Basic of Mathematics JEE Main previous year questions with solutions

5 solved JEE Main questions on Basic of Mathematics, free to read — no sign-in needed. The full chapter has 20 questions; sign in to attempt the remaining 15 in the exam simulator.

1. Q1JEE Main 2026 (04 Apr, Shift 1)Inequalities
   If the set of all solutions of $|x^2 + x - 9| = |x| + |x^2 - 9|$ is $[\alpha, \beta] \cup [\gamma, \infty)$, then $(\alpha^2 + \beta^2 + \gamma^2)$ is equal to:
   1. A.$9$
   2. B.$18$
   3. C.$36$
   4. D.$72$
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   Answer: (B)

   The given equation is $|x^2 + x - 9| = |x| + |x^2 - 9|$. Let $A = x$ and $B = x^2 - 9$. Then $A + B = x^2 + x - 9$. The equation is of the form $|A + B| = |A| + |B|$, which holds true if and only if $A \cdot B \ge 0$. Substituting the values of $A$ and $B$: $x(x^2 - 9) \ge 0$ $x(x - 3)(x + 3) \ge 0$ The critical points are $-3, 0, 3$. Using the wavy curve method, the solution set is: $x \in [-3, 0] \cup [3, \infty)$ Comparing this with the given solution set $[\alpha, \beta] \cup [\gamma, \infty)$, we get: $\alpha = -3$, $\beta = 0$, $\gamma = 3$ We need to find the value of $\alpha^2 + \beta^2 + \gamma^2$: $\alpha^2 + \beta^2 + \gamma^2 = (-3)^2 + 0^2 + 3^2 = 9 + 0 + 9 = 18$ Answer: $18$
2. Q2JEE Main 2025 (22 Jan, Shift 1)Logarithm
   The product of all solutions of the equation $\mathrm{e}^{5\left(\log _{\mathrm{e}} x\right)^2+3}=x^8, x\gt0$, is :
   1. A.$e^{8 / 5}$
   2. B.$e^{6 / 5}$
   3. C.$\mathrm{e}^2$
   4. D.e
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   Answer: (A)

   $\begin{aligned} & \mathrm{e}^{5(\ln x)^2+3}=\mathrm{x}^8 \\ & \Rightarrow \ell \mathrm{ne}^{5(\ln x)^2+3}=\ell \mathrm{nn}^8 \\ & \Rightarrow 5(\ln \mathrm{x})^2+3=8 \ell \mathrm{nx} \\ & (\ell \mathrm{nx}=\mathrm{t}) \\ & \Rightarrow 5 \mathrm{t}^2-8 \mathrm{t}+3=0 \\ & \quad \mathrm{t}_1+\mathrm{t}_2=\frac{8}{5} \\ & \quad \ln \mathrm{x}_1 \mathrm{x}_2=\frac{8}{5} \\ & \mathrm{x}_1 \mathrm{X}_2=\mathrm{e}^{8 / 5}\end{aligned}$
3. Q3JEE Main 2026 (24 Jan, Shift 1)Inequalities
   The number of the real solutions of the equation: $x|x+3|+|x-1|-2=0$ is
   1. A.2
   2. B.4
   3. C.3
   4. D.5
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   Answer: (C)

   Critical points: $x=-3$ and $x=1$. Case $x \lt -3$: $x^2+4x+1=0 \Rightarrow x=-2\pm\sqrt{3}$. Only $x=-2-\sqrt{3} \approx -3.73$ lies in $(-\infty,-3)$. (1 solution) Case $-3 \le x \lt 1$: $x^2+2x-1=0 \Rightarrow x=-1\pm\sqrt{2}$. Both $x=-1+\sqrt{2} \approx 0.41$ and $x=-1-\sqrt{2} \approx -2.41$ lie in $[-3,1)$. (2 solutions) Case $x \ge 1$: $x^2+4x-3=0 \Rightarrow x=-2\pm\sqrt{7}$. Neither root $\ge 1$. (0 solutions) Total number of real solutions $= 3$.
4. Q4JEE Main 2026 (08 Apr, Shift 2)Logarithm
   The sum of squares of all the real solutions of the equation $\log_{(x+1)}(2x^2+5x+3) = 4 - \log_{(2x+3)}(x^2+2x+1)$ is equal to ________.
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   Answer: 2

   For the logarithms to be defined, we must satisfy the following conditions: 1. Base of the first logarithm: $x+1 \gt 0 \Rightarrow x \gt -1$ and $x+1 \neq 1 \Rightarrow x \neq 0$. 2. Base of the second logarithm: $2x+3 \gt 0 \Rightarrow x \gt -\dfrac{3}{2}$ and $2x+3 \neq 1 \Rightarrow x \neq -1$. 3. Arguments must be positive: $2x^2+5x+3 \gt 0$ and $x^2+2x+1 \gt 0$. Taking the intersection of all these conditions, the domain of the equation is $x \in (-1, 0) \cup (0, \infty)$. Now, factorizing the arguments of the logarithms: $2x^2+5x+3 = (2x+3)(x+1)$ $x^2+2x+1 = (x+1)^2$ Substitute these into the given equation: $\log_{(x+1)}((2x+3)(x+1)) = 4 - \log_{(2x+3)}((x+1)^2)$ Using the properties of logarithms: $\log_{(x+1)}(2x+3) + \log_{(x+1)}(x+1) = 4 - 2\log_{(2x+3)}(x+1)$ $\log_{(x+1)}(2x+3) + 1 = 4 - \dfrac{2}{\log_{(x+1)}(2x+3)}$ Let $t = \log_{(x+1)}(2x+3)$. The equation becomes: $t + 1 = 4 - \dfrac{2}{t}$ $t - 3 + \dfrac{2}{t} = 0$ $t^2 - 3t + 2 = 0$ $(t-1)(t-2) = 0$ $\Rightarrow t = 1$ or $t = 2$ Case 1: $t = 1$ $\log_{(x+1)}(2x+3) = 1$ $2x+3 = x+1$ $x = -2$ This value is rejected because $x = -2$ does not fall in the domain $x \gt -1$. Case 2: $t = 2$ $\log_{(x+1)}(2x+3) = 2$ $2x+3 = (x+1)^2$ $2x+3 = x^2+2x+1$ $x^2 = 2$ $x = \pm\sqrt{2}$ Since $x \gt -1$, $x = -\sqrt{2}$ is rejected. The only valid solution is $x = \sqrt{2}$. The sum of squares of all the real solutions is $(\sqrt{2})^2 = 2$. Answer: $2$
5. Q5JEE Main 2023 (10 Apr, Shift 1)Logarithm
   Let $a,b,c$ be the three distinct positive real numbers such that ${\left(2a\right)}^{{\log }_{e}a}={\left(bc\right)}^{{\log }_{e}b}$ and ${b}^{{\log }_{e}2}={a}^{{\log }_{e}c}$ Then $6a+5bc$ is equal to ______.
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   Answer: 8

   Given equation: ${\left(2a\right)}^{\ln \left(a\right)}={\left(bc\right)}^{\ln \left(b\right)}$ $\Rightarrow \ln \left(a\right)\cdot \left(\ln \left(2a\right)\right)=\ln \left(b\right)\cdot \ln \left(bc\right)$ $\Rightarrow \ln \left(a\right)\cdot \left(\ln \left(2\right)+\ln \left(a\right)\right)=\ln \left(b\right)\cdot \left(\ln \left(b\right)+\ln \left(c\right)\right)$ let $\ln \left(a\right)=x,\ln \left(b\right)=y,\ln \left(c\right)=z,x\neq y\neq z$ $\Rightarrow x\left(\ln \left(2\right)+x\right)=y\left(y+z\right)$ $\Rightarrow x\ln \left(2\right)={y}^{2}-{x}^{2}+yz$ ........(i) Similarly, from second equation $\ln \left(2\right)\cdot \ln \left(b\right)=\ln \left(c\right)\cdot \ln \left(a\right)$ $\Rightarrow \ln \left(2\right)=\dfrac{xz}{y}$ ......(ii) Substitute $eq\left(ii\right)\text{in}eq\left(i\right)$, We get, $\Rightarrow {y}^{3}-y{x}^{2}+{y}^{2}z={x}^{2}z$ $\Rightarrow {y}^{2}\left(y+z\right)-{x}^{2}\left(y+z\right)=0$ $\Rightarrow \left({y}^{2}-{x}^{2}\right)\left(y+z\right)=0$ $\Rightarrow \left(x-y\right)\left(x+y\right)\left(y+z\right)=0$ $∵x\neq y\Rightarrow \left(x+y\right)\left(y+z\right)=0$ Now $x=-y,y=-z\text{and}x=z$ But $\ln \left(2\right)=\dfrac{xz}{y}$ $\Rightarrow \ln \left(2\right)=\dfrac{x\left(x\right)}{-x}$ $\Rightarrow \ln \left(2\right)=-\ln \left(a\right)$ $=-\ln 2\Rightarrow a=\dfrac{1}{2}$ and $\Rightarrow y=-z$ $\Rightarrow \ln \left(b\right)=-\ln \left(c\right)$ $\Rightarrow bc=1$ Now $6a+5bc=6\left(\dfrac{1}{2}\right)+5\left(1\right)=8$ Hence this is the required answer.