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

Question 14 Marks

The sum of the series
$2 \times 1 \times{ }^{20} \mathrm{C}_{4}-3 \times 2 \times{ }^{20} \mathrm{C}_{5}+4 \times 3 \times{ }^{20} \mathrm{C}_{6}-5 \times 4 \times$ ${ }^{20} \mathrm{C}_{7}+\ldots+18 \times 17 \times{ }^{20} \mathrm{C}_{20}$, is equal to __________ .

Answer

34
$(1-\mathrm{x})^{20}={ }^{20} \mathrm{C}_{0}-{ }^{20} \mathrm{C}_{1} \mathrm{x}+{ }^{20} \mathrm{C}_{2} \mathrm{x}^{2} \ldots . .+{ }^{20} \mathrm{C}_{20} \mathrm{x}^{20}$
$\frac{(1-\mathrm{x})^{20}}{\mathrm{x}^{2}}=\frac{{ }^{20} \mathrm{C}_{0}}{\mathrm{x}^{2}}-\frac{{ }^{20} \mathrm{C}_{1}}{\mathrm{x}}+{ }^{20} \mathrm{C}_{2}-{ }^{20} \mathrm{C}_{3} \mathrm{x}+{ }^{20} \mathrm{C}_{4} \mathrm{x}^{2} \ldots$.
Diff twice and put $\mathrm{x}=1$
$=6-{ }^{20} \mathrm{C}_{1}(2)+\mathrm{A}$
$\mathrm{A}=40-6=34$

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Question 24 Marks

Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b , respectively, and one focus and the corresponding directrix of this hyperbola be $(-5,0)$ and $5 x+9=0$, respectively. If the product of the focal distances of a point $(\alpha, 2 \sqrt{5})$ on the hyperbola is $p$, then $4 p$ is equal to __________ .

Answer

189
$\mathrm{PF}_{1} \cdot \mathrm{PF}_{2}=(\mathrm{e} \alpha-\mathrm{a})(\mathrm{e} \alpha+\mathrm{a})$
$\mathrm{P}=\mathrm{e}^{2} \alpha^{2}-\mathrm{a}^{2}=\frac{25}{9} \cdot 9 \cdot \frac{9}{4}-9=\frac{189}{4}$
$\left.\begin{array}{c}a e=5 \\ \frac{a}{e}=9 / 5\end{array}\right\} \begin{gathered}a=3 \\ e=5 / 3 \\ b=4\end{gathered}$
$\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$
$\frac{\alpha^{2}}{9}-\frac{4(5)}{16}=1 \Rightarrow \alpha^{2}=9\left(\frac{36}{16}\right)$

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Question 34 Marks

For $t>-1$, let $\alpha_{t}$ and $\beta_{t}$ be the roots of the equation $\left((t+2)^{\frac{1}{7}}-1\right) x^{2}+\left((t+2)^{\frac{1}{6}}-1\right) x+\left((t+2)^{\frac{1}{21}}-1\right)=0$. If $\lim _{t \rightarrow-1^{+}} \alpha_{t}=a$ and $\lim _{t \rightarrow-1^{+}} \beta_{t}=b$, then $72(a+b)^{2}$ is equal to __________ .

Answer

98
$a+b=\lim _{t \rightarrow-1^{+}}(\alpha+\beta)=\lim _{t \rightarrow-1^{+}}-\frac{(t+2)^{\frac{1}{6}}-1}{(t+2)^{\frac{1}{7}}-1}$
let $\mathrm{t}+2=\mathrm{y}$
$a+b=\lim _{y \rightarrow 1^{+}} \frac{y^{1 / 6}-1}{y^{1 / 7}-1}=\frac{7}{6}$
$72(a+b)^{2}=72 \frac{49}{36}=98$

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Question 44 Marks

If $\int\left(\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}^{3}}\right)\left(\sqrt[23]{3 \mathrm{x}^{-24}+\mathrm{x}^{-26}}\right) \mathrm{dx}$ $=-\frac{\alpha}{3(\alpha+1)}\left(3 x^{\beta}+x^{\gamma}\right)^{\frac{\alpha+1}{\alpha}}+C, x>0$, $(\alpha, \beta, \gamma \in Z)$, where $C$ is the constant of integration, then $\alpha+\beta+\gamma$ is equal to __________ .

Answer

19
$\int\left(\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{x}^{4}}\right)\left(\frac{3}{\mathrm{x}}+\frac{1}{\mathrm{x}^{3}}\right)^{\frac{1}{23}} \mathrm{dx}$
using $t=\frac{3}{x}+\frac{1}{x^{3}} \Rightarrow d t=-3\left(\frac{1}{x^{2}}+\frac{1}{x^{4}}\right) d x$
$\int \frac{t^{1 / 23} d t}{-3}=\frac{t^{24 / 23}}{\left(\frac{24}{23}\right)(-3)}+C$
$\Rightarrow \alpha=23 \beta=-1 \gamma=-3$
$\alpha+\beta+\gamma=19$

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Question 54 Marks

If the function $f(x)=\frac{\tan (\tan x)-\sin (\sin x)}{\tan x-\sin x}$ is continuous at $\mathrm{x}=0$, then $f(0)$ is equal to __________ .

Answer

2
$\lim _{x \rightarrow 0} \frac{\frac{\tan (\tan x)-\tan x}{\tan ^{3} x} \frac{\tan { }^{3} x}{x^{3}}+\frac{\tan x-\sin x}{x^{3}}+\frac{\sin x-\sin (\sin x)}{\sin ^{3} x} \frac{\sin ^{3} x}{x^{3}}}{\frac{\tan x-\sin x}{x^{3}}}$ $=\frac{\frac{1}{3}+\frac{1}{2}+\frac{1}{6}}{\frac{1}{2}}=2$

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JEE SECTION - B [MATHS - NUMERIC]

SECTION - B [MATHS - NUMERIC]

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