Exercise 2.13 - Chapter 2 Basic Algebra 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 2.13
Choose the correct or the most suitable questions.
Question 1 .
If $|x+2| \leq 9$, then $x$ belongs to
(a) $(-\infty,-7)$
(b) $[-11,7]$
(c) $(-\infty,-7) \cup[11, \infty)$
(d) $(-11,7)$
Solution:
(b) $[-11,7]$
Hint:
$
\begin{aligned}
& -x-2 \leq 9 x+2 \leq 9 \\
& -x<9+2=11 \mathrm{x} \leq 9-2=7 \\
& \Rightarrow x \geq-11 \\
& \text { so } x \in[-11,7]
\end{aligned}
$
Question 2.
Given that $x, y$ and $b$ are real numbers $x<y, b \geq 0$, then
(a) $x b<y b$ (b) $x b>y b$
(c) $\mathrm{xb} \leq \mathrm{vb}$
(d) $x l b \geq$ ylb
Solution:
(a) $\mathrm{xb}<\mathrm{yb}$
Hint:
$
\begin{aligned}
& x<y, \quad(b>0) \\
& \text { (i.e.,) } x b<y b
\end{aligned} \Rightarrow \frac{x}{b}<\frac{y}{b}
$
Question 3.
If $\frac{|x-2|}{x-2} \geq 0$ then $x$ belongs to
(a) $[2, \infty]$
(b) $(2, \infty)$
(c) $(-\infty, 2)$
(d) $(-2, \infty)$
Solution:

(b) $(2, \infty)$
Hint:
$
\begin{aligned}
& \frac{|x-2|}{x-2} \geq 0 \\
& \therefore x>2 \\
& x \in(2, \infty)
\end{aligned}
$
Question 4.
The solution of $5 x-1<24$ and $5 x+1>-24$ is
(a) $(4,5)$
(b) $(-5,-4)$
(c) $(-5,5)$
(d) $(-5,4)$
Solution:
(c) $(-5,5)$
Hint:

Question 5.
The solution set of the following inequality $|x-1| \geq|\mathrm{x}-3|$ is ......
(a) $[0,2]$
(b) $(2, \infty)$
(c) $(0,2)$
(d) $(-\infty, 2)$
Solution:
(b) $(2, \infty)$
Question 6.
The value of $\log _{\sqrt{2}} 512$ is
(a) 16
(b) 18
(c) 9
(d) 12
Solution:
(b) 18
Hint:
$
\begin{aligned}
\log _{\sqrt{2}} 512=x \Rightarrow 512=\sqrt{2} x \\
2^9=\left(2^{\frac{1}{2}}\right)^x=2^{\frac{x}{2}} \\
\Rightarrow \frac{x}{2}=9 \Rightarrow x=18
\end{aligned}
$
Question 7.
The value of $\log _3 \frac{1}{81}$ is

(a) -2
(b) -8
(c) -4
(d) -9
Solution:
(c) -4
Hint:
$
\begin{aligned}
& \log _3 \frac{1}{81}= x \Rightarrow \frac{1}{81}=3^x \\
& 3^{-4}=3^x \Rightarrow x=-4
\end{aligned}
$
Question 8.
If $\log _{\sqrt{x}} 0.25=4$ then the value of $x$ is
(a) 0.5
(b) 2.5
(c) 1.5
(d) 1.25
Solution:
(a) 0.5
Hint:
$
\Rightarrow \quad \begin{aligned}
& 0.25=(\sqrt{x})^4=x^2 \\
& (0.5)^2=x^2 \Rightarrow x=0.5
\end{aligned}
$
Question 9.
The value of $\log _a b \log _{\mathrm{b}} c \log _c a$ is
(a) 2
(b) 1
(c) 3
(d) 4

Solution:
(b) 1
Hint
$
\Rightarrow \quad \log _b a \log _c b \log _a c=\log _a a=1
$
Question 10.
If 3 is the logarithm of 343 , then the base is .....
(a) 5
(b) 7
(c) 6
(d) 9
Solution:
(b) 7
Hint.
$
\begin{aligned}
& \Rightarrow \log _x 343=3 \Rightarrow 343=x^3 \\
& \text { (.i.e.) } 7^3=x^3 \Rightarrow x=7 \\
& \Rightarrow \mathrm{x}=7
\end{aligned}
$
Question 11.
Find a so that the sum and product of the roots of the equation $2 x^2+(a-3) x+3 a-5=0$ are equal is ........
(a) 1
(b) 2
(c) 0
(d) 4
Solution:
(b) 2
Hint:

Question 12 .
If $\mathrm{a}$ and $\mathrm{b}$ are the roots of the equation $\mathrm{x}^2-\mathrm{kx}+16=0$ and satisfy $\mathrm{a}^2+\mathrm{b}^2=32$, then the value of $\mathrm{k}$ is
(a) 10
(b) -8
(c) $(-8,8)$
(d) 6
Solution:
(c) $-8,8$
Hint:
$
\begin{aligned}
& a+b=k \ldots(1) a b=16 \\
& a^2+b^2=(a+b)^2-2 a b=32 \\
& k^2-32=32 \Rightarrow k^2=64 \Rightarrow k= \pm 8
\end{aligned}
$
Question 13.
The number of solutions of $x^2+|x-1|=1$ is
(a) 1
(b) 0
(c) 2
(d) 3
Solution:
(c) 2

We have two solutions 0,1
Question 14.
The equations whose roots are numerically equal but opposite in sign to the roots of $3 x^2-5 x-7=$ 0 is..$\ldots$
(a) $3 x^2-5 x-7=0$
(b) $3 x^2+5 x-7=0$
(c) $3 x^2-5 x+7=0$
(d) $3 x^2+x-7=0$
Solution:
(b) $3 x^2+5 x-7=0$
Hint:
$
\alpha+\beta=5 / 3, \alpha \beta=-7 / 3
$
With roots $-\alpha,-\beta$
$
\begin{aligned}
& \text { Sum }=-\alpha-\beta=-(\alpha+\beta)=-5 / 3 \\
& \text { Product }=(-\alpha)(-\beta)=\alpha \beta=-7 / 3
\end{aligned}
$
Equation is $x^2-(-5 / 3) x+(-7 / 3)=0$
$
\text { (i.e.,) } 3 x^2+5 x-7=0
$
Question 15.
If 8 and 2 are the roots of $x^2+a x+c=0$ and 3,3 are the roots of $x^2+a x+b=0$, then the roots of the equation $\mathrm{x}^2+\mathrm{ax}+\mathrm{b}=0$ are
(a) 1,2
(b) $-1,1$
(c) 9,1
(d) $-1,2$
Solution:
(c) 9,1
Hint:

$
\begin{aligned}
& \text { Sum }=8+2=10=-\mathrm{a} \Rightarrow \mathrm{a}=-10 \\
& \text { Product }=3 \times 3=9=\mathrm{b} \Rightarrow \mathrm{b}=9 \\
& \text { Now the equation } \mathrm{x}^2+\mathrm{ax}+\mathrm{b}=0 \\
& \Rightarrow \mathrm{x}^2-10 \mathrm{x}+9=0 \\
& \Rightarrow(\mathrm{x}-9)(\mathrm{x}-1)=0 \\
& \mathrm{x}=1 \text { or } 9
\end{aligned}
$
Question 16.
If $\mathrm{a}$ and $\mathrm{b}$ are the real roots of the equation $\mathrm{x}^2-\mathrm{kx}+\mathrm{c}=0$, then the distance between the points $(\mathrm{a}, 0)$ and $(\mathrm{b}, 0)$ is .......
(a) $\sqrt{k^2-4 c}$
(b) $\sqrt{4 k^2-c}$
(c) $\sqrt{4 c-k^2}$
(d) $\sqrt{k-8 c}$
Solution:
(a) $\sqrt{k^2-4 c}$
Hint:
$
\mathrm{a}+\mathrm{b}=\mathrm{k}, \mathrm{ab}=\mathrm{c}
$
Distance between $(a, 0)$ and $(b, 0)$ is $\sqrt{(a-b)^2}=\sqrt{(a+b)^2+4 a b}=\sqrt{k^2-4 c}$
Question 17.
If $\frac{k x}{(x+2)(x-1)}=\frac{2}{x+2}+\frac{1}{x-1}$ then the value of $k$ is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:

(c) 3
$
\begin{aligned}
\frac{k x}{(x+2)(x-1)} & =\frac{2(x-1)+1(x+2)}{(x+2)(x-1)} \\
& =\frac{3 x}{(x+2)(x-1)} \Rightarrow k x=3 x \Rightarrow k=3
\end{aligned}
$
Question 18.
If $\frac{1-2 x}{3+2 x-x^2}=\frac{\mathrm{A}}{3-x}+\frac{\mathrm{B}}{x+1}$ then the value of $\mathrm{A}+\mathrm{B}$ is
(a) $-1 / 2$
(b) $-2 / 3$
(c) $1 / 2$
(d) $2 / 3$
Solution:
(a) $-1 / 2$
Hint:
$
\frac{1-2 x}{3+2 x-x^2}=\frac{A(x+1)+\mathrm{B}(3-x)}{(3-x)(x+1)}
$
Put $x=-1$
$
\begin{aligned}
& 3=\mathrm{A}(0)+\mathrm{B}(4) \Rightarrow \mathrm{B}=3 / 4 \\
& \text { put } x=3 \\
& 1-6=\mathrm{A}(4)+\mathrm{B}(0) \Rightarrow 4 \mathrm{~A}=-5 \Rightarrow \mathrm{A}=-5 / 4 \\
& \therefore \mathrm{A}+\mathrm{B}=\frac{-5}{4}+\frac{3}{4}=\frac{-2}{4}=\frac{-1}{2}
\end{aligned}
$
Question 19.
The number of real roots of $(x+3)^4+(x+5)^4=16$ is
(a) 4
(b) 2
(c) 3
(d) 0
Solution:
(a) 4
Hint:
The equation is $(x+3)^4+(x+5)^4=16$

$
(x+3)^4+(x+5)^4=2^4
$
This is biquadratic equation. It has 4 roots.
Question 20 .
The value of $\log _3 11 . \log _{11} 13 . \log _{13} 15 . \log _{15} 27 . \log _{27} 81$ is ......
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(d) 4
Hint.
$
\log _3 81=\log _3 3^4=4 \log _3 3=4 \times 1=4
$