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

Question 14 Marks

Let $y ^2=12 x$ the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP) $(S Q)=\frac{147}{4}$. Let C be the circle described taking PQ as a diameter. If the equation of a circle $C$ is $64 x^2+64 y^2-\alpha x-64 \sqrt{3} y=\beta$, then $\beta-\alpha$ is equal to _________

Answer

(1328)
Sol. $\quad y^2=12 x \quad a=3 \quad S P \times S Q=\frac{147}{4}$
Let $P \left(3 t ^2, 6 t \right)$ and $t _1 t _2=-1$
(ends of focal chord)
So, $Q\left(\frac{3}{t^2}, \frac{-6}{t}\right)$
$
\begin{array}{l}
S(3,0)\\
SP \times SQ=PM_1 \times QM_2
\end{array}
$
(dist. from directrix)
$
\begin{array}{l}
=\left(3+3 t^2\right)\left(3+\frac{3}{t^2}\right)=\frac{147}{4} \\
\Rightarrow \frac{\left(1+t^2\right)^2}{t^2}=\frac{49}{12} \\
t^2=\frac{3}{4}, \frac{4}{3} \\
t= \pm \frac{\sqrt{3}}{2}, \pm \frac{2}{\sqrt{3}} \\
\text { considering } t=\frac{-\sqrt{3}}{2}
\end{array}
$
$
P\left(\frac{9}{4},-3 \sqrt{3}\right) \text { and } Q(4,4 \sqrt{3})
$
Hence, diametric circle:
$\begin{array}{l}
(x-4)\left(x-\frac{9}{4}\right)+(y+3 \sqrt{3})(y-4 \sqrt{3})=0 \\
\Rightarrow x^2+y^2-\frac{25}{4} x-\sqrt{3} y-27=0 \\
\Rightarrow \alpha=400, \beta=1728 \\
\beta-\alpha=1328
\end{array}
$

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Question 24 Marks

Let integers $a , b \in[-3,3]$ be such that $a + b \neq 0$. Then the number of all possible ordered pairs (a, b), for which $\left|\frac{z-a}{z+b}\right|=1$ and $\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|$$=1, z \in C$, where $\omega$ and $\omega^2$ are the roots of $x^2+x+$ $1=0$, is equal to _________

Answer

(10)
Sol.
$
\begin{array}{l}
a, b \in I,-3 \leq a, b \leq 3, a+b \neq 0 \\
|z-a|=|z+b| \\
\left|\begin{array}{ccc}
z+1 & \omega & \omega^2 \\
\omega & z+\omega^2 & 1 \\
\omega^2 & 1 & z+\omega
\end{array}\right|=1 \\
\Rightarrow\left|\begin{array}{ccc}
z & z & z \\
\omega & z+\omega^2 & 1 \\
\omega^2 & 1 & z+\omega
\end{array}\right|=1 \\
\Rightarrow\left|\begin{array}{ccc}
1 & 1 & 1 \\
\omega & z+\omega^2 & 1 \\
\omega^2 & 1 & z+\omega
\end{array}\right|=1
\end{array}
$
$\begin{array}{l}
\Rightarrow z\left|\begin{array}{ccc}
1 & 0 & 0 \\
\omega & z+\omega^2-\omega & 1-\omega \\
\omega^2 & 1-\omega^2 & z+\omega-\omega^2
\end{array}\right|=1 \\
\Rightarrow z^3=1 \\
\Rightarrow z=\omega, \omega^2, 1
\end{array}
$
Now
$
\begin{array}{l}
|1-a|=|1+b| \\
\Rightarrow 10 \text { pairs }
\end{array}
$

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Question 34 Marks

Let $a_1, a_2, \ldots, a_{204}$ be an Arithmetic Progression such that $a _1+\left( a _5+ a _{10}+ a _{19}+\ldots+ a _{2000}\right)+ a _{2254}=$ 2233. Then $a_1+a_2+a_3+\ldots+a_{3034}$ is equal to _________

Answer

(11132)
Sol. $a_1+a_5+a_{10}+\ldots \ldots+a_{2000}+a_{2024}=2233$
In an A.P. the sum of terms equidistant from ends is equal.$
\begin{array}{l}
a_1+a_{204}=a_5+a_{3000}=a_{10}+a_{2015} \ldots \ldots \\
\Rightarrow 203 \text { pairs } \\
\Rightarrow 203\left(a_1+a_{304}\right)=2233
\end{array}
$
Hence,
$
\begin{array}{l}
S_{2124}=\frac{2024}{2}\left(a_1+a_{2024}\right) \\
=1012 \times 11 \\
=11132
\end{array}
$

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Question 44 Marks

If $\lim _{t \rightarrow 0}\left(\int_0^1(3 x+5)^t d x\right)^{\frac{1}{t}}=\frac{\alpha}{5 e }\left(\frac{8}{5}\right)^{\frac{2}{3}}$, then $\alpha$ is equal to _________ .

Answer

(64)
Sol. $1^{\infty}$ form
$
\begin{array}{l}
\text { Now } L=e^{t \rightarrow 0} \frac{1}{t}\left(\left.\frac{(3 x+5)^{t+1}}{3(t+1)}\right|_0 ^1-1\right) \\
=e^{t \rightarrow 0} \frac{8^{t+1}-5^{t+1}-3 t-3}{3 t(t+1)} \\
=e \frac{8 \ell n 8-5 \ell n 5-3}{3} \\
=\left(\frac{8}{5}\right)^{2 / 3}\left(\frac{64}{5}\right)=\frac{\alpha}{5 e}\left(\frac{8}{5}\right)^{2 / 3}
\end{array}
$
On comparing
$\alpha=64
$

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Question 54 Marks

If $24 \int_0^{\frac{\pi}{4}}\left(\sin \left|4 x-\frac{\pi}{12}\right|+[2 \sin x]\right) d x=2 \pi+\alpha$, where [.] denotes the greatest integer function, then $\alpha$ is equal to _________

Answer

(12)
$\begin{array}{l}Sol. = 24 \int_0^{\frac{\pi}{48}}-\sin \left(4 x-\frac{\pi}{12}\right)+\int_{\pi / 48}^{\pi / 4} \sin \left(4 x-\frac{\pi}{12}\right) \\ +\int_0^{\frac{\pi}{6}}[0] dx +\int_{\pi / 6}^{\pi / 4}[2 \sin x] dx \\ = 24\left[\frac{\left(1-\cos \frac{\pi}{12}\right)}{4}-\frac{\left(-\cos \frac{\pi}{12}-1\right)}{4}\right]+\frac{\pi}{4}-\frac{\pi}{6} \\ = 24\left(\frac{1}{2}+\frac{\pi}{12}\right)=2 \pi+12 \\ \alpha=12\end{array}$

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JEE SECTION - B [MATHS - NUMERIC]

SECTION - B [MATHS - NUMERIC]

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