JEE MAIN Exponential Series Previous Year Question Paper and Solution with PDF

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JEE MAIN Previous Year Questions (PYQs)

JEE MAIN Exponential Series PYQ

Qus : 1 JEE MAIN PYQ

The sum of all the real roots of the equation $({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$ is

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2022 (24 June Evening Shift) PYQ

Solution

Qus : 2 JEE MAIN PYQ

The number of real roots of the equation ${e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0$ is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (25 February Morning Shift) PYQ

Solution

Qus : 3 JEE MAIN PYQ

For $x\ge0$, the least value of $K$ for which $4^{,1+x}+4^{,1-x},\ \dfrac{K}{2},\ 16^{x}+16^{-x}$ are three consecutive terms of an A.P. is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (5 April Evening Shift) PYQ

Solution

Qus : 4 JEE MAIN PYQ

Let $S=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\cdots$ (13 terms) If $13S=\frac{2^k}{n!}$, then $n+k$ is equal to:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Morning Shift) PYQ

Solution

$S=\frac{1}{26!}\left({}^{26}C_1+{}^{26}C_3+\cdots+{}^{26}C_{25}\right)$

$= \frac{2^{25}}{26!}$

$13S=\frac{2^{24}}{25!}$

$\Rightarrow n=25,; k=24$

$n+k=49$

Qus : 5 JEE MAIN PYQ

$ \left( \frac{1}{3} + \frac{4}{7} \right) + \left( \frac{1}{3^2} + \frac{1}{3} \times \frac{4}{7} + \frac{4^2}{7^2} \right) + \left( \frac{1}{3^3} + \frac{1}{3^2} \times \frac{4}{7} + \frac{1}{3} \times \frac{4^2}{7^2} + \frac{4^3}{7^3} \right) + \ldots \text{ upto infinite terms is equal to }$

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (24 January Evening Shift) PYQ

Solution

Let $ a = \frac{4}{7},; b = \frac{1}{3} $

Multiply $ N $ and $ D $ by $ (a - b) = \frac{4}{7} - \frac{1}{3} = \frac{5}{21} $

$ \frac{1}{a - b} \left[ (a^2 - b^2) + (a^3 - b^3) + (a^4 - b^4) + \ldots \right] $

$ = \frac{1}{a - b} \left[ \frac{a^2}{1 - a} - \frac{b^2}{1 - b} \right] $

$ = \frac{21}{5} \left[ \frac{16/49}{1 - 4/7} - \frac{1/9}{1 - 1/3} \right] $

$ = \frac{21}{5} \left[ \frac{16/49}{3/7} - \frac{1/9}{2/3} \right] $

$ = \frac{21}{5} \left[ \frac{16}{21} - \frac{1}{6} \right] $

$ = \frac{21}{5} \cdot \frac{96 - 21}{126} $

$ = \frac{21}{5} \cdot \frac{75}{126} $

$ = \frac{15}{6} = \frac{5}{2} $

Qus : 6 JEE MAIN PYQ

The value of $ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{k(k+1)}{k!} $ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Morning Shift) PYQ

Solution

$ T_k = (-1)^{k+1} \frac{k(k+1)}{k!} = (-1)^{k+1} \frac{k(k-1) + 2k}{k!} $ $ = (-1)^{k+1} \left[ \frac{k(k-1)}{k!} + \frac{2k}{k!} \right] $ $ = (-1)^{k+1} \left[ \frac{1}{(k-2)!} + \frac{2}{(k-1)!} \right] $ $ \Rightarrow \sum = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{(k-2)!} + 2 \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{(k-1)!} $ $ = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} = \frac{1}{e} $

Qus : 7 JEE MAIN PYQ

The number of real roots of the equation ${e^{4x}} + 2{e^{3x}} - {e^x} - 6 = 0$ is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (31 August Morning Shift) PYQ

Solution

Qus : 8 JEE MAIN PYQ

The number of solutions of the equation $e^{\sin x}-2e^{-\sin x}=2$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (31 January Evening Shift) PYQ

Solution

Qus : 9 JEE MAIN PYQ

The sum of all real values of $x$ satisfying the equation $(x^{2} - 5x + 5)^{x^{2} + 4x - 60} = 1$ is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2016 (Offline) PYQ

Solution

Qus : 10 JEE MAIN PYQ

Let $S=\Big\{x\in\mathbb{R}:(\sqrt3+\sqrt2)^{x}+(\sqrt3-\sqrt2)^{x}=10\Big\}$. Then the number of elements in $S$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2024 (1 February Morning Shift) PYQ

Solution

Qus : 11 JEE MAIN PYQ

$ \frac{6}{3^{26}} + \frac{10.1}{3^{25}} + \frac{10.2}{3^{24}} + \frac{10.2^2}{3^{23}} + \ldots + \frac{10.2^{24}}{3} $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ S = \frac{6}{3^{26}} + \frac{10}{3^{25}}\left[ \frac{(6)^{25} - 1}{6 - 1} \right] $

$ S = \frac{6}{3^{26}} + \frac{10}{3^{25}} \cdot \frac{6^{25} - 1}{5} $

$ S = \frac{2}{3^{25}} + 2\left[ 2^{25} - \frac{1}{3^{25}} \right] $

$ S = 2^{26} $

Qus : 12 JEE MAIN PYQ

$ \sum_{r=1}^{25} \frac{r}{r^2 + r + 1} = \frac{p}{q} $, where $ p $ and $ q $ are integers such that $ \gcd(p, q) = 1 $, then $ p + q $ is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2026 (28 January Evening Shift) PYQ

Solution

$ S = \sum \frac{r}{(r^2 + r + 1)} $

$ = \frac{1}{2} \sum \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) $

$ = \frac{1}{2} \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{7}\right) + \ldots + \left(\frac{1}{601} - \frac{1}{651}\right) \right] $

$ = \frac{1}{2}\left(1 - \frac{1}{651}\right) = \frac{325}{651} $

$ \Rightarrow p = 325,; q = 651 \Rightarrow p + q = 976 $

Qus : 13 JEE MAIN PYQ

$ \text{The equation } e^{4x}+8e^{3x}+13e^{2x}-8e^{x}+1=0,\ x\in\mathbb{R}\ \text{ has:} $

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (31 January Evening Shift) PYQ

Solution

Qus : 14 JEE MAIN PYQ

The sum $1+\dfrac{1+3}{2!}+\dfrac{1+3+5}{3!}+\dfrac{1+3+5+7}{4!}+\cdots$ up to $\infty$ terms is equal to

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2025 (3 April Evening Shift) PYQ

Solution

Qus : 15 JEE MAIN PYQ

The number of real roots of the equation ${e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0$ is :

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (25 July Morning Shift) PYQ

Solution

Qus : 16 JEE MAIN PYQ

The value of $\displaystyle \frac{1}{1\cdot 50!}+\frac{1}{3\cdot 48!}+\frac{1}{5\cdot 46!}+\cdots+\frac{1}{49\cdot 2!}+\frac{1}{51\cdot 1!}$ is:

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2023 (1 February Morning Shift) PYQ

Solution

Qus : 17 JEE MAIN PYQ

The lowest integer which is greater than ${\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$ is ______________.

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JEE MAIN JEE Mains PYQ JEE MAIN JEE Main 2021 (25 July Evening Shift) PYQ

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