SECTION - B [MATHS - NUMERIC] - JEE STD 11 Science Questions

Question 14 Marks

For $\mathrm{n} \geq 2$, let $S_{n}$ denote the set of all subsets of $\{1,2 \ldots \ldots ., n\}$ with no two consecutive numbers. For example $\{1,3,5\} \in \mathrm{S}_{6}$, but $\{1,2,4\} \notin \mathrm{S}_{6}$. Then $n\left(S_{5}\right)$ is equal to __________.

Answer

13
$\mathrm{A}=\{1,2,3,4,5 \ldots \ldots \mathrm{n}\}$
No. of subsets having $r$ elements such that no two are consecutive is $={ }^{n-r+1} C_{r}$
for $\mathrm{n}=5$, no. of ways $={ }^{6-r} \mathrm{C}_{\mathrm{r}}$
Subsets having no element $=1$
Subsets having exactly 1 element $={ }^{5} \mathrm{C}_{1}=5$
Subsets having exactly 2 element $={ }^{4} \mathrm{C}_{2}=6$
Subsets having exactly 3 element $={ }^{3} \mathrm{C}_{3}=1$
$\Rightarrow 5+6+1+1=13$

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Question 24 Marks

The number of singular matrices of order 2, whose elements are from the set $\{2,3,6,9\}$ is

Answer

36
$\left|\begin{array}{ll}\mathrm{a} & \mathrm{d} \\ \mathrm{b} & \mathrm{c}\end{array}\right|=\mathrm{ad}-\mathrm{bc} \Rightarrow \mathrm{ad}=\mathrm{bc}$
Case-I Exactly 1 no. is used
$\Rightarrow$ All singular $={ }^{4} C_{1}$
Case-II Exactly 2 no. is used
$\Rightarrow{ }^{4} \mathrm{C}_{2} \times 2 \times 2$
Case-III Exactly 3 no. is used
None will be singular
Case-IV Exactly 4 No. is used
$\mathrm{ad}=\mathrm{bc}$
$\Rightarrow 2 \times 9=3 \times 6$
$\left|\begin{array}{cc}9 & - \\ - & 2\end{array}\right| \Rightarrow{ }^{4} \mathrm{C}_{1} \times 21$
Total $=36$

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Question 34 Marks

Consider the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ having one of its focus at $\mathrm{P}(-3,0)$. If the latus ractum through its other focus subtends a right angle at P and $a^{2} b^{2}=\alpha \sqrt{2}-\beta, \alpha, \beta \in \mathbb{N}$.

Answer

1994
$\mathrm{f}_{1} \equiv(-\mathrm{ae}, 0) \equiv \mathrm{P}(-3,0)$
$\Rightarrow \mathrm{ae}=3$

$\tan 45^{\circ}=\frac{\mathrm{b}^{2} / \mathrm{a}}{2 \mathrm{ae}}$
$2 \mathrm{a}=\frac{\mathrm{b}^{2}}{\mathrm{a}}$
$b^{2}=6 \mathrm{a}$
Also $\mathrm{a}^{2} \mathrm{e}^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}$
$9=a^{2}+6 a$
$a^{2}+6 a-9=0$
$a=-3 \pm 3 \sqrt{2}=-3(1 \pm \sqrt{2})$
$\therefore \quad a^{2} b^{2}=a^{2} .6 a=6 a^{3}$
$=6(135 \sqrt{2}-189)$
$\alpha=810$ and $\beta=1134$
$\therefore \alpha+\beta=1944$

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Question 44 Marks

The number of relations on the set $\mathrm{A}=\{1,2,3\}$ containing at most 6 elements including (1, 2), which are reflexive and transitive but not symmetric, is __________.

Answer

5
$A=\{1,2,3\}$
$(1,1),(2,2),(3,3),(1,2) \in R$
Remaining elements are
$(2,1),(2,3),(1,3),(3,1),(3,2)$
(1) If relation contains exactly 4 elements $=1$ way
(2) if relation contains exactly 5 elements
It can be $(1,3),(3,2) \Rightarrow 2$ ways
(3) If relation contain exactly 6 elements
It can be
$((2,3),(1,3)),((1,3),(3,2)),((3,1),(3,2))$
$\Rightarrow 3$ ways.
Total $=6$ ways

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Question 54 Marks

The number of points of discontinuity of the function $f(\mathrm{x})=\left[\frac{\mathrm{x}^{2}}{2}\right]-[\sqrt{\mathrm{x}}], \mathrm{x} \in[0,4]$, where $[\cdot]$ denotes the greatest integer function is __________.

Answer

8
Check for $\left[\frac{x^{2}}{2}\right]$ and $[\sqrt{x}]$ becomes integers. $\{0,1, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4\}$
Continuous at $0^{+}$, continuous at $4^{-}$ $\left[\frac{x^{2}}{2}\right]=[\sqrt{x}]$, occurs at $x=\sqrt{2}$
$\Rightarrow$ Not continuous
$\therefore$ function is discontinuous at 8 points.

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JEE SECTION - B [MATHS - NUMERIC]

SECTION - B [MATHS - NUMERIC]

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