Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $x^{2}-4 \sqrt{2} x+6=0$ Solution: We have been given $x^{2}-4 \sqrt{2} x+6=0$ $x^{2}-3 \sqrt{2} x-\sqrt{2} x+6=0$ $x(x-3 \sqrt{2})-\sqrt{2}(x-3 \sqrt{2})=0$ $(x-\sqrt{2})(x-3 \sqrt{2})=0$ Therefore, $x-\sqrt{2}=0$ $x=\sqrt{2}$ or, $x-3 \sqrt{2}=0$ $x=3 \sqrt{2}$ Hence, $x=\sqrt{2}$ or $x=3 \sqrt{2}$....
Read More →When p(x) = x3 + ax2 + 2x + a is divided by
Question: When $p(x)=x^{3}+a x^{2}+2 x+a$ is divided by $(x+a)$, the remainder is (a) 0 (b) $-1$ (c) $-15$ (d) 21 Solution: (c) a $x+a=0 \Rightarrow x=-a$ By the remainder theorem, we know that whenp(x) is divided by (x+a), the remainder isp(a).Thus, we have: $p(-a)=(-a)^{3}+a \times(-a)^{2}+2 \times(-a)+a$ $=-a^{3}+a^{3}-2 a+a$ $=-a$...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\left(x-\frac{1}{2}\right)^{2}=4$ Solution: We have been given $\left(x-\frac{1}{2}\right)^{2}=4$ $x^{2}+\frac{1}{4}-x=4$ $x^{2}-x-\frac{15}{4}=0$ $4 x^{2}-4 x-15=0$ Now we solve the above quadratic equation using factorization method. Therefore, $4 x^{2}-10 x+6 x-15=0$ $2 x(2 x-5)+3(2 x-5)=0$ $(2 x+3)(2 x-5)=0$ Now, one of the products must be equal to zero for the whole product to be zero. Hence we equate both the products to...
Read More →Solve the following
Question: If |z+ 4| 3, then the greatest and least values of |z+ 1| are _______ and ____________. Solution: Given |z+ 4| 3 here |z+ 1| = |z+ 1 + 3 3| = |z+ 4 + (3)| Since |a+b| |a| + |b| |z+ 4| + |3| = |z+ 4| + 3 3 + 3 (given) hence maximum value of |z+ 1| is 6 |z+ 1| = |z+ 43| Since |a b| ||a| |b|| |a|+ |b| ⇒ |z+ 1| |z+ 4| + 3 Since |z+ 4| 3 ⇒ |z+ 4| 3 i.e |z+ 1| |z+ 4| + 3 3 + 3 = 0 Hence, minimum value of|z+ 1| is 0....
Read More →When p(x) = x3 – ax2
Question: When $p(x)=x^{3}-a x^{2}+x$ is divided by $(x-a)$, the remainder is (a) 0 (b) $a$ (c) $2 a$ (d) $3 a$ Solution: By remainder theorem, when $p(x)=x^{3}-a x^{2}+x$ is divided by $(x-a)$, then the remainder $=p(a)$. Putting $x=a$ in $p(x)$, we get $p(a)=a^{3}-a \times a^{2}+a=a^{3}-a^{3}+a=a$ $\therefore$ Remainder $=a$ Hence, the correct answer is option (b)....
Read More →For any non-zero complex number
Question: For any non-zero complex number $z, \arg (z)+\arg (\bar{z})=$ ___________________ Solution: For complex numberz Say $z=x+i y=r e^{i \theta}$ where $r=$ modulus of $z, \theta=$ argument of $\bar{z}=x-i y$ $\Rightarrow \bar{z}=r e^{-i \theta}$ Let us arg z = Since $\arg \bar{z}=-\arg z$ $\Rightarrow \arg z+\arg \bar{z}=\theta+(-\theta)$ i.e $\arg z+\arg (\bar{z})=0$...
Read More →For any non-zero complex number
Question: For any non-zero complex number $z, \arg (z)+\arg (\bar{z})=$ ___________________ Solution: For complex numberz Say $z=x+i y=r e^{i \theta}$ where $r=$ modulus of $z, \theta=$ argument of $\bar{z}=x-i y$ $\Rightarrow \bar{z}=r e^{-i \theta}$ Let us arg z = Since $\arg \bar{z}=-\arg z$ $\Rightarrow \arg z+\arg \bar{z}=\theta+(-\theta)$ i.e $\arg z+\arg (\bar{z})=0$...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $a x^{2}+\left(4 a^{2}-3 b\right) x-12 a b=0$ Solution: We have been given $a x^{2}+\left(4 a^{2}-3 b\right) x-12 a b=0$ $a x^{2}+4 a^{2} x-3 b x-12 a b=0$ $a x(x+4 a)-3 b(x+4 a)=0$ $(a x-3 b)(x+4 a)=0$ Therefore, $a x-3 b=0$ $a x=3 b$ $x=\frac{3 b}{a}$ or $x+4 a=0$ $x=-4 a$ Hence, $x=\frac{3 b}{a}$ or $x=-4 a$....
Read More →Solve the following
Question: Ifz1andz2are two complex numbers such thatz1+z2is a real number, thenz2= ____________. Solution: Given for two complex numbers,z1andz2, we havez1+z2is real number Let $z_{1}=x_{1}+i y_{1}$ $z_{2}=x_{2}+i y_{2}$ $\Rightarrow z_{1}+z_{2}=x_{1}+i y_{1}+x_{2}+i y_{2}$ i. e $z_{1}+z_{2}=\left(x_{1}+x_{2}\right)+i\left(y_{1}+y_{2}\right)$ Since $z_{1}+z_{2}$ is real $\Rightarrow y_{1}+y_{2}=0$ i. e $y_{1}=-y_{2}$ i.e $z_{2}=x_{2}+i y_{2}=x_{2}-i y_{1}$...
Read More →Solve the following
Question: Ifz1andz2are two complex numbers such thatz1+z2is a real number, thenz2= ____________. Solution: Given for two complex numbers,z1andz2, we havez1+z2is real number Let $z_{1}=x_{1}+i y_{1}$ $z_{2}=x_{2}+i y_{2}$ $\Rightarrow z_{1}+z_{2}=x_{1}+i y_{1}+x_{2}+i y_{2}$ i. e $z_{1}+z_{2}=\left(x_{1}+x_{2}\right)+i\left(y_{1}+y_{2}\right)$ Since $z_{1}+z_{2}$ is real $\Rightarrow y_{1}+y_{2}=0$ i. e $y_{1}=-y_{2}$ i.e $z_{2}=x_{2}+i y_{2}=x_{2}-i y_{1}$...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $4 x^{2}+4 b x-\left(a^{2}-b^{2}\right)=0$ Solution: We have been given $4 x^{2}+4 b x-\left(a^{2}-b^{2}\right)=0$ $4 x^{2}+2(a+b) x-2(a-b) x-\left(a^{2}-b^{2}\right)=0$ $2 x(2 x+a+b)-(a-b)(2 x+a+b)=0$ $(2 x-(a-b))(2 x+a+b)=0$ Therefore, $2 x-(a-b)=0$ $2 x=a-b$ $x=\frac{a-b}{2}$ or, $2 x+a+b=0$ $2 x=-(a+b)$ $x=\frac{-(a+b)}{2}$ Hence, $x=\frac{a-b}{2}$ or $x=\frac{-(a+b)}{2}$....
Read More →Show that the logarithmic function f :
Question: Solution: $f: R^{+} \rightarrow R$ given by $f(x)=\log _{a} x, a0$ Injectivity:Letxandybe any two elements in the domain (N), such thatf(x)= f(y). $f(x)=f(y)$ $\log _{a} x=\log _{a} y$ $\Rightarrow x=y$ Surjectivity:Letybe any element in the co-domain(R),such thatf(x) = yfor some elementxinR+(domain). $f(x)=y$ $\log _{a} x=y$ $\Rightarrow x=a^{y} \in R^{+}$ So, for every element in the co-domain, there exists some pre-image in the domain. $\Rightarrow f$ is onto. Since $f$ is one-one a...
Read More →Solve the following
Question: If $|z|=4$ and $\arg (z)=\frac{5 \pi}{6}$, then $z=$ ___________________ Solution: If $|z|=4$ and $\arg z=\frac{5 \pi}{6}=\frac{\pi-\pi}{6}$ $\therefore z=|z|\left[\cos \left(\frac{5 \pi}{6}\right)+i \sin \left(\frac{5 \pi}{6}\right)\right]$ $=4\left[\cos \left(\pi-\frac{\pi}{6}\right)+i \sin \left(\pi-\frac{\pi}{6}\right)\right]$ [Since $\cos (\pi-\theta)=-\cos \theta$ and $\sin (\pi-\theta)=\sin \theta]$ $\therefore z=4\left[-\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right]$ Since $\co...
Read More →the remainder is
Question: When $p(x)=4 x^{3}-12 x^{2}+11 x-5$ is divided by $(2 x-1)$, the remainder is (a) 0 (b) $-5$ (c) $-2$ (d) 2 Solution: (c) 2 $2 x-1=0 \Rightarrow x=\frac{1}{2}$ By the remainder theorem, we know that when $p(x)$ is divided by $(2 x-1)$, the remainder is $p\left(\frac{1}{2}\right)$. Now, we have: $p\left(\frac{1}{2}\right)=4 \times\left(\frac{1}{2}\right)^{3}-12 \times\left(\frac{1}{2}\right)^{2}+11 \times \frac{1}{2}-5$ $=\frac{1}{2}-3+\frac{11}{2}-5$ $=-2$...
Read More →The multiplicative inverse of (1 + i) is
Question: The multiplicative inverse of (1 +i) is ____________. Solution: For z = 1 +i Let us suppose multiplicative inverse of 1 +iisa+ib then (1 +i) (a+ib) = 1 i.ea+ib+ai+i2b= 1 i.ea+ib+iab= 1 i.e $a-b+i(a+b)=1+i 0$ On comparing, real and imaginary part, we get a b= 1 anda+b= 0 i.eab= 1 anda= b i.ea+a= 1 i. e $a=\frac{1}{2}$ $b=-\frac{1}{2}$ i.e multiplicative inverse of $1+i$ is $\frac{1}{2}-\frac{i}{2}$...
Read More →Solve this
Question: When $p(x)=x^{3}-3 x^{2}+4 x+32$ is divided by $(x+2)$, the remainder is (a) 0 (b) 32 (c) 36 (d) 4 Solution: (d) 4 $x+2=0 \Rightarrow x=-2$ By the remainder theorem, we know that whenp(x) is divided by (x+ 2), the remainder isp(-2).Now, we have: $p(-2)=(-2)^{3}-3 \times(-2)^{2}+4 \times(-2)+32$ $=-8-12-8+32$ $=4$...
Read More →The sum of the series
Question: The sum of the seriesi+i2+i3+_____ upto 1000 terms is ____________. Solution: i+i2+i3+_____ 1000 terms i.ei+i2+i3+i4_____ +i1000 =i+i2i +1+i5_____ +i1000 =i 1 i+ 1 +i5+ _______ +i1000 $=0+i^{5}+i^{6}+\quad+i^{100}$ Similarly, sum of next form terms is also zero. (∵ 1000 = 4(250) i.e multiple of 4 Hence,i+i2+i3+_____i1000= 0...
Read More →When p(x)
Question: When $p(x)=x^{4}+2 x^{3}-3 x^{2}+x-1$ is divided by $(x-2)$, the remainder is (a) 0 (b) $-1$ (c) $-15$ (d) 21 Solution: (d) 21 $x-2=0 \Rightarrow x=2$ By the remainder theorem, we know that when $p(x)$ is divided by $(x-2)$, the remainder is $p(2)$. Thus, we have: $p(2)=2^{4}+2 \times 2^{3}-3 \times 2^{2}+2-1$ $=16+16-12+1$ $=21$...
Read More →The complex number
Question: The complex number $\frac{(1-i)^{3}}{1-i^{3}}$ in polar form is _________________ Solution: $\frac{(1-i)^{3}}{1-i^{3}}=\frac{(1-i)^{3}}{1+i} \quad\left(\because i^{2}=-1\right)$ $=\frac{(1-i)^{3}}{1+i} \times \frac{1-i}{1-i}$ $=\frac{(1-i)^{4}}{1-i^{2}}$ $=\frac{1}{2}\left\{(1-i)^{2}(1-i)^{2}\right\}$ $=\frac{1}{2}\left\{\left(1+i^{2}-2 i\right)\left(1+i^{2}-2 i\right)\right\}$ $=\frac{1}{2}\{(1-1-2 i)(1-1-2 i)\}$ $=\frac{1}{2}\{+2 i \times 2 i\}$ $=\frac{1}{2} \times 4 i^{2}$ $=-2=-2(...
Read More →Solve this
Question: If $(x+1)$ is a factor of $\left(2 x^{2}+k x\right)$, then $k=?$ (a) 4 (b) $-3$ (c) 2 (d) $-2$ Solution: (c) 2. $(x+1)$ is a factor of $2 x^{2}+k x$ So, $-1$ is a zero of $2 x^{2}+k x$. Thus, we have : $2 \times(-1)^{2}+k \times(-1)=0$ $\Rightarrow 2-k=0$ $\Rightarrow k=2$...
Read More →The value of
Question: The value of $\sqrt{-25} \times \sqrt{-9}$ is _________________ Solution: $\sqrt{-25} \times \sqrt{-9}$ $=5 i \times 3 i \quad(\because \sqrt{-1}=i)$ $=15 i^{2}$ $=-15$ value of $\sqrt{-25} \times \sqrt{-9}$ is $-15$...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0$ Solution: We have been given $x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0$ $x^{2}-\sqrt{3} x-x+\sqrt{3}=0$ $x(x-\sqrt{3})-1(x-\sqrt{3})=0$ $(x-1)(x-\sqrt{3})=0$ Therefore, $x-1=0$ $x=1$ or $x-\sqrt{3}=0$ $x=\sqrt{3}$ Hence, $x=1$ or $x=\sqrt{3}$....
Read More →The polar form of
Question: The polar form of $\left(1^{25}\right)^{3}$ is __________________ Solution: $\left(i^{25}\right)^{3}$ $=\left(i^{24} \cdot i\right)^{3}$ Since $i^{24}=1$ $=(i)^{3}=i^{2} \cdot i=-i$ i.e $\left(i^{25}\right)^{3}=-i$ Here modulus $r=\sqrt{0^{2}+1^{2}}=1$ and argument $\theta=\tan ^{-1}\left|\frac{-1}{0}\right|=\infty=-\frac{\pi}{2}$ $\Rightarrow-i$ has polar form $=\cos \left(\frac{-\pi}{2}\right)+i \sin \left(\frac{-\pi}{2}\right)$ Since $\cos (-\theta)=\cos \theta$ and $\sin (-\theta)=...
Read More →Show that the exponential function $f: R ightarrow R$, given by $f(x)=e^{x}$,
Question: Solution: $f: R \rightarrow R$, given by $f(x)=\mathrm{e}^{x}$ Injectivity:Letxandybe any two elements in the domain (R), such thatf(x) = f(y) $f(x)=f(y)$ $\Rightarrow e^{x}=e^{y}$ $\Rightarrow x=y$ So,fis one-one. Surjectivity:We know that range ofexis (0, ) =R+Co-domain =RBoth are not same.So,fis not onto. If the co-domain is replaced byR+, then the co-domain and range become the same and in that case,fis onto and hence, it is a bijection....
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $x^{2}-(\sqrt{2}+1) x+\sqrt{2}=0$ Solution: We have been given $x^{2}-(\sqrt{2}+1) x+\sqrt{2}=0$ $x^{2}-\sqrt{2} x-x+\sqrt{2}=0$ $x(x-\sqrt{2})-1(x-\sqrt{2})=0$ $(x-1)(x-\sqrt{2})=0$ Therefore, $x-1=0$ $x=1$ or, $x-\sqrt{2}=0$ $x=\sqrt{2}$ Hence, $x=1$ or $x=\sqrt{2}$....
Read More →