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Limits Continuity and Differentiability question

2025 · 7 Apr · Shift 2 · Q50
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Limits Continuity and Differentiability question

2025 · 7 Apr · Shift 2 · Q50

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1
For t>−1\mathrm{t}\gt -1t>−1, let αt\alpha_{\mathrm{t}}αt​ and βt\beta_{\mathrm{t}}βt​ be the roots of the equation ((t+2)1/7−1)x2+((t+2)1/6−1)x+((t+2)1/21−1)=0. If lim⁡t→−1+αt=a and lim⁡t→−1+βt=b, \left((\mathrm{t}+2)^{1 / 7}-1\right) x^2+\left((\mathrm{t}+2)^{1 / 6}-1\right) x+\left((\mathrm{t}+2)^{1 / 21}-1\right)=0 \text {. If } \lim \limits_{\mathrm{t} \rightarrow-1^{+}} \alpha_{\mathrm{t}}=\mathrm{a} \text { and } \lim \limits_{\mathrm{t} \rightarrow-1^{+}} \beta_{\mathrm{t}}=\mathrm{b} \text {, }((t+2)1/7−1)x2+((t+2)1/6−1)x+((t+2)1/21−1)=0. If t→−1+lim​αt​=a and t→−1+lim​βt​=b,  then 72(a+b)272(a+b)^272(a+b)2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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