Solution: Left Hand Limit: $\lim _{x \rightarrow 5-} f(x)=\lim _{x \rightarrow 5-} \frac{x^{2}-25}{x-5}$ $=\lim _{x \rightarrow 5-} \frac{(x+5)(x-5)}{x-5}$ [By middle term splitting]...
Solution: Left Hand Limit: $\lim _{x \rightarrow 3-} f(x)=\lim _{x \rightarrow 3-} \frac{x^{2}-x-6}{x-3}$ $=\lim _{\mathrm{x} \rightarrow 3-} \frac{(\mathrm{x}+2)(\mathrm{x}-3)}{\mathrm{x}-3}$ [By...
Solution: L.H.L.: $\lim _{\mathrm{x} \rightarrow 1-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 1^{-}} \mathrm{x}^{2}+3 \mathrm{x}+4$ $=7$ R.H.L.: $\lim _{x \rightarrow 1^{-}} f(x)=\lim...
Solution: L.H.L.: $\lim _{x \rightarrow 2-} f(x)=\lim _{x \rightarrow 2-} x^{2}$ $=4$ R.H.L.: $\lim _{\mathrm{x} \rightarrow 2^{*}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 2^{*}}...
Solution: We need to find: $-x, y, z$ The given set of lines are : - $\begin{array}{l} \frac{2}{x}-\frac{3}{y}+\frac{3}{z}=10 \\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=10 \\...
Solution: It is given, $\begin{array}{l} A=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right], B=\left[\begin{array}{ccc} 7 & 2 & -6 \\...
Solution: It is given, $\begin{array}{l} A=\left[\begin{array}{ccc} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{array}\right] \\ A^{-1}=\frac{1}{|A|} \operatorname{adj}(A)...
Solution: It is given, $A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]$ $\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj}(A)$...
Solution: We need to find: - $x , y , z$ The given set of lines are : $\begin{array}{l} 4 x+3 y+2 z=60 \\ x+2 y+3 z=45 \\ 6 x+2 y+3 z=70 \end{array}$ Now converting the following equations in matrix...
Solution: We need to find: $-\mathrm{x}, \mathrm{y}, \mathrm{z}$ The given set of lines are : - $\begin{array}{l} x-2 y+z=0 \\ y-z=2 \\ 2 x-3 z=10 \end{array}$ Now converting the following equations...
Solution: We need to find: $-\mathrm{x}, \mathrm{y}, \mathrm{z}$ The given set of lines are : - $\begin{array}{l} 5 x-y=-7 \\ 2 x+3 z=1 \\ 3 y-z=5 \end{array}$ Now, converting the following...
Solution: We need to find: $-\mathrm{x}, \mathrm{y}, \mathrm{z}$ The given set of lines are : - $\begin{array}{l} x-y-2 z=3 \\ x+y=1 \\ x+z=-6 \end{array}$ Now converting the following equations in...
Solution: We need to find: - $x , y , z$ The given set of lines are : - $x + y - z = 1$ $\begin{array}{l} 3 x+y-2 z=3 \\ x-y-z=-1 \end{array}$ Now, converting the following equations in matrix form,...
Solution: We need to find: - $x , y , z$ The given set of lines are : - $\begin{array}{l} 3 x-4 y+2 z=-1 \\ 2 x+3 y+5 z=7 \\ x+z=2 \end{array}$ Now, converting the following equations in matrix...
Solution: We need to find: $-\mathrm{x}, \mathrm{y}, z$ The given set of lines are : - $\begin{array}{l} 6 x-9 y-20 z=-4 \\ 4 x-15 y+10 z=-1 \\ 2 x-3 y-5 z=-1 \end{array}$ Now, converting the...
Solution: We need to find: $-\mathrm{x}, \mathrm{y}, \mathrm{z}$ The given set of lines are: $\begin{array}{l} 4 x-5 y-11 z=12 \\ x-3 y+z=1 \\ 2 x+3 y-7 z=2 \end{array}$ Now, converting the...
Solution: We need to find: - $x, y, z$ The given set of lines are : - $\begin{array}{l} 2 x+3 y+3 z=5 \\ x-2 y+z=-4 \\ 3 x-y-2 z=3 \end{array}$ Now, converting the following equations in matrix...
$\begin{array}{l} x^{2}-4 a x+4 a^{2}-b^{2}=0 \\ \Rightarrow x^{2}-4 a x+(2 a+b)(2 a-b)=0 \\ \Rightarrow x^{2}-[(2 a+b)+(2 a-b)] x+(2 a+b)(2 a-b)=0 \\ \Rightarrow x^{2}-(2 a+b) x-(2 a-b) x+(2 a+b)(2...
$\begin{array}{l} x^{2}+6 x-\left(a^{2}+2 a-8\right)=0 \\ \Rightarrow x^{2}+6 x-(a+4)(a-2)=0 \\ \Rightarrow x^{2}+[(a+4)-(a-2)] x-(a+4)(a-2)=0 \\ \Rightarrow x^{2}+(a+4) x-(a-2) x-(a+4)(a-2)=0 \\...
$\begin{array}{l} x^{2}+5 x-\left(a^{2}+a-6\right)=0 \\ \Rightarrow x^{2}+5 x-(a+3)(a-2)=0 \\ \Rightarrow x^{2}+[(a+3)-(a-2)] x-(a+3)(a-2)=0 \\ \Rightarrow x^{2}+(a+3) x-(a-2) x-(a+3)(a-2)=0 \\...
$\begin{array}{l} 4 x^{2}+4 b x-\left(a^{2}-b^{2}\right)=0 \\ \Rightarrow 4 x^{2}+4 b x-(a-b)(a+b)=0 \\ \Rightarrow 4 x^{2}+2[(a+b)-(a-b)] x-(a-b)(a+b)=0 \\ \Rightarrow 4 x^{2}+2(a+b) x-2(a-b)...
$\begin{array}{l} 4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}=0 \\ \Rightarrow 4 \sqrt{3} x^{2}+8 x-3 x-2 \sqrt{3}=0 \\ \Rightarrow 4 x(\sqrt{3} x+2)-\sqrt{3}(\sqrt{3} x+2)=0 \\ \Rightarrow(\sqrt{3} x+2)(4...
$\begin{array}{l} \sqrt{3} x^{2}-2 \sqrt{2} x-2 \sqrt{3}=0 \\ \Rightarrow \sqrt{3} x^{2}-3 \sqrt{2} x+\sqrt{2} x-2 \sqrt{3}=0 \\ \Rightarrow \sqrt{3} x(x-\sqrt{6})+\sqrt{2}(x-\sqrt{6})=0 \\...
$\begin{array}{l} \sqrt{3} x^{2}+10 x-8 \sqrt{3}=0 \\ \Rightarrow \sqrt{3} x^{2}+12 x-2 x-8 \sqrt{3}=0 \\ \Rightarrow \sqrt{3} x(x+4 \sqrt{3})-2(x+4 \sqrt{3})=0 \\ \Rightarrow(x+4 \sqrt{3})(\sqrt{3}...
$\begin{array}{l} 3 x^{2}+5 \sqrt{5} x-10=0 \\ \Rightarrow 3 x^{2}+6 \sqrt{5} x-\sqrt{5} x-10=0 \\ \Rightarrow 3 x(x+2 \sqrt{5})-\sqrt{5}(x+2 \sqrt{5})=0 \\ \Rightarrow(x+2 \sqrt{5})(3 x-\sqrt{5})=0...
$\begin{array}{l} 2 x^{2}+a x-a^{2}=0 \\ \Rightarrow 2 x^{2}+2 a x-a x-a^{2}=0 \\ \Rightarrow 2 x(x+a)-a(x+a)=0 \\ \Rightarrow(x+a)(2 x-a)=0 \\ \Rightarrow x+a=0 \text { or } 2 x-a=0 \\ \Rightarrow...
$\begin{array}{l} x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0 \\ \Rightarrow x^{2}-\sqrt{3} x-x+\sqrt{3}=0 \\ \Rightarrow x(x-\sqrt{3})-1(x-\sqrt{3})=0 \\ \Rightarrow(x-\sqrt{3})(x-1)=0 \\ \Rightarrow...
It is given that the quadratic equation $9 x^{2}-3 k x+k=0$ has equal roots. $\begin{array}{l} \therefore D=0 \\ \Rightarrow(-3 k)^{2}-4 \times 9 \times k=0 \\ \Rightarrow 9 k^{2}-36 k=0 \\...
It is given that the quadratic equation $x^{2}-4 k x+k=0$ has equal roots. $\begin{array}{l} \therefore D=0 \\ \Rightarrow(-4 k)^{2}-4 \times 1 \times k=0 \\ \Rightarrow 16 k^{2}-4 k=0 \\...
It is given that the roots of the quadratic equation $p x^{2}-2 p x+6=0$ are equal. $\begin{array}{l} \therefore D=0 \\ \Rightarrow(-2 p)^{2}-4 \times p \times 6=0 \\ \Rightarrow 4 p^{2}-24 p=0 \\...
Let $\alpha$ and $\beta$ be the roots of the equation $3 x^{2}-10 x+k=0$. $\therefore \alpha=\frac{1}{\beta} \quad$ (Given) $\Rightarrow \alpha \beta=1$ $\Rightarrow \frac{k}{3}=1 \quad \quad$...
Let the other zero of the given polynomial be $\alpha$. Now, Sum of the zeroes of the given polynomial $=\frac{-(-4)}{1}=4$ $\begin{array}{l} \therefore \alpha+(2+\sqrt{3})=4 \\ \Rightarrow...
It is given that $y=1$ is a root of the equation $a y^{2}+a y+3=0$. $\begin{array}{l} \therefore a \times(1)^{2}+a \times 1+3=0 \\ \Rightarrow a+a+3=0 \\ \Rightarrow 2 a+3=0 \\ \Rightarrow...
It is given that the quadratic equation $p x^{2}-2 \sqrt{5} p x+15=0$ has two equal roots. $\begin{array}{l} \therefore D=0 \\ \Rightarrow(-2 \sqrt{5} p)^{2}-4 \times p \times 15=0 \\ \Rightarrow 20...
It is given that the roots of the quadratic equation $2 x^{2}+8 x+k=0$ are equal. $\begin{array}{l} \therefore D=0 \\ \Rightarrow 8^{2}-4 \times 2 \times k=0 \\ \Rightarrow 64-8 k=0 \\ \Rightarrow...
The given quadratic equation is $3 \sqrt{3} x^{2}+10 x+\sqrt{3}=0$ $\begin{array}{l} 3 \sqrt{3} x^{2}+10 x+\sqrt{3}=0 \\ \Rightarrow 3 \sqrt{3} x^{2}+9 x+x+\sqrt{3}=0 \\ \Rightarrow 3 \sqrt{3}...
Solution: We need to find: $-\mathrm{x}, \mathrm{y}, \mathrm{z}$ The given set of lines are : - $x+y+z=6$ $\begin{array}{l} x+2 z=7 \\ 3 x+y+z=12 \end{array}$ Now converting following equations in...
Solution: We need to find: $-\mathrm{x}, \mathrm{y}, \mathrm{z}$ The given set of lines are : - $\begin{array}{l} 2 x-3 y+5 z=11 \\ 3 x+2 y-4 z=-5 \\ x+y-2 z=-3 \end{array}$ Now converting the...
Solution: We need to find: $-\mathrm{x}, \mathrm{y}, \mathrm{z}$ The given set of lines are : - $\begin{array}{l} x+y+z=4 \\ 2 x-y+z=-1 \\ 2 x+y-3 z=-9 \end{array}$ Now, converting the following...
Solution: We need to find: - $x, y$ The given set of lines are : - $\begin{array}{l} 2 \mathrm{x}-3 \mathrm{y}+1=0 \\ \mathrm{x}+4 y+3=0 \end{array}$ On converting the following equations in matrix...
Solution: We need to find: $-\mathrm{x}, \mathrm{y}$ The given set of lines are : - $\begin{array}{l} 5 \mathrm{x}+7 \mathrm{y}+2=0 \\ 4 \mathrm{x}+6 \mathrm{y}+3=0 \end{array}$ On converting the...
Solution: We need to prove: Set of given lines are inconsistent. The given set of lines are : - $\begin{array}{l} 3 x-y-2 z=2 \\ 2 y-z=-1 \\ 3 x-5 y=3 \end{array}$ Now, converting the following...
Solution: We need to prove: Set of given lines are inconsistent. The given set of lines are : - $\begin{array}{l} 2 x-y+3 z=1 \\ 3 x-2 y+5 z=-4 \\ 5 x-4 y+9 z=14 \end{array}$ On converting the...
Solution: We need to prove: Set of given lines are inconsistent. The given set of lines are: $\begin{array}{l} 2 x+3 y=5 \\ 6 x+9 y=10 \end{array}$ Now, converting the following equations in matrix...
Solution: We need to prove: Set of given lines are inconsistent. The given set of lines are: - $\begin{array}{l} x+2 y=9 \\ 2 x+4 y=7 \end{array}$ Now, converting the following equations in matrix...
Solution: Given matrix as $A=\left(\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right)$ Find: the adjoint of the...
Solution: Given matrix as $A=\left(\begin{array}{ccc}9 & 7 & 3 \\ 5 & -1 & 4 \\ 6 & 8 & 2\end{array}\right)$. Find: the adjoint of the matrix given. Step: 1 Find the minor...
Solution: Given matrix as $A=\left(\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right)$ Find: the adjoint of the matrix given. Step: 1 Find the minor...
Solution: Given matrix as $A=\left(\begin{array}{ll}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right)$. Find: the adjoint of the matrix given. Step: 1 Find the minor...
Solution: Given matrix as $A=\left(\begin{array}{cc}3 & -5 \\ -1 & 2\end{array}\right)$ Find: the adjoint of the matrix given. Step: 1 Find the minor matrix of $A$....
Solution: Given matrix as $A=\left(\begin{array}{ll}2 & 3 \\ 5 & 9\end{array}\right)$. Find: the adjoint of the matrix given. Step: 1 Find the minor matrix of $A$....
$(v)$ For multiplicative inverse of every element of $A, a^{*} b=$ e where $a, b \in A$. $\begin{array}{l} 1 \times b_{1}=1 \\ b_{1}=1 \\ -1 \times b_{2}=1 \end{array}$ $\begin{array}{l}...
$(i)$ A is said to be closed on * if all the elements of $a^{*} b \in A$. composition table is $\left(\text { as } i^{2}=-1\right)$ As table contains all elements from set $A, A$ is close for...
$(iii)$ For associative, $\begin{array}{l} (a, b)^{*}((c, d) *(e, f))=(a, b) *(c+e, d+f) \\ =(a+c+e, b+d+f) \\ ((a, b) *(c, d))^{*}(e, f)=(a+c, b+d) *(e, f) \\ =(a+c+e, b+d+f) \end{array}$ $(iv)$...
$(v)$ for a binary operation * if e is identity element then it is invertible with respect to * if for an element $b, a^{*} b=e=b^{*} a$ where $b$ is called inverse of $*$ and denoted by $a^{-1}$....
$(i)$ $*$ is an operation as $a^{*} b=a+b+$ ab where $a, b \in Q-\{-1\} .$ Let $a=1$ and $b=\frac{-3}{2}$ two rational numbers. $\mathrm{a}^{*} \mathrm{~b}=1+\frac{-3}{2}+1 \cdot \frac{-3}{2}...
(iii) For associative binary operation, $a^{*}\left(b^{*} c\right)=\left(a^{*} b\right) * c$. $\begin{array}{l} a^{*}\left(b^{*} c\right)=a^{*} \frac{1}{2}(b+c) \Rightarrow...
(iii) For associative binary operation, $a^{*}\left(b^{*} c\right)=\left(a^{*} b\right) * c$. $\begin{array}{l} a^{*}(b * c)=a^{*} \frac{b c}{2} \Rightarrow \frac{a \cdot \frac{b c}{2}}{2}=\frac{a b...
(i) $^{*}$ is an operation as $\mathrm{a}^{*} \mathrm{~b}=\frac{\mathrm{ab}}{2}$ where $\mathrm{a}, \mathrm{b} \in \mathrm{Q}^{+} .$Let $\mathrm{a}=\frac{1}{2}$ and $\mathrm{b}=2 \mathrm{two}$...
(i) For commutative binary operation, $a^{*} b=b^{*} a$. $\mathrm{a}^{*} \mathrm{~b}=\frac{\mathrm{ab}}{4}$ and $\mathrm{b}^{*} \mathrm{a}=\frac{\mathrm{ba}}{4}$ as multiplication is commutative $a...
$*$ is an operation as $a^{*} b=a+b-a b$ where $a, b \in Z$. Let $a=\frac{1}{2}$ and $b=2$ two integers. $a * b=\frac{1}{2} * 2=\frac{1}{2}+2-\frac{1}{2} \cdot 2 \Rightarrow...
$*$ is an operation as $m^{*} n=\operatorname{LCM}(m, n)$ where $m, n \in N$. Let $m=2$ and $b=3$ two natural numbers. $\begin{array}{l} m^{*} n=2^{*} 3 \\ =\operatorname{LCM}(2,3) \\ =6 \in...
The given quadratic equation is $2 x^{2}-x-6=0$ $\begin{array}{l} 2 x^{2}-x-6=0 \\ \Rightarrow 2 x^{2}-4 x+3 x-6=0 \\ \Rightarrow 2 x(x-2)+3(x-2)=0 \\ \Rightarrow(x-2)(2 x+3)=0 \\ \Rightarrow x-2=0...
It is given that $x=\frac{-1}{2}$ is a solution of the quadratic equation $3 x^{2}+2 k x-3=0$ $\begin{array}{l} \therefore 3 \times\left(\frac{-1}{2}\right)^{2}+2 k...
The given equation is $3 x^{2}+13 x+14=0$. Putting $x=-2$ in the given equation, we get $L H S-3 \times(-2)^{2}+13 \times(-2)+14=12-26+14=0=R H S$ $\therefore x=-2$ is a solution of the given...
The given equation is $x^{2}+6 x+9=0$ Putting $x=-3$ in the given equation, we get $L H S=(-3)^{2}+6 \times(-3)+9=9-18+9=0=R H S$ $\therefore x=-3$ is a solution of the given equation.
Let the required natural numbers be $x$ and $(8-x)$. It is given that the product of the two numbers is $15 .$ $\begin{array}{l} \therefore x(8-x)=15 \\ \Rightarrow 8 x-x^{2}=15 \\ \Rightarrow...
Answer is (b) $2, \frac{-3}{2}$ The given quadratic equation is $2 x^{2}-x-6=0$. $\begin{array}{l} 2 x^{2}-x-6=0 \\ \Rightarrow 2 x^{2}-4 x+3 x-6=0 \\ \Rightarrow 2 x(x-2)+3(x-2)=0 \\...
Let the breadth of the rectangular field be $x \mathrm{~m}$. $\therefore$ Length of the rectangular field $=(x+8) m$ Area of the rectangular field $=240 \mathrm{~m}^{2}$ $\therefore(x+8) \times...
Answer is (c) $16 \mathrm{~m}$ Let the length and breadth of the rectangle be $l$ and $b$. Perimeter of the rectangle $=82 \mathrm{~m}$ $\begin{array}{l} \Rightarrow 2 \times(l+b)=82 \\ \Rightarrow...
Answer is (a) $\frac{5}{4}$ or $\frac{4}{5}$ Let the required number be $x$. According to the question: $\begin{array}{l} x+\frac{1}{x}=\frac{41}{20} \\ \Rightarrow \frac{x^{2}+1}{x}=\frac{41}{20}...
Answer is (b) $k \geq \frac{-9}{2}$ It is given that the roots of the equation $\left(k x^{2}-6 x-2=0\right)$ are real. $\begin{array}{l} \therefore D \geq 0 \\ \Rightarrow\left(b^{2}-4 a c\right)...
Answer is c) $-2<\mathrm{k}<2$ It is given that the equation $x^{2}-k x+1=0$ has no real roots. $\begin{array}{l} \therefore\left(b^{2}-4 a c\right)<0 \\ \Rightarrow(-k)^{2}-4 \times 1...
Answer is (c) $\frac{-8}{5}<k<\frac{8}{5}$ It is given that the equation $\left(x^{2}+5 k x+16=0\right)$ has no real roots. $\begin{array}{l} \therefore\left(b^{2}-4 a c\right)<0 \\...
Answer is (d) either $k>2 \sqrt{5}$ or $k<-2 \sqrt{5}$ It is given that the roots of the equation $\left(5 x^{2}-k+1=0\right)$ are real and distinct. $\begin{array}{l} \therefore\left(b^{2}-4...
Answer is (b) real, unequal and irrational $\begin{array}{l} \because D=\left(b^{2}-4 a c\right) \\ =(-6)^{2}-4 \times 2 \times 3 \\ =36-24 \\ =12 \end{array}$ 12 is greater than 0 and it is not a...
Answer is (d) imaginary $\begin{array}{l} \because D=\left(b^{2}-4 a c\right) \\ =(-6)^{2}-4 \times 2 \times 7 \\ =36-56 \\ =-20<0 \end{array}$ Thus, the roots of the equation are...
Answer is (b) real and unequal We know that when discriminant, $D>0$, the roots of the given quadratic cquation are real and uncqual.
Answer is $(a)>0$ The roots of the equation are real and unequal when $\left(b^{2}-4 a c\right)>0$.
Answer is $(\mathrm{d}) \pm \frac{4}{3}$ It is given that the roots of the equation $\left(4 x^{2}-3 k x+1=0\right)$ are equal. $\begin{array}{l} \therefore\left(b^{2}-4 a c\right)=0 \\...
Answer is (a) 1 or 4 It is given that the roots of the equation $\left(x^{2}+2(k+2) x+9 k=0\right)$ are equal. $\begin{array}{l} \therefore\left(b^{2}-4 a c\right)=0 \\ \Rightarrow\{2(k+2)\}^{2}-4...
Answer is (c) 2 or $-2$ It is given that the roots of the equation $\left(9 x^{2}+6 k x+4=0\right)$ are equal. $\begin{array}{l} \therefore\left(b^{2}-4 a c\right)=0 \\ \Rightarrow(6 k)^{2}-4 \times...
Answer is (d) $\frac{b^{2}}{4 a}$ It is given that the roots of the equation $\left(a x^{2}+b x+c=0\right)$ are equal. $\begin{array}{l} \therefore\left(b^{2}-4 a c\right)=0 \\ \Rightarrow b^{2}=4 a...
Answer is (c) $\mathrm{c}=\mathrm{a}$ Let the roots of the equation $\left(a x^{2}+b x+c=0\right)$ be $\alpha$ and $\frac{1}{\alpha}$. $\therefore$ Product of the roots $=\alpha \times...
Answer is (c) $-4$ It is given that $\alpha$ and $\beta$ are the roots of the equation $3 x^{2}+8 x+2=0$ $\therefore \alpha+\beta=-\frac{8}{3}$ and $\alpha \beta=\frac{2}{3}$...
Answer is (a) $x^{2}-6 x+6=0$ Given: Sum of roots $=6$ Product of roots $=6$ Thus, the equation is: $x^{2}-6 x+6=0$
Answer is (b) $x^{2}-3 x-10=0$, It is given that the roots of the quadratic equation are 5 and $-2$. Then, the equation is: $\begin{array}{l} x^{2}-(5-2) x+5 \times(-2)=0 \\ \Rightarrow x^{2}-3...
Answer is (d) $\frac{-2}{3}$ Given: $k x^{2}+2 x+3 k=0$ Sum of the roots $=$ Product of the roots $\begin{array}{l} \Rightarrow \frac{-2}{k}=\frac{3 k}{k} \\ \Rightarrow 3 k=-2 \\ \Rightarrow...
Answer is (d)5 Let the roots of the equation $\frac{-2}{3}$ be $\alpha$ and $\frac{1}{\alpha}$. $\therefore$ Product of the roots $=\frac{c}{a}$ $\Rightarrow \alpha \times...
Answer is (d)3. Given: $3 x^{2}-10 x+3=0$ One root of the equation is $\frac{1}{3}$. Let the other root be $\alpha$. Product of the roots $=\frac{c}{a}$ $\begin{array}{l} \Rightarrow \frac{1}{3}...
Answer is (d) $2: 3$ Given: $7 x^{2}-12 x+18=0$ $\therefore \alpha+\beta=\frac{12}{7}$ and $\beta=\frac{18}{7}$, where $\alpha$ and $\beta$ are the roots of the equation $\therefore$ Ratio of the...
Answer is (c) 8 It is given that the product of the roots of the equation $x^{2}-3 x+k=10$ is $-2$ The equation can be rewritten as: $x^{2}-3 x+(k-10)=0$ Product of the roots of a quadratic equation...
Answer is (b) -2 Sum of the roots of the equation $x^{2}-6 x+2=0$ is $\alpha+\beta=\frac{-b}{a}=\frac{-(-6)}{1}=6$, where $\alpha$ and $\beta$ are the roots of the equation.
Solution: (i) $\mathrm{f}(\pi)$ Here, $x=\Pi$, which is irrational $f(\pi)=-1$ (ii) $f(2+\sqrt{3})$ Here, $x=2+\sqrt{3}$, which is irrational $\therefore f(2+\sqrt{3})=-1$
Answer is (b) $-7$ It is given that one root of the equation $2 x^{2}+a x+6=0$ is 2 . $\begin{array}{l} \therefore 2 \times 2^{2}+a \times 2+6=0 \\ \Rightarrow 2 a+14=0 \\ \Rightarrow a=-7...
Answer is (b) $-11$ It is given that $x=3$ is a solution of $3 x^{2}+(k-1) x+9=0$; therefore, we have: $\begin{array}{l} 3(3)^{2}+(k-1) \times 3+9=0 \\ \Rightarrow 27+3(k-1)+9=0 \\ \Rightarrow...
Solution: For domain $\left(1+x^{2}\right) \neq 0$ $\begin{array}{l} \Rightarrow x^{2} \neq-1 \\ \Rightarrow \operatorname{dom}(f)=R \end{array}$ For the range of $\mathrm{x}$ : $\begin{array}{l}...
Answer is (c) $(\sqrt{2} x+3)^{2}=2 x^{2}+6$ $\begin{array}{l} \because(\sqrt{2} x+3)^{2}=2 x^{2}+6 \\ \Rightarrow 2 x^{2}+9+6 \sqrt{2} x=2 x^{2}+6 \end{array}$ $\Rightarrow 6 \sqrt{2} x+3=0$, which...
Answer is (b) $x^{3}-x^{2}=(x-1)^{3}$ $\because x^{3}-x^{2}=(x-1)^{3}$ $\Rightarrow x^{3}-x^{2}=x^{3}-3 x^{2}+3 x-1$ $\Rightarrow 2 x^{2}-3 x+1=0$, which is a quadratic equation
Solution: For a relation to be a function each element of first set should have different image in the second set(Range) (i) f = {( - 1, 2), (1, 8), (2, 11), (3, 14)} Here, each of the first set...
Answer is (d) $2 x^{2}-5 x=(x-1)^{2}$ A quadratic equation is the equation with degree 2 . $\because 2 x^{2}-5 x=(x-1)^{2}$ $\Rightarrow 2 x^{2}-5 x=x^{2}-2 x+1$ $\Rightarrow 2 x^{2}-5 x-x^{2}+2...
Solution: As the function $f(x)$ can accept any values as per the given domain $R$, so, the domain of the function $f(x)=x^{2}+1$ is $R$ The minimum value of $\mathrm{f}(\mathrm{x})=1$ $\Rightarrow$...
Let the shortest side be $x \mathrm{~m}$. Therefore, according to the question: Hypotenuse $=(2 x-1) m$ Third side $=(x+1) m$ On applying Pythagoras theorem, we get: $\begin{array}{l} (2...
Solution: $\begin{array}{l} f(n)=\left\{\begin{array}{l} \frac{1}{2}(n-1), \text { when } n \text { is odd } \\ -\frac{1}{2} n, \text { when } n \text { is even } \end{array}\right. \\ f(1)=0 \\...
Let the base and altitude of the right-angled triangle be $x$ and $y \mathrm{~cm}$, respectively Therefore, the hypotenuse will be $(x+2) \mathrm{cm}$. $\therefore(x+2)^{2}=y^{2}+x^{2}$ Again, the...
Let one side of the right-angled triangle be $x \mathrm{~m}$ and the other side be $(x+4) \mathrm{m}$. On applying Pythagoras theorem, we have: $\begin{array}{l} 20^{2}=(x+4)^{2}+x^{2} \\...
Let the base be $x \mathrm{~m}$. Therefore, the altitude will be $(x+7) m$ $\begin{array}{l} \text { Area of a triangle }=\frac{1}{2} \times \text { Base } \times \text { Altitude } \\ \therefore...
Let the altitude of the triangle be $x \mathrm{~m}$. Therefore, the base will be $3 x \mathrm{~m}$. $\begin{array}{l} \text { Area of a triangle }=\frac{1}{2} \times \text { Base } \times \text {...
Solution: In the range of $\mathrm{N} \mathrm{f}(\mathrm{x})$ is monotonically increasing. $\therefore f(n)=n^{2}+n+1$ is one one. But Range of $f(n)=[0.75, \infty) \neq N($ codomain $)$ Thus,...
Let the altitude of the triangle be $x \mathrm{~cm}$ Therefore, the base of the triangle will be $(x+10) \mathrm{cm}$ $\begin{array}{l} \text { Area of triangle }=\frac{1}{2} x(x+10)=600 \\...
Solution: (i) $f: N \rightarrow N: f(x)=x^{3}$ is one - one into. $f(x)=x^{3}$ As the function $f(x)$ is monotonically increasing from the domain $N \rightarrow N$ $\therefore f(x)$ is one -one...
Let the length and breadth of the rectangular garden be $x$ and $y$ meter, respectively. Given: $x y=180 \mathrm{sq} \mathrm{m}$$\ldots(i)$ and $\begin{array}{l} 2 y+x=39 \\ \Rightarrow x=39-2 y...
Let the breadth of rectangle be $x \mathrm{~cm}$. According to the question: Side of the square $=(x+4) \mathrm{cm}$ Length of the rectangle $=\{3(x+4)\} \mathrm{cm}$ It is given that the areas of...
Let the length of the side of the first and the second square be $x$ and $y \cdot$ respectively. According to the question: $x^{2}+y^{2}=640$ Also, $\begin{array}{l} 4 x-4 y=64 \\ \Rightarrow x-y=16...
Solution: (i) $f: N \rightarrow N: f(x)=x^{2}$ is one - one into. As the function $f(x)$ is monotonically increasing from the domain $N \rightarrow N$ $\therefore f(x)$ is one -one Range of...
Let the width of the path be $x \mathrm{~m}$. $\therefore$ Length of the field including the path $=16+x+x=16+2 x$ Breadth of the field including the path $=10+x+x=10+2 x$ Now, (Area of the field...
Let the length and breadth of the rectangular plot be $x$ and $y$ meter, respectively. Therefore, we have: $\begin{array}{l} \text { Perimeter }=2(x+y)=62 \quad \ldots . .(i) \text { and } \\ \text...
Solution: $f: \left[0, \frac{\pi}{2}\right] \rightarrow \mathrm{R}$ for given function $\mathrm{f}(\mathrm{x})=\sin$ Recalling the graph for $\sin \mathrm{x}$, we realise that for any two values on...
Let the breath of the rectangular hall be $x$ meter. Therefore, the length of the rectangular hall will be $(x+3)$ meter. According to the question: $\begin{array}{l} x(x+3)=238 \\ \Rightarrow...
Let the length and breadth of the rectangle be $3 x \mathrm{~m}$ and $x \mathrm{~m}$, respectively. According to the question: $\begin{array}{l} 3 x \times x=147 \\ \Rightarrow 3 x^{2}=147 \\...
Let the length and breadth of the rectangle be $2 x \mathrm{~m}$ and $x \mathrm{~m}$, respectively. According to the question: $\begin{array}{l} 2 x \times x=288 \\ \Rightarrow 2 x^{2}=288 \\...
Let the tap of smaller diameter fill the tank in $x$ hours. $\therefore$ Time taken by the tap of larger diameter to fill the tank $=(x-9) h$ Suppose the volume of the tank be $V$. Volume of the...
Solution: We need to show that $f: R \rightarrow R$ given by $f(x)=x s$ is one-one and onto. A function which is onto has every element of co-domain mapped to the at least one element of Domain....
Let the time taken by one pipe to fill the tank be $x$ minutes. $\therefore$ Time taken by the other pipe to fill the tank $=(x+5) \min$ Suppose the volume of the tank be $V$. Volume of the tank...
Solution: We need to show that f: $\mathrm{R} \rightarrow \mathrm{R}$ given by $\mathrm{f}(\mathrm{x})=\mathrm{x} 4$ is many-one into. A function which is not onto is into. A function where more...
Let one pipe fills the cistern in $x$ mins. Therefore, the other pipe will fill the cistern in $(x+3)$ mins. Time taken by both, running together, to fill the cistern $=3 \frac{1}{13} \min...
Let B takes $x$ days to complete the work. Therefore, A will take $(x-10)$ days. $\begin{array}{l} \therefore \frac{1}{x}+\frac{1}{(x-10)}=\frac{1}{12} \\ \Rightarrow...
Let the speed of the stream be $x \mathrm{~km} / \mathrm{hr}$. $\therefore$ Downstream speed $=(9+x) \mathrm{km} / \mathrm{hr}$ Upstream speed $=(9-x) \mathrm{km} / \mathrm{hr}$ Distance covered...
Speed of the boat in still water $=8 \mathrm{~km} / \mathrm{hr}$. Let the speed of the stream be $x \mathrm{~km} / \mathrm{hr}$. $\therefore$ Speed upstream $=(8-x) \mathrm{km} / \mathrm{hr}$ Speed...
Solution: We need to show that $f: R \rightarrow R$ given by $f(x) = 1 + x^2$ is many-one into. A function which is not onto is into. A function where more than one element in Set A maps to one...
Solution: (i) $\mathrm{f}(-1)$ $x=-1$, it is satisfying the condition $-2 \leq x \leq 3$ Therefore, $f(x)=x_{2}-2$ $\begin{aligned} \begin{aligned} \therefore \mathrm{f}(-1) =(-1)_{2}-2 \\ =1-2 \\...
Let the speed of the stream be $x \mathrm{~km} / \mathrm{hr}$. Given: Speed of the boat $=18 \mathrm{~km} / \mathrm{hr}$ $\therefore$ Speed downstream $=(18+x) \mathrm{km} / h r$ Speed upstream...
Let the speed of the Deccan Queen be $x \mathrm{~km} / \mathrm{hr}$. According to the question: Speed of another train $=(x-20) \mathrm{km} / \mathrm{hr}$ $\begin{array}{l} \therefore...
Let the usual speed $x \mathrm{~km} / \mathrm{hr}$. According to the question: $\begin{array}{l} \frac{300}{x}-\frac{300}{(x+5)}=2 \\ \Rightarrow \frac{300(x+5)-300 x}{x(x+5)}=2 \\ \Rightarrow...
Solution: (i) $\mathrm{f}(2)$ $x=2$, it is satisfying the condition $-2 \leq x \leq 3$ Therefore, $f(x)=x_{2}-2$ $\begin{aligned} \therefore \mathrm{f}(2) =22-2 \\ =4-2 \\ =2 \\ \therefore...
Let the original speed of the train be $x \mathrm{~km} / \mathrm{hr}$. According to the question: $\frac{90}{x}-\frac{90}{(x+15)}=\frac{1}{2}$ $\begin{array}{l} \Rightarrow \frac{90(x+15)-90...
Solution: (i) Neither one-one nor onto $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ given by $\mathrm{f}(\mathrm{x})=|\mathrm{x}|=\left\{\begin{array}{l}\mathrm{x}, \text { if } \mathrm{x} \geq 0...
Let the speed of the train be xkmph The time taken by the train to travel $180 \mathrm{~km}$ is $\frac{180}{\mathrm{x}} \mathrm{h}$ The increased speed is $\mathrm{x}+9$ The time taken is...
Solution: (i) One-One but not Onto $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$ be a mapping given by $\mathrm{f}(\mathrm{x})=\mathrm{x} 2$ For one-one $\begin{array}{l} f(x)=f(y) \\ x_{2}=y z \\...
Let the first speed of the train be $x \mathrm{~km} / \mathrm{h}$. Time taken to cover $54 \mathrm{~km}=\frac{54}{x} h .$ New speed of the train $=(x+6) \mathrm{km} / \mathrm{h}$ $\therefore$ Time...
Let the usual speed of the train be $x \mathrm{~km} / \mathrm{h}$. $\therefore$ Reduced speed of the train $=(x-8) \mathrm{km} / \mathrm{h}$ Total distance to be covered $=480 \mathrm{~km}$ Time...
Solution: (i)Bijective function: It is, also known as one-one onto function and is a function where for every element of set A, there is exactly one image in set B, such that no element is set B is...
Solution: A function is stated as the relation between the two sets, where there is exactly one element in set B, for every element of set A. A function is represented as f: A → B, which means ‘f’...
Solution: $\begin{array}{l} \mathrm{A}=\{1,2,3,4\} \text { and } \mathrm{R}=\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,3) \\ (3,2)\} \text { (Given) } \end{array}$ $\mathrm{R}$ is reflexive if $\mathrm{a}...
Solution: $A=\{1,2,3\}$ and $\bar{R}=\{(1,1),(2,2),(3,3),(1,2),(2,3)\}$ (Given) $\mathrm{R}$ is reflexive if $\mathrm{a} \in \mathrm{A}$ and $(\mathrm{a}, \mathrm{a}) \in \mathrm{R}$ Here,...
Solution: $\mathrm{R}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}=\mathrm{b}_{2}\right\}$ for all $\mathrm{a}, \mathrm{b} \in \mathrm{N}$ (As given) Non-Reflexivity: Assume $a$ be an arbitrary...
Solution: If $R$ is Reflexive, Symmetric and Transitive, then $R$ is an equivalence relation. Reflexivity: Suppose $a$ and $\mathrm{b}$ be an arbitrary element of $\mathrm{N} \times \mathrm{N}$...
Solution: $\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{Z}$ and $(\mathrm{a}-\mathrm{b})$ is divisible by 5$\}$ (As given) If $R$ is Reflexive, Symmetric and Transitive,...
Solution: Suppose $\mathrm{R}=\left\{\left(\Delta 1, \Delta_{2}\right): \Delta_{1} \sim \Delta_{2}\right\}$ be a relation defined on A. (As given) If $\mathrm{R}$ is Reflexive, Symmetric and...
Solution: (i) Reflexivity: Suppose $\mathrm{p}$ is an arbitrary element of $\mathrm{S}$. So now, $\mathrm{p} \leq \mathrm{p}$ $\Rightarrow(p, p) \in R$ Therefore, $\mathrm{R}$ is reflexive. (ii)...
Solution: (i) $A \subset B$ and $(A, B) \in R$ (As given) i.e. $\mathrm{A}$ is a proper subset of $\mathrm{B}$ But, B cannot be a proper subset of A Since, B can contain atleast one element that is...
Solution: Since a< 4, $a = 1, 2, 3$ Therefore, R $=$ {(1, 6), (2, 7), (3, 8)} As a result, Domain(R) = {1, 2, 3} and Range(R) = {6, 7, 8}
Solution: Since 1<a< 5, $a = 2, 3, 4$ Therefore, R $= {\{(2,{\frac12}), (3,{\frac13}), (4,{\frac14})}\}$ As a result, Domain(R) = {2, 3, 4} and Range(R) =...
Solution: As $|a| < 3$, $a = −2$, $−1$, $0$, $1$, $2$ Therefore, R = {(−2, 3), (−1, 2), (0, 1), (1, 0), (2, 1)} As a result, Domain(R) = {-2, -1, 0, 1, 2} and Range(R) = {3, 2, 1, 0}
Solution: $x + 2y = 8$ (given) $x = 8 – 2y$ Putting $y = 1$ $x = 8 – 2(1) = 6$ Putting $y = 2$ $x = 8 – 2(2) = 4$ Putting $y = 3$ $x = 8 – 2(3) = 2$ Putting $y = 4$ $x = 8 – 2(4) = 0$; which is not...
Solution: R = {(2, 8), (3, 27) Therefore, Range of R = {8 27}
For a binary operation if a*e = a, then e s called the right identity If $\mathrm{e}^{*} \mathrm{a}=\mathrm{a}$ then $\mathrm{e}$ is called the left identity For the given binary operation,...
$\text { let } a=1, b=2 \in N$ $a * b=1^{3}+2^{3}=9$ And $b^{*} a=2^{3}+1^{3}=9$ => ${ }^{*}$ is commutative. Let $c=3$ $\begin{array}{l} (a * b)^{*} c=9 * c=9^{3}+3^{3} \\ a *(b * c)=a...
let $\mathrm{a}=1, \mathrm{~b}=0 \in \mathrm{R}-\{-1\}$ $a * b=\frac{1}{0+1}=1$ And $b * a=\frac{0}{1+1}=0$ => $^{*}$ is not commutative. Let $c=3$ $\begin{array}{l} (a * b) * c=1^{*}...
(iii) let $c=3$. $(a * h) * c=1.5 *^{*} c=\frac{1}{-}(15+2)=275$ $a *(b * c)=a * \frac{1}{2}(2+3)=1 * 2.5=\frac{1}{2}(1+2.5)=1.75$ hence * is not associative.
(iii)let $\mathrm{x} \in \mathrm{N}$ and $\mathrm{x} * 1=\operatorname{lcm}(x, 1)=\mathrm{x}=\operatorname{lcm}(1, \mathrm{x})$ 1 is the identity element. (iv) let there exist $y$ in $n$ such that...
To prove: $*$ is neither commutative nor associative Let us assume that * is commutative $\Rightarrow a^{b}=b^{a}$ for all $a, b \in N$ This is valid only for $a=b$ For example take $a=1, b=2$...
To prove: $*$ is not a binary operation Since, a and $b$ are defined on positive integer set And $\mathrm{a}^{*} \mathrm{~b}=|\mathrm{a}-\mathrm{b}|$ $\Rightarrow a^{*} b=(a-b)$, when $a>b$...
$2 * 4$ since, $a^{*} b=a+3 b^{2}$ $\Rightarrow 2 * 4=\left(2+3 \times 4^{2}\right)=2+48=50$
$3 * 5$ and $5 * 3$ $a * b=(2 a-b)^{2}$ $\Rightarrow 3 * 5=(6-5)^{2}=1$ $5 * 3=(10-3)^{2}=49$ $\Rightarrow 3 * 5$ is not equal to $5 * 3$
To find: value of $x$ Since, $a * b=\frac{a b}{5}$ $\Rightarrow \mathrm{x} * 5=\frac{5 \mathrm{x}}{5}=\mathrm{x}$ Now $(2 * x)=\frac{2 x}{5}$ $\Rightarrow \frac{2 \mathrm{x}}{5}=10 \Rightarrow...
$4 * 5$ $a * b=3 a+4 b-2$ Since, $a=4$ and $b=5$ $\begin{array}{l} \Rightarrow 4 * 5=3 \times 4+4 \times 5-2=12+20-2=30 \\ \Rightarrow 4 * 5=30 \end{array}$
$\begin{array}{l} \sin \left(\frac{\pi}{2}+\frac{\pi}{3}\right) \\ =\sin \left(\frac{5 \pi}{6}\right) \\ =\sin \left(\pi-\frac{\pi}{6}\right) \\ =\sin \frac{\pi}{6} \\ =\frac{1}{2} \end{array}$
$\cos \left\{\pi-\frac{\pi}{6}+\frac{\pi}{6}\right\}$ $\begin{array}{l} =\cos \{\pi\} \\ =\cos \left(\frac{\pi}{2}+\frac{\pi}{2}\right) \\ =-1 \end{array}$
(iii) Let $\tan ^{-1}(-\sqrt{3})=x$ $\Rightarrow-\tan ^{-1}(\sqrt{3})=x\left[\right.$ Formula: $\left.\tan ^{-1}(-x)=-\tan ^{-1}(x)\right]$ $\Rightarrow \sqrt{3}=-\tan x$ $\therefore...
(vii) Let $\operatorname{cosec}^{-1}(\sqrt{2})=x$ $\Rightarrow \sqrt{2}=\operatorname{cosec} x$ $\therefore \mathrm{x}=\frac{\pi}{4}$
(v) Let $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=x$ $\Rightarrow \frac{1}{\sqrt{3}}=\tan \mathrm{x}$ [We know which value of $\mathrm{x}$ $\therefore \mathrm{x}=\frac{\pi}{6}$ (vi) Let $\sec...
(iii) Let $\cos ^{-1}\left(\frac{1}{2}\right)=x$ $\Rightarrow \frac{1}{2}=\cos x$ [We know which value of $x$ when put in this expression will give us this result] $\therefore...
(i) Let $\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=x$ $\Rightarrow \frac{\sqrt{3}}{2}=\sin x$ => $\therefore \mathrm{x}=\frac{\pi}{3}$ (ii) Let $\sin ^{-1}\left(\frac{1}{2}\right)=x$ $\Rightarrow...
(a) 4 (b) 5 (c) 3 (d) 2 Answer: (c) 3 We have: a5 = 20 and a7 + a11 = 64. Let a be the first term and d be the common difference of the AP. Then, Thus, the common difference of the AP is...
(a) 50 (b) -50 (c) 30 (d) -30 Answer: (b) -50
(a) (3n + 8) (b) (4n – 7) (c) (15 – 2n) (d) (2n – 15) Answer: (d) (2n – 15)
(a) (6n - 2) (b) (7n – 3) (c) (8n – 2) (d) (8n + 2) Answer: (c) (8n – 2)
(a) ( 5 - 2n) (b) ( 6 – 2n) (c) (2n – 5) (d) (2n – 6) Answer: (b) ( 6 – 2n)
(a) 6 (b) 9 (c) 15 (d) -3 Solution: (a) b
(a) 6n+3 (b) 15 (c) 12 (d) 21 Solution: (b) 15
(a) 19 (b) 23 (c) 22 (d) cannot be determined Solution: (c) 22
$ (a)\,\sqrt{84} $ $ (b)\,\sqrt{98} $ $ (c)\,\sqrt{70} $ $ (d)\,\sqrt{112} $ Solution: $ (d)\,\sqrt{112} $ The AP is $ \sqrt{7},\sqrt{28},\sqrt{63},...... $ $ =\sqrt{7},\sqrt{4\times...
(a)1/3 (b)-1/3 (c) b (d) –b Solution: (d) -b
(a) p (b) –p (c) -1 (d) 1 Solution: (c) -1
(vii) (viii)
(iv)