Factorise:
Question: Factorise: $x^{2}+7 \sqrt{6} x+60$ Solution: $x^{2}+7 \sqrt{6} x+60$ $=x^{2}+5 \sqrt{6} x+2 \sqrt{6} x+60$ $=x(x+5 \sqrt{6})+2 \sqrt{6}(x+5 \sqrt{6})$ $=(x+5 \sqrt{6})(x+2 \sqrt{6})$...
Read More →Show that the equation 2(a2+b2)x2+2(a+b)x+1=0 has no real roots,
Question: Show that the equation $2\left(a^{2}+b^{2}\right) x^{2}+2(a+b) x+1=0$ has no real roots, when $a \neq b$. Solution: The quadric equation is $2\left(a^{2}+b^{2}\right) x^{2}+2(a+b) x+1=0$ Here, $a=2\left(a^{2}+b^{2}\right), b=2(a+b)$ and,$c=1$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=2\left(a^{2}+b^{2}\right), b=2(a+b)$ and, $c=1$ $D=\{2(a+b)\}^{2}-4 \times 2\left(a^{2}+b^{2}\right) \times 1$ $=4\left(a^{2}+2 a b+b^{2}\right)-8\left(a^{2}+b^{2}\right)$ $=4 a^{2}+8 a b+4 b...
Read More →State with reasons whether the following functions have inverse:
Question: State with reasons whether the following functions have inverse: (i) $f:\{1,2,3,4\} \rightarrow\{10\}$ with $f=\{(1,10),(2,10),(3,10),(4,10)\}$ (ii) $g:\{5,6,7,8\} \rightarrow\{1,2,3,4\}$ with $g=\{(5,4),(6,3),(7,4),(8,2)\}$ (iii) $h:\{2,3,4,5\} \rightarrow\{7,9,11,13\}$ with $h=\{(2,7),(3,9),(4,11),(5,13)\}$ Solution: (i) $f:\{1,2,3,4\} \rightarrow\{10\}$ with $f=\{(1,10),(2,10),(3,10),(4,10)\}$ We have: $f(1)=f(2)=f(3)=f(4)=10$ $\Rightarrow f$ is not one-one. $\Rightarrow f$ is not a...
Read More →Solve each of the following system of equations in R. 7. 2x + 5 ≤ 0, x − 3 ≤ 0
Question: Solve each of the following system of equations in R. 7. 2x+ 5 0,x 3 0 Solution: We have, $2 x+5 \leq 0$ $\Rightarrow 2 x \leq-5$ $\Rightarrow x \leq \frac{-5}{2}$ $\Rightarrow x \in\left(-\infty, \frac{-5}{2}\right] \quad \ldots(\mathrm{i})$ Also, $x-3 \leq 0$ $\Rightarrow x \leq 3$ $\Rightarrow x \in(-\infty, 3] \quad \ldots \quad$ (ii) Thus, the solution of the given set of inequalities is the intersection of (i) and (ii). $\left(-\infty, \frac{-5}{2}\right] \cap(-\infty, 3]=\left(-...
Read More →Factorise:
Question: Factorise: $x^{2}-4 x+3$ Solution: $x^{2}-4 x+3$ $=x^{2}-3 x-x+3$ $=x(x-3)-1(x-3)$ $=(x-1)(x-3)$...
Read More →Solve each of the following system of equations in R. 6. 2x − 3 < 7,
Question: Solve each of the following system of equations in R. 6. 2x 3 7, 2x 4 Solution: $2 x-37$ $\Rightarrow 2 x7+3$ $\Rightarrow x5$ $\Rightarrow x \in(-\infty, 5) \quad \ldots(\mathrm{i})$ Also, $2 x-4$ $\Rightarrow x-2$ $\Rightarrow x \in(-2, \infty) \quad \ldots$ (ii) Thus, the solution of the given set of inequalities is the intersection of (i) and (ii). $(-\infty, 5) \cap(-2, \infty)=(-2,5)$ Thus, the solution of the given set of inequalities is $(-2,5)$....
Read More →If the roots of the equation (c2−ab)x2−2(a2−bc)x+b2−ac=0 are equal,
Question: If the roots of the equation $\left(c^{2}-a b\right) x^{2}-2\left(a^{2}-b c\right) x+b^{2}-a c=0$ are equal, prove that either $a=0$ or $a^{3}+b^{3}+c^{3}=3 a b c$. Solution: The given quadric equation is $\left(c^{2}-a b\right) x^{2}-2\left(a^{2}-b c\right) x+\left(b^{2}-a c\right)=0$, and roots are equal. Then prove that either $a=0$ or $a^{3}+b^{3}+c^{3}=3 a b c$ Here, $a=\left(c^{2}-a b\right), b=-2\left(a^{2}-b c\right)$ and,$c=\left(b^{2}-a c\right)$ As we know that $D=b^{2}-4 a ...
Read More →Factorise:
Question: Factorise: $x^{2}-22 x+120$ Solution: $x^{2}-22 x+120$ $=x^{2}-12 x-10 x+120$ $=x(x-12)-10(x-12)$ $=(x-10)(x-12)$...
Read More →If f, g : R → R be two functions defined as f(x) = |x| + x and g(x)
Question: If $f, \mathrm{~g}: \mathrm{R} \rightarrow \mathrm{R}$ be two functions defined as $f(x)=|x|+x$ and $\mathrm{g}(x)=|x|-x, \forall x \in \mathrm{R}$. Then find fog and gof. Hence find fog(-3), fog(5) and gof $(-2)$. Solution: Given: $f(x)=|x|+x$ and $g(x)=|x|-x, \forall x \in \mathrm{R}$ $f o g=f(g(x))=|g(x)|+g(x)$ $=|| x|-x|+(|x|-x)$ Therefore, $f(g(x))= \begin{cases}0 x \geq 0 \\ 4 x x0\end{cases}$ $f o g= \begin{cases}4 x x0 \\ 0 x \geq 0\end{cases}$ $g o f=g(f(x))=|f(x)|-f(x)$ $=|| ...
Read More →Solve each of the following system of equations in R. 5. 3x − 6 > 0,
Question: Solve each of the following system of equations in R. 5. 3x 6 0, 2x 5 0 Solution: $3 x-60$ $\Rightarrow 3 x6$ $\Rightarrow x2$ $\Rightarrow x \in(2, \infty) \ldots(\mathrm{i})$ Also, $2 x-50$ $\Rightarrow 2 x5$ $\Rightarrow x\frac{5}{2}$ $\Rightarrow x \in\left(\frac{5}{2}, \infty\right) \ldots$ (ii) Solution of the given set of inequalities is the intersection of (i) and (ii). $(2, \infty) \cap\left(\frac{5}{2}, \infty\right)=\left(\frac{5}{2}, \infty\right)$ Thus, the solution of the...
Read More →Factorise:
Question: Factorise: $x^{2}-21 x+90$ Solution: $x^{2}-21 x+90$ $=x^{2}-15 x-6 x+90$ $=x(x-15)-6(x-15)$ $=(x-6)(x-15)$...
Read More →Factorise:
Question: Factorise: $x^{2}+2 \sqrt{3} x-24$ Solution: $x^{2}+2 \sqrt{3} x-24$ $=x^{2}+4 \sqrt{3} x-2 \sqrt{3} x-24$ $=x(x+4 \sqrt{3})-2 \sqrt{3}(x+4 \sqrt{3})$ $=(x+4 \sqrt{3})(x-2 \sqrt{3})$...
Read More →If p, q are real and p ≠ q, then show that the roots of the equation
Question: If $p, q$ are real and $p \neq q$, then show that the roots of the equation $(p-q) x^{2}+5(p+q) x-2(p-q)=0$ are real and unequal. Solution: The quadric equation is $(p-q) x^{2}+5(p+q) x-2(p-q)=0$ Here, $a=(p-q), b=5(p+q)$ and, $c=-2(p-q)$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=(p-q), b=5(p+q)$ and, $c=-2(p-q)$ $D=\{5(p+q)\}^{2}-4(p-q)(-2(p-q))$ $=25\left(p^{2}+2 p q+q^{2}\right)+8\left(p^{2}-2 p q+q^{2}\right)$ $=25 p^{2}+50 p q+25 q^{2}+8 p^{2}-16 p q+8 q^{2}$ $=33 p^...
Read More →2x + 6 ≥ 0, 4x − 7 < 0
Question: 2x+ 6 0, 4x 7 0 Solution: We have, $2 x+60$ $\Rightarrow 2 x \geqslant-6$ $\Rightarrow x \geqslant-3$ $\Rightarrow x \in[-3, \infty) \ldots(\mathrm{i})$ Also, $4 x-70$ $\Rightarrow 4 x7$ $\Rightarrow x\frac{7}{4}$ $\Rightarrow x \in\left(-\infty, \frac{7}{4}\right) \ldots(\mathrm{ii})$ Thus, the solution of the given inequations is the intersection of (i) and (ii). $[-3, \infty) \cap\left(-\infty \frac{7}{4}\right)=\left[-3, \frac{7}{4}\right)$ Thus, the solution of the given inequatio...
Read More →Solve each of the following system of equations in R. 3. x − 2 > 0,
Question: Solve each of the following system of equations in R. 3.x 2 0, 3x 18 Solution: We have, $x-20$ $\Rightarrow x2$ $\Rightarrow x \in(2, \infty) \ldots$ (i) Also, $3 x18$ $\Rightarrow x6$ $\Rightarrow x \in(-\infty, 6) \ldots$ (ii) Solution of the given set of the inequations is intersection of (i) and (ii) $(2, \infty) \cap(-\infty, 6)=(2,6)$ Thus, $(2,6)$ is the solution of the given set of inequalities....
Read More →Factorise:
Question: Factorise: $x^{2}+7 x-98$ Solution: $x^{2}+7 x-98$ $=x^{2}+14 x-7 x-98$ $=x(x+14)-7(x+14)$ $=(x+14)(x-7)$...
Read More →If the roots of the equations ax2 + 2bx + c = 0 and bx2−2ac−−√x+b=0
Question: If the roots of the equations $a x^{2}+2 b x+c=0$ and $b x^{2}-2 \sqrt{a c} x+b=0$ are simultaneously real, then prove that $b^{2}=a c$. Solution: The given equations are $a x^{2}+2 b x+c=0 \ldots \ldots$ (1) $b x^{2}-2 \sqrt{a c} x+b=0 \ldots \ldots$ (2) Roots are simultaneously real Then prove that $b^{2}=a c$. Let $D_{1}$ and $D_{2}$ be the discriminants of equation (1) and (2) respectively, Then, $D_{1}=(2 b)^{2}-4 a c$ $=4 b^{2}-4 a c$ And $D_{2}=(-2 \sqrt{a c})^{2}-4 \times b \ti...
Read More →Solve each of the following system of equations in R. 2. 2x − 7 > 5 − x,
Question: Solve each of the following system of equations in R. 2. 2x 7 5 x, 11 5x 1 Solution: We have, $2 x-75-x$ $\Rightarrow 2 x+x5+7$ $\Rightarrow 3 x12$ $\Rightarrow x4$ $\Rightarrow x \in(4, \infty) \ldots$ (i) Also, $11-5 x \leq 1$ $\Rightarrow 5 x \geqslant 11-1$ $\Rightarrow x \geqslant 2$ $\Rightarrow x \in[2, \infty) \ldots$ (ii) $S$ olution set of the given set of inequations is intersection of (i) and (ii) $(4, \infty) \cap[2, \infty)=(4, \infty)$ Thus, the solution set of the given...
Read More →Let f(x)
Question: Let $f(x)=\left\{\begin{array}{ll}1+x, 0 \leq x \leq 2 \\ 3-x, 2x \leq 3\end{array}\right.$. Find fof Solution: $f(x)= \begin{cases}1+x, 0 \leq x \leq 2 \\ 3-x, 2x \leq 3\end{cases}$ It can be written as, $f(x)=\left\{\begin{array}{c}1+x, 0 \leq x \leq 1 \\ 1+x, 1x \leq 2 \\ 3-x, 2x \leq 3\end{array}\right.$ When, $0 \leq x \leq 1$ Now when, $0 \leq x \leq 1$ then, $1 \leq x+1 \leq 2$ When, $1x \leq 2$ Now when, $1x \leq 2$ then, $2x+1 \leq 3$ When, $2x \leq 3$ Now when , $2x \leq 3$ t...
Read More →Solve this
Question: $x^{2}+19 x-150$ Solution: $x^{2}+19 x-150$ $=x^{2}+25 x-6 x-150$ $=x(x+25)-6(x+25)$ $=(x+25)(x-6)$...
Read More →Solve each of the following system of equations in R.
Question: Solve each of the following system of equations in R. 1.x+ 3 0, 2x 14 Solution: $x+30$ $\Rightarrow x-3$ $\Rightarrow x \in(-3, \infty) \ldots(\mathrm{i})$ Also, $2 x14$ $\Rightarrow x7 \quad$ [Dividing both the sides by 2$]$ $\Rightarrow x \in(-\infty, 7) \ldots$ (ii) Thus, the solution of the given set of inequalities is the intersection of (i) and (ii). $(-3, \infty) \cap(-\infty, 7)=(-3,7)$ $\therefore x \in(-3,7)$ Thus, the solution of the given set of inequalities is $(-3,7)$....
Read More →If the roots of the equation (a2 + b2)x2 − 2 (ac + bd)x + (c2 + d2) = 0
Question: If the roots of the equation $\left(a^{2}+b^{2}\right) x^{2}-2(a c+b d) x+\left(c^{2}+d^{2}\right)=0$ are equal, prove that $\frac{a}{b}=\frac{c}{d}$. Solution: The given quadric equation is $\left(a^{2}+b^{2}\right) x^{2}-2(a c+b d) x+\left(c^{2}+d^{2}\right)=0$, and roots are real Then prove that $\frac{a}{b}=\frac{c}{d}$. Here, $a=\left(a^{2}+b^{2}\right), b=-2(a c+b d)$ and, $c=\left(c^{2}+d^{2}\right)$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=\left(a^{2}+b^{2}\right...
Read More →Factorise:
Question: Factorise: $x^{2}+20 x-69$ Solution: $x^{2}+20 x-69$ $=x^{2}+23 x-3 x-69$ $=x(x+23)-3(x+23)$ $=(x+23)(x-3)$...
Read More →Solve the following
Question: $\frac{x}{x-5}\frac{1}{2}$ Solution: $\frac{x}{x-5}\frac{1}{2}$ $\Rightarrow \frac{x}{x-5}-\frac{1}{2}0$ $\Rightarrow \frac{2 x-x+5}{2(x-5)}0$ $\Rightarrow \frac{x+5}{2(x-5)}0$ $\Rightarrow \frac{x+5}{x-5}0$ $\therefore x \in(-\infty,-5) \cup(5, \infty)$...
Read More →Solve the following
Question: $\frac{7 x-5}{8 x+3}4$ Solution: We have, $\frac{7 x-5}{8 x+3}4$ $\Rightarrow \frac{7 x-5}{8 x+3}-40$ $\Rightarrow \frac{7 x-5-4(8 x+3)}{8 x+3}0$ $\Rightarrow \frac{7 x-5-32 x-12}{8 x+3}0$ $\Rightarrow \frac{-25 x-17}{8 x+3}0$ $\Rightarrow \frac{25 x+17}{8 x+3}0 \quad$ (Multiplying by $-1$ to make the coefficient of $x$ in the LHS positive) $\mathrm{x} \in\left(\frac{-17}{25}, \frac{-3}{8}\right)$...
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