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← JEE Main 2022 · 27 Jun · Shift 2
MathematicsInteger+4 / -1Question 25 of 90

JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 25

Let [t][t] denote the greatest integer ≤t\le t and {t}\{t\} denote the fractional part of tt. The integral value of α\alpha for which the left hand limit of the function
f(x)=[1+x]+α2[x]+{x}+[x]−12[x]+{x}f(x)=[1+x]+\frac{\alpha^{2[x]+\{x\}}+[x]-1}{2[x]+\{x\}}
at x=0x=0 is equal to α−43\alpha-\dfrac43, is ______.

Answer and worked solution

Official answer: 3

Left of 0. [x]=−1[x]=-1, {x}→1\{x\}\to1, [1+x]=0[1+x]=0, so 2[x]+{x}→−12[x]+\{x\}\to-1.

Limit. α−1−2−1=2−1α=α−43\dfrac{\alpha^{-1}-2}{-1}=2-\dfrac1\alpha=\alpha-\dfrac43, i.e. 3α2−10α+3=03\alpha^2-10\alpha+3=0: α=3\alpha=3 or 13\frac13.

Integer value 3.

Answer: 3