JEE MAIN 2022 Previous Year Questions (PYQs) – Page 9 of 31
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$
\text{Let } \alpha, \beta \text{ be the roots of the equation } x^{2} - \sqrt{2}x + \sqrt{6} = 0
\text{ and } \dfrac{1}{\alpha^{2}} + 1, ; \dfrac{1}{\beta^{2}} + 1 \text{ be the roots of the equation }
x^{2} + ax + b = 0.
$
$\text{Then the roots of the equation } x^{2} - (a+b-2)x + (a+b+2) = 0 \text{ are :}$
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Solution
If the shortest distance between the lines $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{\lambda}$ and $\dfrac{x-2}{1} = \dfrac{y-4}{4} = \dfrac{z-5}{5}$ is $\dfrac{1}{\sqrt{3}}$, then the sum of all possible values of $\lambda$ is:
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Solution
$S = \{\, x \in [-6,3] \setminus \{-2,2\} \;:\; \dfrac{|x+3|-1}{|x|-2} \geq 0 \,\}$
$T = \{\, x \in \mathbb{Z} \;:\; x^{2} - 7|x| + 9 \leq 0 \,\}$
Then the number of elements in $S \cap T$ is :
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Solution
If the shortest distance between the lines $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{\lambda}$ and $\dfrac{x-2}{1} = \dfrac{y-4}{4} = \dfrac{z-5}{5}$ is $\dfrac{1}{\sqrt{3}}$, then the sum of all possible values of $\lambda$ is:
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Solution
$ \text{Let } A \text{ and } B \text{ be any two } 3\times 3 \text{ symmetric and skew-symmetric matrices respectively. Then which of the following is NOT true?} $
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Solution
If $y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2} < {x^3} < {{3\pi } \over 2}$, then
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Solution
$ \text{Let } f(x)=ax^{2}+bx+c \text{ be such that } f(1)=3,\ f(-2)=\lambda \text{ and } f(3)=4. $
$ \text{If } f(0)+f(1)+f(-2)+f(3)=14,\ \text{then } \lambda \text{ is equal to:} $
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Solution
Let $\lambda^*$ be the largest value of $\lambda$ for which the function $f_\lambda(x) = 4\lambda x^3 - 36\lambda x^2 + 36x + 48$ is increasing for all $x \in \mathbb{R}$. Then $f_{\lambda^*}(1) + f_{\lambda^*}(-1)$ is equal to:
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Solution
The function $f : \mathbb{R} \to \mathbb{R}$ defined by
$$
f(x) = \lim_{n \to \infty} \frac{\cos(2 \pi x) - x^{2n} \sin(x-1)}{1 + x^{2n+1} - x^{2n}}
$$
is continuous for all $x$ in :
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Solution
The value of $\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {(1 + {{\cos }^2}x)({e^{\cos x}} + {e^{ - \cos x}})}}dx} $ is equal to:
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Solution
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