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

Question 14 Marks

Let $r$ be the radius of the circle, which touches x -axis at point $(\mathrm{a}, 0), \mathrm{a}<0$ and the parabola $\mathrm{y}^{2}=9 \mathrm{x}$ at the point $(4,6)$. Then $r$ is equal to __________

Answer

(30)

$(x-a)^{2}+(y-r)^{2}=r^{2}$
$(4-a)^{2}+(6-r)^{2}=r^{2}$
$16+a^{2}-8 a+36+r^{2}-12 r=r^{2}$
$a^{2}-8 a-12 r+52=0$
Tangent to parabola at $(4,6)$ is
$6.4=9 .\left(\frac{x+4}{2}\right)$ i.e. $3 x-4 y+12=0$
This is also tangent to the circle
$\therefore \mathrm{CP}=\mathrm{r}$
$\frac{3 a-4 r+12}{5}= \pm r$
$3 a +12=4 r \pm 5 r \left\{_{- r }^{ ar }\right.\quad\ldots(1)$
equation of circle is
$(x-a)^2+(y-r)^2=r^2$
satsty $P(4,6) \Rightarrow a^{2}-8 a-12 r+52=0\quad\ldots(2)$
From equation (1)
If $\mathrm{a}+4=3 \mathrm{r}$ then $\mathrm{a}=+6$ (rejected)
If $3 a+12=-r$ then $a=-14$ and $r=30$

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Question 24 Marks

The product of the last two digits of $(1919)^{1919}$ is __________

Answer

(63)
$(1919)^{1919}=(1920-1)^{1919}$
$={ }^{1919} \mathrm{C}_{0}(1920)^{1919}-{ }^{1919} \mathrm{C}_{1}(1920){ }^{1918}+\ldots .$.
$+{ }^{1919} \mathrm{C}_{1918}(1920){ }^{1}-{ }^{1919} \mathrm{C}_{1919}$
$=100 \lambda+1919 \times 1920-1$
$=100 \lambda+3684480-1$
$=100 \lambda+\ldots \ldots \ldots . .79$ (last two digit)
$\Rightarrow$ Number having last two digit 79
$\therefore$ Product of last two digit 63

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Question 34 Marks

Let the area of the triangle formed by the lines $x+2=y-1=z, \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}$ and $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}$ be A. Then $A^{2}$ is equal to __________

Answer

(56)
$\mathrm{L}_{1}: \mathrm{x}+2=\mathrm{y}-1=\mathrm{z}=\ell$
$\mathrm{L}_{2}: \frac{\mathrm{x}-3}{5}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}=\mathrm{m}$
$L_{3}: \frac{\mathrm{x}}{-3}=\frac{\mathrm{y}-3}{5}=\frac{\mathrm{z}-2}{1}=\mathrm{n}$
Point of intersection of $L_{1}$ and $L_{2}$
$\left.\begin{array}{r}\ell-2=5 \mathrm{~m}+3 \\ \ell+1=-\mathrm{m} \\ \ell=\mathrm{m}+1\end{array}\right\} \ell=0, \mathrm{~m}=-1 \quad \mathrm{~A}(-2,1,0)$
Point of intersection of $L_{2}$ and $L_{3}$
$\left.\begin{array}{l}5 m+3=-3 n \\ -m=3 n+3 \\ m+1=n+2\end{array}\right\} m=0, n=-1, B(3,0,1)$
Point of intersection $L_{3}$ and $L_{4}$
$\left.\begin{array}{r}-3 \mathrm{n}=\ell-2 \\ 3 \mathrm{n}+3=\ell+1 \\ \mathrm{n}+2=\ell\end{array}\right\} \ell=2, \mathrm{n}=0, \mathrm{C}(0,3,2)$

$\mathrm{A}=\frac{1}{2}|\hat{\mathrm{i}}(4)-\hat{\mathrm{j}}(-8)+\hat{\mathrm{k}}(-12)|$
$A=\frac{1}{2} \sqrt{16+64+144}=\sqrt{56}$
$\mathrm{A}^{2}=56$

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Question 44 Marks

Let the domain of the function
$f(x)=\cos ^{-1}\left(\frac{4 x+5}{3 x-7}\right)$ be $[\alpha, \beta]$ and the domain of
$\mathrm{g}(\mathrm{x})=\log _{2}\left(2-6 \log _{27}(2 \mathrm{x}+5)\right)$ be $(\gamma, \delta)$.
Then $|7(\alpha+\beta)+4(\gamma+\delta)|$ is equal to __________

Answer

(96)
$f(x)=\cos ^{-1}\left(\frac{4 x+5}{3 x-7}\right)$
$\Rightarrow-1 \leq\left(\frac{4 x+5}{3 x-7}\right) \leq 1$
$\left(\frac{4 x+5}{3 x-7}\right) \geq-1$
$\frac{4 \mathrm{x}+5+3 \mathrm{x}-7}{3 \mathrm{x}-7} \geq 0$
$\Rightarrow \frac{7 \mathrm{x}-2}{3 \mathrm{x}-7} \geq 0$

$\mathrm{x} \in\left(-\infty, \frac{2}{7}\right] \cup\left(\frac{7}{3}, \infty\right)$
$\& \frac{4 x+5}{3 x-7} \leq 1 \Rightarrow \frac{x+12}{3 x-7} \leq 0$

$\therefore$ Domain of $\mathrm{f}(\mathrm{x})$ is
$\left[-12, \frac{2}{7}\right] \alpha=-12, \beta=\frac{2}{7}$
$\mathrm{g}(\mathrm{x})=\log _{2}\left(2-6 \log _{27}(2 \mathrm{x}+5)\right)$
Domain
$2-6 \log _{27}(2 x+5)>0$
$\Rightarrow \quad 6 \log _{27}(2 x+5)<2$
$\Rightarrow \quad \log _{27}(2 x+5)<\frac{1}{3}$
$\Rightarrow \quad 2 \mathrm{x}+5<3$
$\Rightarrow \quad x<-1$
$\& 2 x+5>0 \Rightarrow x>-\frac{5}{2}$
Domain is $x \in\left(-\frac{5}{2},-1\right)$
$\gamma=-\frac{5}{2}, \delta=-1$
$|7(\alpha+\beta)+4(\gamma+\delta)|=\left\lvert\, 7\left(\left.-12+\frac{2}{7}+4\left(-\frac{5}{2}-1\right) \right\rvert\,\right.\right.$
$|-82-14|=96$

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Question 54 Marks

Let the area of the bounded region $\left\{(x, y): 0 \leq 9 x \leq y^{2}, y \geq 3 x-6\right\}$ be A. Then $6 A$ is equal to __________

Answer

(15)
$0 \leq 9 x \leq y^{2} \& y \geq 3 x-6$

$A=$ Required Area $=\left[\int_{0}^{1}(-3 \sqrt{x}) d x-\int_{0}^{1}(3 x-6) d x\right]$
$A=-\left.3\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)\right|_{0} ^{1}-\left.\left(\frac{3 x^{2}}{2}-6 x\right)\right|_{0} ^{1}$
$\mathrm{A}=-2[1-0]\left[\frac{3}{2}-6\right]$
$\mathrm{A}=-2-\frac{3}{2}+6=\frac{5}{2}$ Sq. unit
$\therefore 6 \mathrm{~A}=6 \times \frac{5}{2}=15$

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JEE SECTION - B [MATHS - NUMERIC]

SECTION - B [MATHS - NUMERIC]

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