c^{2}$ is equal to:

If $\mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \frac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}$,then $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module