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

Question 14 Marks

Let $\overrightarrow{ c }$ be the projection vector of $\overrightarrow{ b }=\lambda \hat{ i }+4 \hat{ k }, \lambda>0$, on the vector $\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k}$. If $|\vec{a}+\vec{c}|=7$, then the area of the parallelogram formed by the vectors $\vec{b}$ and $\overrightarrow{ c }$ is ___________ .

Answer

16
$
\begin{array}{l}Sol.
\overrightarrow{c}=\left(\frac{\overrightarrow{b} \cdot \overrightarrow{a}}{|\overrightarrow{a}|}\right) \frac{\overrightarrow{a}}{|\overrightarrow{a}|} \\
=\left(\frac{\lambda+8}{9}\right)(\hat{i}+2 \hat{j}+2 \hat{k}) \\
|\overrightarrow{a}+\overrightarrow{c}|-=7 \Rightarrow \lambda=4
\end{array}
$
Area of parallelogram
$\begin{array}{l}=|\overrightarrow{ b } \times \overrightarrow{ c }|=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ \frac{4}{3} & \frac{8}{3} & \frac{8}{3} \\ 4 & 0 & 4\end{array}\right| \\ =16\end{array}$

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Question 24 Marks

Let $L_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and
$L_2: \frac{x-2}{2}-\frac{y}{0}-\frac{z+4}{\alpha}, \alpha \in R$, be two lines, which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A (1,1,-1)$ on $L _2$, then the value of $26 \alpha(PB)^2$ is ._________

Answer

216
$
\begin{array}{l}Sol. Point B\$3 \lambda+1,-\lambda+1,-1) \equiv(2 \mu+2,0, \alpha \mu-4) \\
3 \lambda+1=2 \mu+2 \\
-\lambda+1=0 \\
-1=\alpha \mu-4 \\
\lambda=1, \mu=1, \alpha=3 \\
B(4,0,-1)
\end{array}
$
Let Point 'P' is $(2 \delta+2,0,3 \delta-4)$
Dr's of AP $< 2 \delta+1,-1,3 \delta-3>$
$
\begin{array}{l}
AP \perp L_2 \Rightarrow \delta=\frac{7}{13} \\
P\left(\frac{40}{13}, 0, \frac{-31}{13}\right) \\
2 \sigma \delta(PB)^2=26 \times 3 \times\left(\frac{144}{169}+\frac{324}{169}\right) \\
=216
\end{array}
$

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Question 34 Marks

Let $A$ be a square matrix of order 3 such that $\operatorname{det}(A)=-2$ and $\operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A)))=2^{n+\pi} \cdot 3^{\operatorname{mn}}$, $m > n$. Then $4 m+2 n$ is equal to $\qquad$ .

Answer

34
$\begin{array}{l}Sol. | A |=-2 \\ \operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A))) \\ =\left.3^3 \operatorname{det}(\operatorname{adj}(-\operatorname{adj}(3 A)))\right) \\ = 3^3(-6)^6(\operatorname{det}(3 A))^4 \\ = 3^{21} \times 2^{10} \\ m+ n =10 \\ mn =21 \\ m=7 ; n =3\end{array}$

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Question 44 Marks

If $\sum_{ r =0}^5 \frac{{ }^{11} C _{2 r+1}}{2 r +2}=\frac{ m }{ n }, \operatorname{gcd}( m , n )=1$, then $m - n$ is equal to $\qquad$ .

Answer

2035
$
\begin{array}{l}
Sol.\int_0^1(1+x)^{11} d x=\left[C_0 x+\frac{C_1 x^2}{2}+\frac{C_2 x^3}{3}+\ldots\right]_0^1 \\
\frac{2^{12}-1}{12}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\frac{C_3}{4}+\ldots \\
\int_{-1}^0(1+x)^{11} d x=\left[C_0 x+\frac{C_1 x^2}{2}+\frac{C_2 x^3}{3}+\ldots\right]_{-1}^0 \\
\frac{1}{12}=C_0-\frac{C_1}{2}+\frac{C_2}{3}-\frac{C_3}{4}+\ldots \\
\frac{2^{12}-2}{12}=2\left(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\ldots\right) \\
\frac{C_1}{2}+\frac{C_3}{4}-\frac{C_3}{6}+\ldots=\frac{2^{11}-1}{12}=\frac{2047}{12}
\end{array}
$

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Question 54 Marks

Let the function,$
f(x)=\left\{\begin{array}{ll}
-3 a x^2-2, & x<1 \\
a^2+b x, & x \geq 1
\end{array}\right.
$
Be differentiable for all $x \in R$, where $a >1, b \in R$. If the area of the region enclosed by $y=f(x)$ and the line $y=-20$ is $\alpha+\beta \sqrt{3}, \alpha, \beta, \in Z$, then the value of $\alpha+\beta$ is $\qquad$

Answer

34
Sol. $f(x)$ is continuous and differentiable<br>$
\begin{array}{l}
\text { at } x=1 ; \quad LHL=RHL, LHD=RHD \\
-3 a-2=a^2+b,-6 a=b \\
a=2,1 ; b=-12 \\
f(x)=\left\{\begin{array}{cc}
-6 x^2-2 & ;
4<1 \\
4-12 x & ; x \geq 1
\end{array}\right.
\end{array}
$

$\begin{array}{l}\text { Area }=\int_{-\sqrt{3}}^1\left(-6 x^2-2+20\right) d x+\int_1^2(4-12 x+20) d x \\ 16+12 \sqrt{3}+6=22+12 \sqrt{3}\end{array}$

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JEE SECTION - B [MATHS - NUMERIC]

SECTION - B [MATHS - NUMERIC]

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