JEE MAIN Mathematics Previous Year Questions (PYQs) – Page 8 of 275
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Let $R=\{(1,2),(2,3),(3,3)\}$ be a relation on the set $\{1,2,3,4\}$. The minimum number of ordered pairs that must be added to $R$ so that it becomes an equivalence relation is:
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Solution
Let the sum of the focal distances of the point P(4,3) on the hyperbola $H:\ \dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ be $8\sqrt{\dfrac{5}{3}}$. If for H, the length of the latus rectum is l and the product of the focal distances of the point P is m, then $9l^{2}+6m$ is equal to:
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Solution
Given sum of focal distances:
$SP_1 + SP_2 = 2ex = 8\sqrt{\frac{5}{3}}$
At $P(4,3)$ ⇒ $2e \cdot 4 = 8\sqrt{\frac{5}{3}}$
⇒ $e=\sqrt{\frac{5}{3}}$
Now, $b^2 = a^2(e^2 - 1) = a^2\left(\frac{5}{3}-1\right)=\frac{2}{3}a^2$
Using point $(4,3)$: $\frac{16}{a^2}-\frac{9}{b^2}=1$
Substitute $b^2=\frac{2}{3}a^2$:
$\frac{16}{a^2}-\frac{9}{(2/3)a^2}=1$
⇒ $\frac{16}{a^2}-\frac{27}{2a^2}=1$
⇒ $\frac{5}{2a^2}=1$
⇒ $a^2=\frac{5}{2},\ b^2=\frac{5}{3}$
Latus rectum: $l=\frac{2b^2}{a}$ ⇒ $l^2=\frac{4b^4}{a^2}$
⇒ $l^2=\frac{4(25/9)}{5/2}=\frac{40}{9}$
⇒ $9l^2=40$
Product of focal distances: $m=(ex+a)(ex-a)=e^2x^2-a^2$
⇒ $m=\frac{5}{3}\cdot16-\frac{5}{2}=\frac{145}{6}$
⇒ $6m=145$
$9l^2+6m=40+145=185$
$\boxed{185}$
At $P(4,3)$ ⇒ $2e \cdot 4 = 8\sqrt{\frac{5}{3}}$
⇒ $e=\sqrt{\frac{5}{3}}$
Now, $b^2 = a^2(e^2 - 1) = a^2\left(\frac{5}{3}-1\right)=\frac{2}{3}a^2$
Using point $(4,3)$: $\frac{16}{a^2}-\frac{9}{b^2}=1$
Substitute $b^2=\frac{2}{3}a^2$:
$\frac{16}{a^2}-\frac{9}{(2/3)a^2}=1$
⇒ $\frac{16}{a^2}-\frac{27}{2a^2}=1$
⇒ $\frac{5}{2a^2}=1$
⇒ $a^2=\frac{5}{2},\ b^2=\frac{5}{3}$
Latus rectum: $l=\frac{2b^2}{a}$ ⇒ $l^2=\frac{4b^4}{a^2}$
⇒ $l^2=\frac{4(25/9)}{5/2}=\frac{40}{9}$
⇒ $9l^2=40$
Product of focal distances: $m=(ex+a)(ex-a)=e^2x^2-a^2$
⇒ $m=\frac{5}{3}\cdot16-\frac{5}{2}=\frac{145}{6}$
⇒ $6m=145$
$9l^2+6m=40+145=185$
$\boxed{185}$
Let $A =
\begin{bmatrix}
2 & b & 1 \\
b & b^2 + 1 & b \\
1 & b & 2
\end{bmatrix}$ where $b > 0$.
Then the minimum value of $\dfrac{\det(A)}{b}$ is –
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Solution
Let $a \in \left(0, \dfrac{\pi}{2}\right)$ be fixed.
If
$\displaystyle \int \dfrac{\tan x + \tan a}{\tan x - \tan a} , dx = A(x)\cos 2a + B(x)\sin 2a + C,$
where $C$ is a constant of integration,
then the functions $A(x)$ and $B(x)$ are respectively:
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Solution
If $\alpha,\beta \ne 0$, and $f(n) = \alpha^{n} + \beta^{n}$ and
$\begin{vmatrix}
3 & 1 + f(1) & 1 + f(2) \\
1+f(1) & 1 + f(2) & 1 + f(3) \\
1+f(2) & 1 + f(3) & 1 + f(4)
\end{vmatrix}
= K(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$, then $K$ is equal to :
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Solution
Let $A = {2,3,5,7,9}$. Let $R$ be the relation on $A$ defined by $xRy$ if and only if $2x \le 3y$. Let $\ell$ be the number of elements in $R$, and $m$ be the minimum number of elements required to be added in $R$ to make it a symmetric relation. Then $\ell + m$ is equal to:
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Solution
$A = {2,3,5,7,9}$
$y \ge \frac{2x}{3}$
$x = 2 \Rightarrow y = 2,3,5,7,9$
$x = 3 \Rightarrow y = 2,3,5,7,9$
$x = 5 \Rightarrow y = 5,7,9$
$x = 7 \Rightarrow y = 5,7,9$
$x = 9 \Rightarrow y = 7,9$
$\ell = 18$
To make symmetric, elements to be added:
${(5,2), (7,2), (9,2), (5,3), (7,3), (9,3), (9,5)}$
$m = 7$
$\ell + m = 25$
If {p} denotes the fractional part of the number p, then $\left\{ {{{{3^{200}}} \over 8}} \right\}$, is equal to :
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Solution
Let $\overrightarrow a = \widehat i + \widehat j + 2\widehat k$ and $\overrightarrow b = - \widehat i + 2\widehat j + 3\widehat k$. Then the vector product $\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\left( {\overrightarrow a \times \left( {\left( {\overrightarrow a - \overrightarrow b } \right) \times \overrightarrow b } \right)} \right) \times \overrightarrow b } \right)$ is equal to :
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Solution
If the maximum value of $a$, for which the function $f_a(x)=\tan^{-1}(2x)-3ax+7$ is non-decreasing in $\left(-\tfrac{\pi}{6},\,\tfrac{\pi}{6}\right)$, is $\bar a$, then $f_{\bar a}\!\left(\tfrac{\pi}{8}\right)$ is equal to :
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Solution
Let $P(S)$ denote the power set of $S=\{1,2,3,\ldots,10\}$. Define the relations $R_{1}$ and $R_{2}$ on $P(S)$ as
$A\,R_{1}\,B \iff (A\cap B^{c})\cup(B\cap A^{c})=\varnothing$ and
$A\,R_{2}\,B \iff A\cup B^{c}=B\cup A^{c}$, for all $A,B\in P(S)$. Then:
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Solution
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