Basic of Mathematics JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Basic of Mathematics, free to read — no sign-in needed. The full chapter has 6 questions; sign in to attempt the remaining 1 in the exam simulator.

1. Q1JEE Advanced Adv 2024 (Paper 1)
   Let $a=3 \sqrt{2}$ and $b=\frac{1}{5^{1 / 6} \sqrt{6}}$. If $x, y \in \mathbb{R}$ are such that $\begin{gathered}3 x+2 y=\log _a(18)^{\frac{5}{4}} \\2 x-y=\log _b(\sqrt{1080})\end{gathered}$ and then $4 x+5 y$ is equal to . ________.
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   Answer: 8

   $\begin{aligned} & 3 x+2 y=\log _{3 \sqrt{2}}(3 \sqrt{2})^{\frac{5}{2}}=\frac{5}{2} \\ & \Rightarrow \quad 6 x+4 y=5 \quad \ldots \ldots(1)\end{aligned}$ $2 x-y=\log _{\frac{1}{5^{1 / 6} \sqrt{6}}}\left(5^{\frac{1}{6}} \sqrt{6}\right)^3=-3$ $\begin{array}{ll}\Rightarrow & 2 x-y=-3 \quad.....(2)\\ & \text { equation (1) }-(2) \\ \Rightarrow \quad & 4 x+5 y=8\end{array}$
2. Q2JEE Advanced Adv 2022 (Paper 2)
   The product of all positive real values of $x$ satisfying the equation ${x}^{\left(16{\left({\log }_{5}x\right)}^{3}-68{\log }_{5}x\right)}={5}^{-16}$ is
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   Answer: 1

   Taking log to the base $5$ on both sides $\left(16{\left({\log }_{5}x\right)}^{3}-68\left({\log }_{5}x\right)\right)\left({\log }_{5}x\right)=-16$ Let $\left({\log }_{5}x\right)=t$ $16{t}^{4}-68{t}^{2}+16=0$ $\Rightarrow 4{t}^{4}-16{t}^{2}-{t}^{2}+4=0$ $\Rightarrow \left(4{t}^{2}-1\right)\left({t}^{2}-4\right)=0$ $\Rightarrow t=\pm \dfrac{1}{2},\pm 2$ So ${\log }_{5}x=\pm \dfrac{1}{2}$ or $\pm 2$ $\Rightarrow x={5}^{\dfrac{1}{2}},{5}^{\dfrac{-1}{2}},{5}^{2},{5}^{-2}$ Hence, the product is $\sqrt{5}\times \dfrac{1}{\sqrt{5}}\times 25\times \dfrac{1}{25}=1$
3. Q3JEE Advanced Adv 2018 (Paper 1)
   The value of ${\left({\left({\log }_{2}⁡9\right)}^{2}\right)}^{\dfrac{1}{{\log }_{2}⁡\left({\log }_{2}⁡9\right)}}\times {\left(\sqrt{7}\right)}^{\dfrac{1}{{\log }_{4}⁡7}}$ is_____.
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   Answer: 8

   ${\log }_{2}⁡{9}^{\dfrac{2}{{\log }_{2}⁡\left({\log }_{2}⁡9\right)}}\times {7}^{\dfrac{\dfrac{1}{2}}{{\log }_{4}⁡7}}$ $={({\log }_{2}9)}^{2{\log }_{({\log }_{2}9)}2}\times {7}^{\dfrac{1}{2}{\log }_{7}4}$ $={({\log }_{2}9)}^{{\log }_{({\log }_{2}9)}{2}^{2}}\times {7}^{{\log }_{7}{4}^{\dfrac{1}{2}}}$ $={2}^{2}\times {4}^{\dfrac{1}{2}}$ $=4\times 2=8$
4. Q4JEE Advanced Adv 2013 (Paper 2)
   If ${3}^{\text{x}}={4}^{\text{x}-1}$, then x =
   1. A.$\dfrac{{\text{2 log}}_{3}2}{{\text{2 log}}_{3}2-1}$
   2. B.$\dfrac{2}{2-{\text{log}}_{2}3}$
   3. C.$\dfrac{1}{1-{\text{log}}_{4}3}$
   4. D.$\dfrac{{\text{2 log}}_{2}3}{{\text{2 log}}_{2}3-1}$
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   Answer: A,B,C

   3x = 4x - 1 x = (x - 1) log3 4 x ( 1 - 2 log3 2) = -2log3 2 $\text{x}=\dfrac{2{\text{log}}_{3}2}{2{\text{log}}_{3}2-1}$ x log2 3 = (x - 1).2 x (log2 3 - 2) = -2 $\text{x}=\dfrac{2}{2{-\text{log}}_{2}3}$ $\text{x}=\dfrac{1}{1-\dfrac{1}{2}{\text{log}}_{2}3}=\dfrac{1}{1-{\text{log}}_{4}3}$
5. Q5JEE Advanced Adv 2011 (Paper 1)
   Let $\left(x_0, y_0\right)$ be the solution of the following equations $(2 x)^{\log 2}=(3 y)^{\log 3}$, $3^{\log x}=2^{\log y}$, then $x_0$ is equal to
   1. A.$\frac{1}{6}$
   2. B.$\frac{1}{2}$
   3. C.$\frac{1}{2}$
   4. D.6
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   Answer: (C)

   Taking log on both sides, $$\log 2 \cdot \log (2 x)=\log 3(\log 3 y)$$ $$\begin{array}{ll} \Rightarrow & \log 2\{\log 2+\log x\} \\ & =\log 3\{\log 3+\log y\} \\ \text { and } & \log x \cdot \log 3=\log y \log 2 \\ & \log y=\frac{\log x \cdot \log 3}{\log 2} \end{array}$$ From Eqs. (i) and (ii), we get $$\begin{aligned} & \log 2\{\log 2+\log x\} \\ &=\log 3 \cdot\left\{\log 3+\frac{\log x \cdot \log 3}{\log 2}\right\} \\ & \Rightarrow \quad(\log 2)^2+\log 2 \cdot \log x \\ &=(\log 3)^2+\frac{(\log 3)^2}{(\log 2)} \cdot \log x \\ & \Rightarrow \quad \log x\left\{\frac{(\log 3)^2}{\log 2}-\log 2\right\} \\ &=(\log 2)^2-(\log 3)^2 \end{aligned}$$ $$\begin{aligned} & \Rightarrow \quad \log x=-\log 2=\log 2^{-1} \\ & \therefore \quad x=\frac{1}{2} \end{aligned}$$