For some a, b , let f(x)=left|begin{array}{ccc}a+frac{sin x}{x} & 1 & b a & 1+frac{sin x

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3 and 4 .Determinants and Matrices

medium

For some $a, b$, let $f(x)=\left|\begin{array}{ccc}a+\frac{\sin x}{x} & 1 & b \\ a & 1+\frac{\sin x}{x} & b \\ a & 1 & b+\frac{\sin x}{x}\end{array}\right|, \quad x \neq 0$, $\lim _{ x \rightarrow 0} f ( x )=\lambda+\mu a + vb$. Then $(\lambda+\mu+v)^2$ is equal to:

A$25$

B$9$

C$36$

D$16$

(JEE MAIN-2025)

Solution

$\lim _{x \rightarrow 0} f(x)=\left|\begin{array}{ccc}a+1 & 1 & b \\ a & 1+1 & b \\ a & 1 & b+1\end{array}\right|$
$=(a+1)(2(b+1)-b)+1(a b-a(b+1))+b a$
$=(a+1)(b+2)-a+a b$
$=b+a+2=\lambda+\mu a+v b$
$\lambda=2, \mu=1, v=1 \Rightarrow(\lambda+\mu+v)^2=16$

Standard 12

Mathematics

If $p{\lambda ^4} + q{\lambda ^3} + r{\lambda ^2} + s\lambda + t = $ $\left| {\,\begin{array}{*{20}{c}}{{\lambda ^2} + 3\lambda }&{\lambda – 1}&{\lambda + 3}\\{\lambda + 1}&{2 – \lambda }&{\lambda – 4}\\{\lambda – 3}&{\lambda + 4}&{3\lambda }\end{array}\,} \right|,$ the value of $t$ is

hard

(IIT-1981)

Let $A = \left[ {\begin{array}{*{20}{c}}
2&b&1 \\
b&{{b^2} + 1}&b \\
1&b&2
\end{array}} \right]$ where $b > 0$. Then the minimum value of $\frac{{\det \left( A \right)}}{b}$ is

hard

(JEE MAIN-2019)

The values of $x,y,z$ in order of the system of equations $3x + y + 2z = 3,$ $2x – 3y – z = – 3$, $x + 2y + z = 4,$ are

medium

The number of real values of $\lambda $ for which the system of linear equations $2x + 4y – \lambda z = 0$ ;$4x + \lambda y + 2z = 0$ ; $\lambda x + 2y+ 2z = 0$ has infinitely many solutions, is

hard

(JEE MAIN-2017)

The equation $\left| {\begin{array}{{20}{c}}{{{(1 + x)}^2}}&{{{(1 – x)}^2}}&{ – \,(2 + {x^2})}\\{2x + 1}&{3x}&{1 – 5x}\\{x + 1}&{2x}&{2 – 3x}\end{array}} \right|$ $+$ $\left| {\begin{array}{{20}{c}}{{{(1 + x)}^2}}&{2x + 1}&{x + 1}\\{{{(1 – x)}^2}}&{3x}&{2x}\\{1 – 2x}&{3x – 2}&{2x – 3}\end{array}} \right|$ $= 0$

normal

For some $a, b$, let $f(x)=\left|\begin{array}{ccc}a+\frac{\sin x}{x} & 1 & b \\ a & 1+\frac{\sin x}{x} & b \\ a & 1 & b+\frac{\sin x}{x}\end{array}\right|, \quad x \neq 0$, $\lim _{ x \rightarrow 0} f ( x )=\lambda+\mu a + vb$. Then $(\lambda+\mu+v)^2$ is equal to:

Solution

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Let $A = \left[ {\begin{array}{*{20}{c}}
2&b&1 \\
b&{{b^2} + 1}&b \\
1&b&2
\end{array}} \right]$ where $b > 0$. Then the minimum value of $\frac{{\det \left( A \right)}}{b}$ is