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

Question 14 Marks

Let $A =\{1,2,3\}$. The number of relations on A, containing (1, 2) and (2, 3), which are reflexive and transitive but not symmetric, is _________.

Answer

(3)
Sol. Transitivity
$
(1,2) \in R,(2,3) \in R \Rightarrow(1,3) \in R
$
For reflexive $(1,1),(2,2)(3,3) \in R$
Now $(2,1),(3,2),(3,1)$
$(3,1)$ cannot be taken
(1) $(2,1)$ taken and $(3,2)$ not taken
(2) $(3,2)$ taken and $(2,1)$ not taken
(3) Both not taken
therefore 3 relations are possible.

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Question 24 Marks

If $\sum_{ r =1}^{30} \frac{ r ^2\left({ }^{30} C _{ r }\right)^2}{{ }^{30} C _{r-1}}=\alpha \times 2^{29}$, then $\alpha$ is equal to _________ .

Answer

Sol.

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Question 34 Marks

Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then $(Q R)^2$ is equal to _________.

Answer

(28)
Sol.

$
PR=\operatorname{cosec} \theta, PQ=4 \sec (30+\theta)
$
For equilateral
$
\begin{array}{l}
d=PR=PQ \\
\Rightarrow \cos \left(\theta+30^{\circ}\right)=4 \sin \theta \\
\Rightarrow \frac{\sqrt{3}}{2} \cos \theta-\frac{1}{2} \sin \theta=4 \sin \theta \\
\Rightarrow \tan \theta=\frac{1}{3 \sqrt{3}} \\
QR^2=d^2=\operatorname{cosec}^2 \theta=28
\end{array}
$

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Question 44 Marks

Let $A (6,8), B (10 \cos \alpha,-10 \quad \sin \alpha)$ and C $(-10 \sin \alpha, 10 \cos \alpha)$, be the vertices of a triangle.If $L(a, 9)$ and $G(h, k)$ be its orthocenter and centroid respectively, then $(5 a-3 h+6 k+100 \sin 2 \alpha)$ is equal to _________.

Answer

(145)
Sol. All the three points $A , B , C$ lie on the circle $x^2+y^2=100$ so circumcentre is $(0,0)$

$
\begin{array}{l}
\frac{a+0}{3}=h \Rightarrow a=3 h \\
\text { and } \frac{9+0}{3}=k \Rightarrow k=3
\end{array}
$
also centroid $\frac{6+10 \cos \alpha-10 \sin \alpha}{3}= h$
$
\begin{array}{l}
\Rightarrow 10(\cos \alpha-\sin \alpha)-3 h-6 ...(i)\\
\text { and } \frac{8+10 \cos \alpha-10 \sin \alpha}{3}=k \\
\Rightarrow 10(\cos \alpha-\sin \alpha)=3 k-8=9-8=1\text { ...(ii)} \\
\text { on squaring } 100(1-\sin 2 \alpha)=1 \\
\Rightarrow 100 \sin 2 \alpha=99
\end{array}
$
from equ. (i) and (ii) we get $h =\frac{7}{3}$
$
\begin{array}{l}
\text { Now } 5 a-3 h+6 k+100 \sin 2 \alpha \\
=15 h-3 h+6 k+100 \sin 2 \alpha \\
=12 \times \frac{7}{3}+18+99 \\
=145
\end{array}
$

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Question 54 Marks

Let $y=f(x)$ be the solution of the differential equation $\frac{d y}{d x}+\frac{x y}{x^2-1}=\frac{x^6+4 x}{\sqrt{1-x^2}}$, -1 < x < 1 such that $f(0)=0$. If $6 \int_{-1 / 2}^{1 / 2} f(x) d x=2 \pi-\alpha$ then $\alpha^2$ is equal to _____________.

Answer

(27)

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JEE SECTION - B [MATHS - NUMERIC]

SECTION - B [MATHS - NUMERIC]

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