Assertion (A) & Reason (B) MCQ

MCQ 11 Mark

Let $a, b \in R$ be such that the function $f$ given by $f(x)=\log |x|+b x^2+a x, x \neq 0$ has extreme values: at $x=-1$ and $x =2$.
Assertion (A): f has local maximum at $x=-1$ and at $x=2$
Reason (R): $a =\frac{1}{2}$ and $b =-\frac{1}{4}$

A
Both A and R are true and R is the correct explanation of A.
B
Both A and R are true but R is not the correct explanation of A.
C
A is true but R is false.
D
A is false but R is true.

Answer

(a) Both A and R are true and R is the correct explanation of A .
Explanation: $f ( x )=\log IxI + bx ^2+ ax , x \neq 0$
$
\Rightarrow f^{\prime}(x)=\frac{1}{x}+2 b x+a, x \neq 0
$
Given $x=-1$ and $x=2$ are extreme values of $f(x)$.
So, $f^{\prime}(-1)=0$ and $f^{\prime}(2)=0$
$
\Rightarrow-1-2 b+a=0 \text { and } \frac{1}{2}+4 b+a=0
$
Solving these equations, we get $a =\frac{1}{2}, b=-\frac{1}{4}$
$\therefore$ Reason is true.
Now, $f ^{\prime}( x ) \frac{1}{x}-\frac{1}{2} x+\frac{1}{2}=\frac{2-x^2+x}{2 x}$
$\Rightarrow f^{\prime}(x)=0 \Rightarrow x=-1$ and $x=2$
$
f^{\prime \prime}(x)=-\frac{1}{x^2}-\frac{1}{2} \Rightarrow f^{\prime \prime}(-1)=\frac{1}{1}-\frac{1}{2}=-\frac{3}{2}<0
$
$\Rightarrow x=-1$ is a point of local maxima.
Also $f^{\prime \prime}(2)=-\frac{1}{4}-\frac{1}{2}=-\frac{3}{4}<0$
$\Rightarrow x =2$ is a point of local maxima.
Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.

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MCQ 21 Mark

Assertion (A): If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]$, then $|3 A|=27|A|$
Reason (R): If $A$ is a square matrix of order $n$, then $|k A|=k^n|A|$.

A
Both A and R are true and R is the correct explanation of A.
B
Both A and R are true but R is not the correct explanation of A.
C
A is true but R is false.
D
A is false but R is true.

Answer

(a) Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
Explanation: We know that if $A$ is square matrix of order $n$, then $|k A|=k^n|A|$ (see properties)
$\therefore$ Reason is true.
$
\begin{array}{l}
\text { Given } A=\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right] \text { (Expand by } C_1 \text { ) } \\
\Rightarrow|A|=1(4-0)-0+0 \\
\Rightarrow|A|=4
\end{array}
$
Now, $3 A=\left[\begin{array}{rrr}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{array}\right]$ (Expand by $C_1$ )
$
\begin{array}{l}
\Rightarrow|3 A|=3(36-0)-0+0 \\
\Rightarrow|3 A|=108 \\
\Rightarrow|3 A|=27 \times 4 \Rightarrow|3 A|=27|A|
\end{array}
$
$\therefore$ Assertion is true.
Hence, both Assertion and Reason are true and Reason is the correct explanation of Assertion.

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Applied Maths Assertion (A) & Reason (B) MCQ

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