Solve the following
Question: $\frac{x}{5}\frac{3 x-2}{4}-\frac{5 x-3}{5}$ Solution: $\frac{x}{5}\frac{3 x-2}{4}-\frac{5 x-3}{5}$ $\Rightarrow 20 \times\left(\frac{x}{5}\right)20 \times\left(\frac{3 x-2}{4}-\frac{5 x-3}{5}\right) \quad$ (Multiplying both the sides by 20$)$ $\Rightarrow 4 x5(3 x-2)-4(5 x-3)$ $\Rightarrow 4 x15 x-10-20 x+12$ $\Rightarrow 4 x-5 x+2$ $\Rightarrow 4 x+5 x2 \quad$ (Transposing $-5 x$ to the LHS) $\Rightarrow 9 \mathrm{x}2$ $\Rightarrow x\frac{2}{9} \quad$ (Dividing both the sides by 9 ) ...
Read More →−(x − 3) + 4 < 5 − 2x
Question: (x 3) + 4 5 2x Solution: $-(x-3)+45-2 x$ $\Rightarrow-x+3+45-2 x$ $\Rightarrow-x+75-2 x$ $\Rightarrow-x+2 x5-7 \quad[$ Transposing $-2 \mathrm{x}$ to LHS and 7 to RHS $]$ $\Rightarrow x-2$ Hence, the solution set of the given inequation is $(-\infty,-2)$....
Read More →Solve this
Question: $x^{4}-625$ Solution: $x^{4}-625$ $=\left(x^{2}\right)^{2}-25^{2}$ $=\left(x^{2}+25\right)\left(x^{2}-25\right) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=\left(x^{2}+25\right)\left(x^{2}-5^{2}\right)$ $=\left(x^{2}+25\right)(x+5)(x-5) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →Solve the following
Question: $\frac{3 x-2}{5} \leq \frac{4 x-3}{2}$ Solution: $\frac{3 x-2}{5} \leq \frac{4 x-3}{2}$ $\Rightarrow 2(3 x-2) \leq 5(4 x-3)$ $\Rightarrow 6 x-4 \leq 20 x-15$ $\Rightarrow-4+15 \leq 20 x-6 x \quad$ (Transposing $6 x$ to the RHS and $-15$ to the LHS) $\Rightarrow 11 \leq 14 x$ $\Rightarrow 14 x \geq 11$ $\Rightarrow x \geq \frac{11}{14} \quad$ (Dividing both the sides by 14 ) Hence, the solution set of the given inequation is $\left[\frac{11}{14}, \infty\right)$....
Read More →Solve the following
Question: $2(3-x) \geq \frac{x}{5}+4$ Solution: $2(3-x) \geq \frac{x}{5}+4$ $\Rightarrow 6-2 x \geq \frac{x}{5}+4$ $\Rightarrow 6-4 \geq \frac{x}{5}+2 x \quad[$ Transposing $-2 x$ to the RHS and 4 to the LHS $]$ $\Rightarrow 2 \geq \frac{11 x}{5}$ $\Rightarrow \frac{11 x}{5} \leq 2$ $\Rightarrow x \leq \frac{10}{11} \quad\left[\right.$ Mltiplying both the sides by $\left.\frac{5}{11}\right]$ Thus, the solution set of the given inequation is $\left(-\infty, \frac{10}{11}\right]$....
Read More →3x + 9 ≥ −x + 19
Question: 3x+ 9 x+ 19 Solution: $3 x+9 \geqslant-x+19$ $\Rightarrow 3 x+x \geqslant 19-9$ (Transposing $-x$ to the LHS and 9 to the RHS) $\Rightarrow 4 x \geq 10$ $\Rightarrow x \geq \frac{5}{2}$(Dividing both the sides by 4 ) Hence, the solution set of the given inequation is $\left[\frac{5}{2}, \infty\right)$....
Read More →x + 5 > 4x − 10
Question: x+ 5 4x 10 Solution: We have, $x+54 x-10$ $\Rightarrow 5+104 x-x$ (Transposing $x$ to the RHS and $-10$ to the LHS) $\Rightarrow 153 \mathrm{x}$ $\Rightarrow 3 x15$ $\Rightarrow x5 \quad$ (Dividing both sides by 3 ) $\Rightarrow x \in(-\infty, 5)$ Hence, the solution set of the given inequation is $(-\infty, 5)$....
Read More →3x − 7 > x + 1
Question: 3x 7 x+ 1 Solution: $3 x-7x+1$ $\Rightarrow 3 x-x1+7$ ( Transposing $x$ to LHS and $-7$ to RHS) $\Rightarrow 2 x8$ $\Rightarrow x4$ ( Dividing both sides by 2 ) Hence, the solution set of the given inequation is $(4, \infty)$...
Read More →Solve: 4x − 2 < 8, when
Question: Solve: 4x 2 8, when(i)x R (ii)x Z (iii)x N Solution: We have, $4 x-28$ $\Rightarrow 4 x8+2$ (Transposing $-2$ to the RHS) $\Rightarrow 4 x10$ $\Rightarrow x\frac{10}{4}$ (Dividing both the sides by 4 ) $\Rightarrow x\frac{5}{2}$ (i) $x \in R$ Then, the solution of the given inequation is $\left(-\infty, \frac{5}{2}\right)$. (ii) $x \in Z$ Then, the solution of the given inequation is $\{\ldots \ldots \ldots-3,-2,-1,0,1,2\}$. (iii) $x \in N$ Then, the solution of the given inequation is...
Read More →Solve: −4x > 30, when
Question: Solve: 4x 30, when(i)x R (ii)x Z (iii)x N Solution: $-4 x30$ $\Rightarrow x-\frac{30}{4}$ (Dividing both the sides by $-4$ ) $\Rightarrow x-\frac{15}{2}$ (i) $x \in R$ Then, the solution of the given inequation is $\left(-\infty,-\frac{15}{2}\right)$. (ii) $x \in Z$ Then, the solution of the given inequation is $\{\ldots \ldots \ldots-9,-8\}$. (iii) $x \in N$ Then, the solution of the given inequation is $\phi$....
Read More →Solve: 12x < 50, when
Question: Solve: 12x 50, when (i)x R (ii)x Z (iii)x N Solution: We have, $12 x50$ $\Rightarrow x\frac{50}{12}$ [Dividing both the sides by 12] $\Rightarrow x\frac{25}{6}$ (i) $x \in R$ Then, the solution of the given inequation is $\left(-\infty, \frac{25}{6}\right)$. (ii) $x \in Z$ Then, the solution of the given inequation is $\{\ldots \ldots \ldots-3,-2,-1,0,1,2,3,4\}$. (iii) $x \in N$ Then, the solution of the given inequation is $\{1,2,3,4\}$....
Read More →If $f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R$ and $g:[-1,1] \rightarrow R$ be defined as
[question] Question. If $f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R$ and $g:[-1,1] \rightarrow R$ be defined as $f(x)=\tan x$ and $g(x)=\sqrt{1-x^{2}}$ respectively, describe fog and gof. [/question] [solution] Solution: $g(x)=\sqrt{1-x^{2}}$ $\Rightarrow x^{2} \geq 0, \forall x \in[-1,1]$ $\Rightarrow-x^{2} \leq 0, \forall x \in[-1,1]$ $\Rightarrow 1-x^{2} \leq 1, \forall x \in[-1,1]$ We know that $1-x^{2} \geq 0$ $\Rightarrow 0 \leq 1-x^{2} \leq 1$ $\Rightarrow$ Range of $g(x)=...
Read More →If $f(x)=\sqrt{1-x}$ and $g(x)=\log _{e} x$ are two real functions, then describe functions fog and $g \circ f$.
[question] Question. If $f(x)=\sqrt{1-x}$ and $g(x)=\log _{e} x$ are two real functions, then describe functions fog and $g \circ f$. [/question] [solution] Solution: $f(x)=\sqrt{1-x}$ For domain, $1-\mathrm{x} \geq 0$ $\Rightarrow x \leq 1$ $\Rightarrow$ domain of $f=(-\infty, 1]$ $\Rightarrow f:(-\infty, 1] \rightarrow(0, \infty)$ $g(x)=\log _{e} x$ Clearly, $g:(0, \infty) \rightarrow R$ Computation of $\mathrm{fog}$ : Clearly, the range of $g$ is not a subset of the domain of $f$. So, we need...
Read More →Let $f(x)=x^{2}+x+1$ and $g(x)=\sin x$. Show that fog $\neq$ gof.
[question] Question. Let $f(x)=x^{2}+x+1$ and $g(x)=\sin x$. Show that fog $\neq$ gof. [/question] [solution] Solution: $(f o g)(x)=f(g(x))=f(\sin x)=\sin ^{2} x+\sin x+1$ and $(g o f)(x)=g(f(x))=g\left(x^{2}+x+1\right)=\sin \left(x^{2}+x+1\right)$ So, $f o g \neq g o f$. [/solution]...
Read More →If $\left(x^{51}+51\right)$ is divided by
[question] Question. If $\left(x^{51}+51\right)$ is divided by $(x+1)$ then the remainder is (a) 0 (b) 1 (c) 49 (d) 50 [/question] [solution] Solution: Let $f(x)=x^{51}+51$ By remainder theorem, when f(x) is divided by (x + 1), then the remainder = f(−1). Putting x = −1 in f(x), we get $f(-1)=(-1)^{51}+51=-1+51=50$ ∴ Remainder = 50 Thus, the remainder when $\left(x^{51}+51\right)$ is divided by $(x+1)$ is 50 Hence, the correct answer is option (d). [/solution]...
Read More →If $p(x)=5 x-4 x^{2}+3$ then $p(-1)=?$
[question] Question. If $p(x)=5 x-4 x^{2}+3$ then $p(-1)=?$ (a) 2 (b) –2 (c) 6 (d) –6 [/question] [solution] Solution: $p(x)=5 x-4 x^{2}+3$ Putting $x=-1$ in $p(x)$, we get $p(-1)=5 \times(-1)-4 \times(-1)^{2}+3=-5-4+3=-6$ Hence, the correct answer is option (d). [/solution]...
Read More →Solve the following
Question: $3 x^{2}-4 x+\frac{20}{3}=0$ Solution: Given: $3 x^{2}-4 x+\frac{20}{3}=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x=c=0$, we get $a=3, b=-4$ and $c=\frac{20}{3}$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\Rightarrow \alpha=\frac{4+\sqrt{16-4 \times 3 \times \frac{20}{3}}}{6}$ and $\beta=\frac{4-\sqrt{16-4 \times 3 \times \frac{20}{3}}}{6}$ $\Rightarrow...
Read More →Solve the following
Question: $x^{2}-2 x+\frac{3}{2}=0$ Solution: $x^{2}-2 x+\frac{3}{2}=0$ $\Rightarrow x^{2}-2 x+1+\frac{1}{2}=0$ $\Rightarrow(x-1)^{2}-\left(\frac{1}{\sqrt{2}} i\right)^{2}=0$ $\Rightarrow\left(x-1+\frac{1}{\sqrt{2}} i\right)\left(x-1-\frac{1}{\sqrt{2}} i\right)=0$ $\Rightarrow\left(x-1-\frac{1}{\sqrt{2}} i\right)=0$ or, $\left(x-1+\frac{1}{\sqrt{2}} i\right)=0$ $\Rightarrow x=1+\frac{1}{\sqrt{2}} i \quad$ or, $\quad x=1-\frac{1}{\sqrt{2}} i$ Hence, the roots of the equation are $1 \pm \frac{1}{\...
Read More →Solve the following
Question: $-x^{2}+x-2=0$ Solution: $-x^{2}+x-2=0$ $\Rightarrow x^{2}-x+2=0$ $\Rightarrow x^{2}-x+\frac{1}{4}+\frac{7}{4}=0$ $\Rightarrow x^{2}-2 \times x \times \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\frac{7}{4} i^{2}=0$ $\Rightarrow\left(x-\frac{1}{2}\right)^{2}-\left(\frac{i \sqrt{7}}{2}\right)^{2}=0$ $\Rightarrow\left(x-\frac{1}{2}+\frac{i \sqrt{7}}{2}\right)\left(x-\frac{1}{2}-\frac{i \sqrt{7}}{2}\right)=0$ $\Rightarrow\left(x-\frac{1}{2}+\frac{\sqrt{7}}{2} i\right)=0$ or, $\left(x-\frac{1...
Read More →Solve the following
Question: $-x^{2}+x-2=0$ Solution: $-x^{2}+x-2=0$ $\Rightarrow x^{2}-x+2=0$ $\Rightarrow x^{2}-x+\frac{1}{4}+\frac{7}{4}=0$ $\Rightarrow x^{2}-2 \times x \times \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\frac{7}{4} i^{2}=0$ $\Rightarrow\left(x-\frac{1}{2}\right)^{2}-\left(\frac{i \sqrt{7}}{2}\right)^{2}=0$ $\Rightarrow\left(x-\frac{1}{2}+\frac{i \sqrt{7}}{2}\right)\left(x-\frac{1}{2}-\frac{i \sqrt{7}}{2}\right)=0$ $\Rightarrow\left(x-\frac{1}{2}+\frac{\sqrt{7}}{2} i\right)=0$ or, $\left(x-\frac{1...
Read More →Solve the following
Question: $\sqrt{5} x^{2}+x+\sqrt{5}=0$ Solution: Given: $\sqrt{5} x^{2}+x+\sqrt{5}=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=\sqrt{5}, b=1$ and $c=\sqrt{5}$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{-1+\sqrt{1-4 \times \sqrt{5} \times \sqrt{5}}}{2 \sqrt{5}}$ and $\quad \beta=\frac{-1-\sqrt{1-4 \times \sqrt{5} \times \sqrt{5}}}{2 \...
Read More →Solve the following
Question: $x^{2}+\frac{x}{\sqrt{2}}+1=0$ Solution: Given equation: $x^{2}+\frac{x}{\sqrt{2}}+1=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=1, b=\frac{1}{\sqrt{2}}$ and $c=1$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{-\frac{1}{\sqrt{2}}+\sqrt{\frac{1}{2}-4 \times 1 \times 1}}{2} \quad$ and $\quad \beta=\frac{-\frac{1}{\sqrt{2}}-\sqrt{...
Read More →Solve the following
Question: $x^{2}+x+\frac{1}{\sqrt{2}}=0$ Solution: Given equation: $x^{2}+x+\frac{1}{\sqrt{2}}=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=1, b=1$ and $c=\frac{1}{\sqrt{2}}$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{-1+\sqrt{1-4 \times \frac{1}{\sqrt{2}}}}{2}$ and $\beta=\frac{-1-\sqrt{1-4 \times \frac{1}{\sqrt{2}}}}{2}$ $\Rightarrow...
Read More →Solve the following
Question: $\sqrt{2} x^{2}+x+\sqrt{2}=0$ Solution: Given: $\sqrt{2} x^{2}+x+\sqrt{2}=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=\sqrt{2}, b=1$ and $c=\sqrt{2}$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{-1+\sqrt{1-4 \times \sqrt{2} \times \sqrt{2}}}{2 \sqrt{2}}$ and $\beta=\frac{-1-\sqrt{1-4 \times \sqrt{2} \times \sqrt{2}}}{2 \sqrt{2...
Read More →Solve the following
Question: $\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}=0$ Solution: Given: $\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=\sqrt{3}, b=-\sqrt{2}$ and $c=3 \sqrt{3}$. Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get: $\alpha=\frac{\sqrt{2}+\sqrt{2-4 \times \sqrt{3} \times 3 \sqrt{3}}}{2 \sqrt{3}}$ and $\beta=\frac{\sqrt{2}-\sqrt{2...
Read More →