Matrices and Determinants Set… | JEE Advanced 2021 P2

JEE Advanced 2021 Paper 2 Mathematics: Paragraph 1

Let

A=\begin{pmatrix}3&-1\\1&1\end{pmatrix}

. Define sequences

(x_n)

and

(y_n)

\begin{pmatrix}x_n\\y_n\end{pmatrix}=A^{n}\begin{pmatrix}1\\0\end{pmatrix}

for

n\ge0

. It is known that the characteristic polynomial of

A

has a repeated root.

Question 1

Single correct option

The value of

y_{10}

Show answer and solution

Answer: (B) $5120$

The characteristic polynomial is

\lambda^2-4\lambda+4=(\lambda-2)^2

, so

A=2I+N

with

N=A-2I=\begin{pmatrix}1&-1\\1&-1\end{pmatrix}

and

N^2=0

. Therefore

A^n=2^nI+n2^{n-1}N

, and

A^{n}\binom10=\binom{2^n+n2^{n-1}}{n2^{n-1}}

. So

y_n=n2^{n-1}

and

y_{10}=10\cdot2^9=5120

. Option 6144 is

x_{10}

, which a student would obtain by reading the wrong component; 2560 comes from using

n2^{n-2}

, and 10240 from

n2^{n}

Single correct option

The smallest positive integer

n

for which

x_n>10^{4}

Show answer and solution

Answer: (B) $11$

From

A^n=2^nI+n2^{n-1}N

with

N=\begin{pmatrix}1&-1\\1&-1\end{pmatrix}

, the first component is

x_n=2^n+n2^{n-1}=2^{n-1}(n+2)

. Evaluate:

x_9=256\cdot11=2816

x_{10}=512\cdot12=6144

x_{11}=1024\cdot13=13312

. Since

x_n

is increasing and

x_{10}=6144<10^4<13312=x_{11}

, the smallest such

n

11

. Option

10

results from mistaking

y_{10}

-type growth, while

12

and

13

are larger than necessary.

Original question written in the JEE Advanced 2021 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake