For t\gt -1 , let \alpha_t and \beta_t be the roots of the equation ((t+2)^1 / 7-1) x^2+((t+2)^1 / 6-1) x+((t+2)^1 / 21-1)=0 . If \lim \limits_t arrow-1^+ \alpha_t=a and \lim \limits_t arrow-1^+ \beta_t=b , then 72(a+b)^2 is equal to \underline\hspace2cm .

Limits Continuity and Differentiability PYQ — JEE Main 2025 · 7 Apr · Shift 2 · Q50 with Solution

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Correct answer: 98

Let

\qquad B_t=(t+2)^{1/6}-1, \qquad C_t=(t+2)^{1/21}-1.$$ The given quadratic is $$A_t x^2+B_t x+C_t=0,$$ with roots $\alpha_t,\beta_t$. We need $a=\lim_{t\to -1^+}\alpha_t$ and $b=\lim_{t\to -1^+}\beta_t$, then find $$72(a+b)^2.$$ --- ## 1. Behavior of coefficients as $t\to -1^+$ As $t\to -1^+$, we have $t+2\to 1^+$, so $$A_t\to 0,\quad B_t\to 0,\quad C_t\to 0.$$ Thus the quadratic becomes a degenerate $0=0$ form, so we must compare the rates at which the coefficients vanish. Let $$u=t+2.$$ Then $u\to 1^+$. Using the standard expansion near $u=1$: $$u^r-1\sim r(u-1) \quad (u\to 1),$$ we get $$A_t\sim \frac{1}{7}(u-1),\qquad B_t\sim \frac{1}{6}(u-1),\qquad C_t\sim \frac{1}{21}(u-1).$$ Hence, dividing the equation by $(u-1)$ (or equivalently taking the limiting ratio), the roots approach the roots of $$\frac{1}{7}x^2+\frac{1}{6}x+\frac{1}{21}=0.$$ Multiplying by $42$: $$6x^2+7x+2=0.$$ --- ## 2. Find the limiting roots Solve $$6x^2+7x+2=0.$$ Factorizing: $$6x^2+7x+2=(3x+2)(2x+1)=0.$$ So the roots are $$x=-\frac{2}{3},\qquad x=-\frac{1}{2}.$$ Therefore, $$\{a,b\}=\left\{-\frac{2}{3},-\frac{1}{2}\right\}.$$ So $$a+b=-\frac{2}{3}-\frac{1}{2}=-\frac{7}{6}.$$ --- ## 3. Compute $72(a+b)^2$ $$72(a+b)^2=72\left(\frac{7}{6}\right)^2=72\cdot \frac{49}{36}=2\cdot 49=98.$$ --- ## 4. Final answer $$\boxed{98}$$ The derived answer matches the stored correct answer.

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View written solutionFree

Correct answer: 98

Let

\qquad B_t=(t+2)^{1/6}-1, \qquad C_t=(t+2)^{1/21}-1.$$ The given quadratic is $$A_t x^2+B_t x+C_t=0,$$ with roots $\alpha_t,\beta_t$. We need $a=\lim_{t\to -1^+}\alpha_t$ and $b=\lim_{t\to -1^+}\beta_t$, then find $$72(a+b)^2.$$ --- ## 1. Behavior of coefficients as $t\to -1^+$ As $t\to -1^+$, we have $t+2\to 1^+$, so $$A_t\to 0,\quad B_t\to 0,\quad C_t\to 0.$$ Thus the quadratic becomes a degenerate $0=0$ form, so we must compare the rates at which the coefficients vanish. Let $$u=t+2.$$ Then $u\to 1^+$. Using the standard expansion near $u=1$: $$u^r-1\sim r(u-1) \quad (u\to 1),$$ we get $$A_t\sim \frac{1}{7}(u-1),\qquad B_t\sim \frac{1}{6}(u-1),\qquad C_t\sim \frac{1}{21}(u-1).$$ Hence, dividing the equation by $(u-1)$ (or equivalently taking the limiting ratio), the roots approach the roots of $$\frac{1}{7}x^2+\frac{1}{6}x+\frac{1}{21}=0.$$ Multiplying by $42$: $$6x^2+7x+2=0.$$ --- ## 2. Find the limiting roots Solve $$6x^2+7x+2=0.$$ Factorizing: $$6x^2+7x+2=(3x+2)(2x+1)=0.$$ So the roots are $$x=-\frac{2}{3},\qquad x=-\frac{1}{2}.$$ Therefore, $$\{a,b\}=\left\{-\frac{2}{3},-\frac{1}{2}\right\}.$$ So $$a+b=-\frac{2}{3}-\frac{1}{2}=-\frac{7}{6}.$$ --- ## 3. Compute $72(a+b)^2$ $$72(a+b)^2=72\left(\frac{7}{6}\right)^2=72\cdot \frac{49}{36}=2\cdot 49=98.$$ --- ## 4. Final answer $$\boxed{98}$$ The derived answer matches the stored correct answer.