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

Question 14 Marks

The roots of the quadratic equation $3 x ^2- px + q =0$ are $10^{\text {th }}$ and $11^{\text {th }}$ terms of an arithmetic progression with common difference $\frac{3}{2}$. If the sum of the first 11 terms of this arithmetic progression is 88 , then $q -2 q$ is equal to __________ .

Answer

$\begin{array}{l}
S_{11}=\frac{11}{2}(2 a+10 d)=88 \\
a+5 d=8 \\
a=8-5 \times \frac{3}{2}=\frac{1}{2}
\end{array}
$
Roots are$
\begin{array}{l}
T_{10}=a+9 d=\frac{1}{2}+9 \times \frac{3}{2}=14 \\
T_{11}=a+10 d=\frac{1}{2}+10 \times \frac{3}{2}=\frac{31}{2} \\
\frac{p}{3}=T_{10}+T_{11}=14+\frac{31}{2}=\frac{59}{2}
\end{array}
$
$\begin{array}{l} p =\frac{177}{2} \\ \frac{ q }{3}= T _{10} \times T _{11}=7 \times 31=217 \\ q =651 \\ q -2 p \\ =651-177 \\ =474\end{array}$

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Question 24 Marks

The variance of the numbers $8,21,34,47, \ldots, 320$, is __________.

Answer

$\operatorname{Var}(8,21,34,47, \ldots \ldots ., 320)$
$\operatorname{Var}(0,13,26,39, \ldots . . ., 312)$
$13^2 . \operatorname{Var}(0,1,2, \ldots . . ., 24)$
$13^2 . \operatorname{Var}(1,2,3, \ldots . . ., 25)$
So, $\sigma^2=13^2 \times\left(\frac{25^2-1}{12}\right)=8788$
Alternate solution
$
\begin{array}{l}
8+(n-1) 13=320 \\
13 n=325 \\
n=25
\end{array}
$
$\begin{array}{l}\text { no. of terms }=25 \\ \text { mean }=\frac{\sum x _{ i }}{ n }=\frac{8+21+\ldots+320}{25}=\frac{\frac{25}{2}(8+320)}{25} \\ \text { variance } \sigma^2=\frac{\sum x _{ i }^2}{ n }-(\text { mean })^2 \\ =\frac{8^2+21^2+\ldots .+320^2}{13}-(164)^2 \\ =8788\end{array}$

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Question 34 Marks

The focus of the parabola $y^2=4 x+16$ is the centre of the circle C of radius 5 . If the values of $\lambda$, for which $C$ passes through the point of intersection of the lines $3 x-y=0$ and $x+\lambda y=4$, are $\lambda_1$ and $\lambda_2, \lambda_1<\lambda_2$, then $12 \lambda_1+29 \lambda_2$ is equal to __________.

Answer

$y^2=4(x+4)$
Equation of circle
$
(x+3)^2+y^2=25
$
Passes through the point of intersection of two lines $3 x-y=0$ and $x+\lambda y=4$ which is $\left(\frac{4}{3 \lambda+1}, \frac{12}{3 \lambda+1}\right)$, after solving with circle, we get
$
\begin{array}{l}
\lambda=-\frac{7}{6}, 1 \\
12 \lambda_1+29 \lambda_2 \\
-14+29=15
\end{array}
$

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Question 44 Marks

Let $\alpha, \beta$ be the roots of the equation $x^2-a x-b=0$ with $\operatorname{Im}(\alpha)<\operatorname{Im}(\beta)$. Let $P_n=\alpha^n-\beta^n$. If $P_3=-5 \sqrt{7} i, \quad P_4=-3 \sqrt{7} i, \quad P_5=11 \sqrt{7} i \quad$ and $P_6=45 \sqrt{7} i$, then $\left|\alpha^4+\beta^4\right|$ is equal to __________ .

Answer

$\begin{array}{l}
\alpha+\beta=a \quad \alpha \beta=-b \\
P_6=a_5+bP_4 \\
45 \sqrt{7} i=a \times 11 \sqrt{7} i+b(-3 \sqrt{7}) i \\
45=11 a-3 b
\end{array}
$
and
$
\begin{array}{l}
P_5=aP_4+bP_3 \\
11 \sqrt{7} i=a(-3 \sqrt{7} i)+b(-5 \sqrt{7} i) \\
11=-3 a-5 b \\
a=3, b=-4 \\
\left|\alpha^4+\beta^4\right|=\sqrt{\left(\alpha^4-\beta^4\right)^2+4 \alpha^4 \beta^4} \\
=\sqrt{-63+4.4^4} \\
=\sqrt{-63+1024}=\sqrt{961}=31
\end{array}
$

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Question 54 Marks

The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is __________.

Answer

A : number of ways that all boys sit together $=5!\times 5!$
B : number of ways if no 2 boys
sit together $=4!\times 5!$
$A \cap B =\phi$
Required no. of ways $=5!\times 5!+4!\times 5!=17280$

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JEE SECTION - B [MATHS - NUMERIC]

SECTION - B [MATHS - NUMERIC]

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