Give examples of two surjective functions
Question: Give examples of two surjective functions $f_{1}$ and $f_{2}$ from $Z$ to $Z$ such that $f_{1}+f_{2}$ is not surjective. Solution: We know that $f_{1}: R \rightarrow R$, given by $f_{1}(x)=x$, and $f_{2}(x)=-x$ are surjective functions. Proving $f_{1}$ is surjective : Let $y$ be an element in the co-domain $(R)$, such that $f_{1}(x)=y$. $f_{1}(x)=y$ $\Rightarrow x=y$, which is in $R$. So, for every element in the co-domain, there exists some pre-image in the domain. So, $f_{1}$ is surj...
Read More →The locus of z satisfying arg
Question: The locus of $z$ satisfying $\arg (z)=\frac{\pi}{3}$ is _______________ Solution: Given arg $(z)=\frac{\pi}{3}$ and for $z=x+i y$ $\frac{\pi}{3}=\tan ^{-1}\left(\frac{y}{x}\right)$ i. e $\tan \frac{\pi}{3}=\frac{y}{x}$ i. e $\sqrt{3}=\frac{y}{x}$ i.e $y=\sqrt{3} x$ in I quadrant except origin...
Read More →The value of
Question: The value of $(-\sqrt{-1})^{4 n-3}$, where $n \in N$, is Solution: $(-\sqrt{-1})^{4 n-3}$ $=(-i)^{4 n-3}$ $=(-i)^{4 n}(-i)^{-3}$ $=1(-i)^{-3}$ $=\left(\frac{1}{-i}\right)^{3}$ $=-\frac{1}{i^{3}}$ $=\frac{-1}{i^{2} \cdot i}$ $=\frac{-1}{-1(i)}$ $=+\frac{1}{i} \times \frac{i}{i}$ $=\frac{i}{i^{2}}$ Hence, $(-\sqrt{-1})^{4 n-3}=-i$...
Read More →The zeros of the polynomial p(x)
Question: The zeros of the polynomial $p(x)=2 x^{2}+5 x-3$ are (a) $\frac{1}{2}, 3$ (b) $\frac{1}{2},-3$ (c) $\frac{-1}{2}, 3$ (d) $1, \frac{-1}{2}$ Solution: The given polynomial is $p(x)=2 x^{2}+5 x-3$. Putting $x=\frac{1}{2}$ in $p(x)$, we get $p\left(\frac{1}{2}\right)=2 \times\left(\frac{1}{2}\right)^{2}+5 \times \frac{1}{2}-3=\frac{1}{2}+\frac{5}{2}-3=3-3=0$ Therefore, $x=\frac{1}{2}$ is a zero of the polynomial $p(x)$. Puttingx= 3 inp(x), we get $p(-3)=2 \times(-3)^{2}+5 \times(-3)-3=18-1...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: Solution: We have been given $\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}$ $x^{2}-x+2 x^{2}-4 x=6\left(x^{2}-x-2 x+2\right)$ $3 x^{2}-13 x+12=0$ Therefore, $3 x^{2}-9 x-4 x+12=0$ $3 x(x-3)-4(x-3)=0$ $(3 x-4)(x-3)=0$ Therefore, $3 x-4=0$ $3 x=4$ $x=\frac{4}{3}$ or, $x-3=0$ $x=3$ Hence, $x=\frac{4}{3}$ or $x=3$....
Read More →Solve the following
Question: If $|z|=2$ and $\arg (z)=\frac{\pi}{4}$, then $z=$ Solution: for $|z|=2=r \arg z=\frac{\pi}{4}$ $z=r(\cos (\arg z)+i \sin (\arg z))$ i. e $z=2\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$ $=2\left(\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right)$ $z=\sqrt{2}+i \sqrt{2}$ hence, $z=\sqrt{2}(1+i)$...
Read More →The real value of 'a' for which
Question: The real value of 'a' for which 3i3 2ai2+ (1 a)i+ 5 is real is ____________. Solution: $3 \hat{}^{3}-2 a i^{2}+(1-a) i+5$ i.e $32^{2} i-2 a(-1)+(1-a) i+5$ i.e 3i+ 2a+ (1 a)i+ 5 i.e 2a+ 5 +i(1 a 3) i.e 2a+ 5 +i(a 2) Since the above expression is given to be real ⇒ a 2 = 0 ⇒a= 2...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: Solution: We have been given $\frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7}$ $7\left(x^{2}+9-6 x-x^{2}-9-6 x\right)=48\left(x^{2}-9\right)$ $48 x^{2}+84 x-432=0$ $4 x^{2}+7 x-36=0$ Therefore, $4 x^{2}+16 x-9 x-36=0$ $4 x(x+4)-9(x+4)=0$ $(4 x-9)(x+4)=0$ Therefore, $4 x-9=0$ $4 x=9$ $x=\frac{9}{4}$ or, $x+4=0$ $x=-4$ Hence, $x=\frac{9}{4}$ or $x=-4$....
Read More →The zeros of the polynomial p(x)
Question: The zeros of the polynomial $p(x)=x^{2}+x-6$ are (a) 2,3 (b) $-2,3$ (c) $2,-3$ (d) $-2,-3$ Solution: The given polynomial is $p(x)=x^{2}+x-6$. Puttingx= 2 inp(x), we get $p(2)=2^{2}+2-6=4+2-6=0$ Therefore,x= 2is a zero of the polynomialp(x).Puttingx= 3 inp(x), we get $p(-3)=(-3)^{2}-3-6=9-9=0$ Therefore,x= 3 is a zero of the polynomialp(x).Thus, 2 and3 are the zeroes of the givenpolynomialp(x).Hence, the correct answer is option (c)....
Read More →Give examples of two one-one functions
Question: Give examples of two one-one functions $f_{1}$ and $f_{2}$ from $R$ to $R$, such that $f_{1}+f_{2}: R \rightarrow R$. defined by $\left(f_{1}+f_{2}\right)(x)=f_{1}(x)+f_{2}(x)$ is not one-one. Solution: We know that $f_{1}: R \rightarrow R$, given by $f_{1}(x)=x$, and $f_{2}(x)=-x$ are one-one. Proving $f_{1}$ is one-one: Let $f_{1}(x)=f_{1}(y)$ $\Rightarrow x=y$ So, $f_{1}$ is one-one. Proving $f_{2}$ is one-one: Let $f_{2}(x)=f_{2}(y)$ $\Rightarrow-x=-y$ $\Rightarrow x=y$ So, $f_{2}$...
Read More →Solve the following
Question: Ifx 0 is a real number, then arg (x) = ____________. Solution: Ifx 0 i.ez=x+i0 andxis negative $\Rightarrow \theta=\tan ^{-1}\left|\frac{0}{x}\right|=\tan ^{-1} 0=0$ Hence, $z$ lies in II quadrant chg $z=\pi$....
Read More →Solve the following
Question: If $z=-1+\sqrt{-} \overline{3}$, then $\arg (z)=$ Solution: $z=-1+i \sqrt{-} \overline{3}$ $x=-1$ and $y=\sqrt{3}$ $\theta=\tan ^{-1} \frac{\sqrt{3}}{-1}$ $=\tan ^{-1}|-\sqrt{3}|$ $\theta=\frac{\pi}{3}$ Sincezlies in IV quadrant. $\Rightarrow$ argument of $z$ is $\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$....
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}$ Solution: We have been given $\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}$ $4\left(x^{2}+3 x-x+x^{2}+2-2 x\right)=17\left(x^{2}-2 x\right)$ $9 x^{2}-34 x-8=0$ $9 x^{2}-36 x+2 x-8=0$ $9 x(x-4)+2(x-4)=0$ $(9 x+2)(x-4)=0$ Therefore, $9 x+2=0$ $9 x=-2$ $x=\frac{-2}{9}$ or, $x-4=0$ $x=4$ Hence, $x=\frac{-2}{9}$ or $x=4$....
Read More →Zero of the polynomial p(x)
Question: Zero of the polynomial $p(x)=2 x+5$ is (a) $\frac{-2}{5}$ (b) $\frac{-5}{2}$ (c) $\frac{2}{5}$ (d) $\frac{5}{2}$ Solution: The zero of the polynomialp(x) can be obtained by puttingp(x) = 0. $p(x)=0$ $\Rightarrow 2 x+5=0$ $\Rightarrow 2 x=-5$ $\Rightarrow x=-\frac{5}{2}$ Hence, the correct answer is option (b)....
Read More →Solve the following
Question: If |z+ 2| = |z 2|, then the locus ofzis ____________. Solution: |z+ 2| = |z 2| forz = x + iy i.e |x+iy+ 2| = |x+iy 2| i.e |(x+ 2) +iy| = |(x 2) +iy| Square both sides, |(x+ 2)| +iy|2= |(x 2)| +iy|2 i.e (x+ 2)2+y2= (x 2)2+y2 i.e $x^{2}+4+4 x+y^{2}=x^{2}+4-4 x+y^{2}$ i.e $f x=0$ i.e $x=0$ Hence, locus is perpendicular bisector of the segment joining (2, 0) and (2, 0)....
Read More →Find the number of all onto functions from the set A = {1, 2, 3, ..., n} to itself.
Question: Find the number of all onto functions from the setA= {1, 2, 3, ...,n} to itself. Solution: We know that every onto function from $A$ to itself is one-one. So, the number of one-one functions $=$ number of bijections $=n !$...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3}$ Solution: We have been given $\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3}$ $\frac{2 x(x-3)+(2 x-5)(x-4)}{x^{2}-7 x+12}=\frac{25}{3}$ $6 x^{2}-18 x+6 x^{2}-24 x-15 x+60=25 x^{2}-175 x+300$ $13 x^{2}-118 x+240=0$ $13 x^{2}-78 x-40 x+240=0$ $13 x(x-6)-40(x-6)=0$ $(x-6)(13 x-40)=0$ Therefore, $x-6=0$ $x=6$ or, $13 x-40=0$ $13 x=40$ $x=\frac{40}{13}$ Hence, $x=6$ or $x=\frac{40}{13}...
Read More →If A = {1, 2, 3}, show that a onto function
Question: IfA= {1, 2, 3}, show that a onto functionf:AAmust be one-one. Solution: A ={1, 2, 3}Possible onto functions fromAtoAcan be the following: (i) {(1, 1), (2, 2), (3, 3)}(ii) {(1, 1), (2, 3), (3, 2)}(iii) {(1, 2 ), (2, 2), (3, 3 )}(iv) {(1, 2), (2, 1), (3, 3)}(v) {(1, 3), (2, 2), (3, 1)}(vi) {(1, 3), (2, 1), (3,2 )} Here, in each function, different elements of the domain have different images.So, all the functions are one-one....
Read More →Solve the following
Question: If |z+ 2i| = |z 2i|, then the locus ofzis ____________. Solution: Given |z+ 2i| = |z 2i| forz= |x+iy| |x+iy+ 2i| = |x+iy 2i| Squaring both sides, |x+i(y+ 2)|2= |x+i(y 2)|2 i.ex2+ (y+ 2)|2=x2+ (y 2)|2 i.ex2+y2+ 4 + 4y=x2+y2+ 4 4y i.e 8y= 0 i.ey= 0 locus is perpendicular bisector of the segment joining (0, 2) and (0, 2)...
Read More →Solve the following
Question: If |z+ 2i| = |z 2i|, then the locus ofzis ____________. Solution: Given |z+ 2i| = |z 2i| forz= |x+iy| |x+iy+ 2i| = |x+iy 2i| Squaring both sides, |x+i(y+ 2)|2= |x+i(y 2)|2 i.ex2+ (y+ 2)|2=x2+ (y 2)|2 i.ex2+y2+ 4 + 4y=x2+y2+ 4 4y i.e 8y= 0 i.ey= 0 locus is perpendicular bisector of the segment joining (0, 2) and (0, 2)...
Read More →If A = {1, 2, 3}, show that a one-one function
Question: IfA= {1, 2, 3}, show that a one-one functionf:AAmust be onto. Solution: A ={1, 2, 3}Number of elements inA= 3 Number of one $-$ one functions $=$ number of ways of arranging 3 elements $=3 !=6$ So, the possible one -one functions can be the following: (i) {(1, 1), (2, 2), (3, 3)}(ii) {(1, 1), (2, 3), (3, 2)}(iii) {(1, 2 ), (2, 2), (3, 3 )}(iv) {(1, 2), (2, 1), (3, 3)}(v) {(1, 3), (2, 2), (3, 1)}(vi) {(1, 3), (2, 1), (3,2 )}Here, in each function, range = {1, 2, 3}, which is same as the...
Read More →(x + 1) is a factor of the polynomial
Question: (x+ 1) is a factor of the polynomial (a) $x^{3}+x^{2}-x+1$ (b) $x^{3}+2 x^{2}-x-2$ (c) $x^{3}+2 x^{2}-x+2$ (d) $x^{4}+x^{3}+x^{2}+1$ Solution: (b) $x^{3}-2 x^{2}-x-2$ Let: $f(x)=x^{3}-2 x^{2}-x-2$ By the factor theorem, (x+ 1) will be a factor off(x) iff(-1) = 0.We have: $f(-1)=(-1)^{3}+2 \times(-1)^{2}-(-1)-2$ $=-1+2+1-2$ $=0$ Hence, $(x+1)$ is a factor of $f(x)=x^{3}+2 x^{2}-x-2$....
Read More →Solve the following
Question: If $\left|\frac{z-2}{z+2}\right|=\frac{\pi}{6}$, then the locus of $z$ is Solution: Given $\left|\frac{z-2}{z+2}\right|=\frac{\pi}{6}$ i. e $|z-2|=\frac{\pi}{6}|z+2|$ i. e $|z-2|^{2}=\frac{\pi^{2}}{6^{2}}|z+2|^{2} \quad$ for $z=x+i y$ i. e $|x+i y-2|^{2}=\frac{\pi^{2}}{36}|x+i y+2|^{2}$ i. e $|(x-2)+i y|^{2}=\frac{\pi^{2}}{36}|(x+2)+i y|^{2}$ i. e $(x-2)^{2}+y^{2}=\frac{\pi^{2}}{36}\left[(x+2)^{2}+y^{2}\right]$ i. e $x^{2}+y^{2}+4-4 x=\frac{\pi^{2}}{36}\left[x^{2}+y^{2}+4 x+4\right]$ i...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{x+3}{x+2}=\frac{3 x-7}{2 x-3}$ Solution: We have been given $\frac{x+3}{x+2}=\frac{3 x-7}{2 x-3}$ $2 x^{2}-3 x+6 x-9=3 x^{2}-7 x+6 x-14$ $x^{2}-4 x-5=0$ $x^{2}-5 x+x-5=0$ $x(x-5)+1(x-5)=0$ $(x+1)(x-5)=0$ Therefore, $x+1=0$ $x=-1$ or, $x-5=0$ $x=5$ Hence, $x=-1$ or $x=5$....
Read More →The modulus and argument of sin
Question: The modulus and argument of $\sin \frac{\pi}{5}+i\left(1-\cos \frac{\pi}{5}\right)$ are _______________and __________________ respectively. Solution: Since complex number is $\sin \frac{\pi}{5}+i\left(1-\cos \frac{\pi}{5}\right)$ Modulus is $\sqrt{\sin ^{2} \frac{\pi}{5}+\left(1-\cos \frac{\pi}{5}\right)^{2}}$ $=\sqrt{\sin ^{2} \frac{\pi}{5}+1+\cos ^{2} \frac{\pi}{5}-2 \cos \frac{\pi}{5}}$ $=\sqrt{-2 \cos \frac{\pi}{5}+1+1} \quad\left(\because \cos ^{2} \frac{\pi}{5}+\sin ^{2} \frac{\p...
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