\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{The complete set of values of }k,\text{ for which the quadratic equation}
\displaystyle x^2-kx+k+2=0\text{ has equal roots, consists of}
\displaystyle \text{(a) }2+\sqrt{12}\qquad \text{(b) }2\pm\sqrt{12}\qquad \text{(c) }2-\sqrt{12}\qquad \text{(d) }-2-\sqrt{2}
\displaystyle \text{Answer: (b) }2\pm\sqrt{12}
\displaystyle \text{For equal roots, the discriminant must be zero.}
\displaystyle D=(-k)^2-4(1)(k+2)=0.
\displaystyle k^2-4k-8=0.
\displaystyle k=\frac{4\pm\sqrt{16+32}}{2}.
\displaystyle k=\frac{4\pm\sqrt{48}}{2} =2\pm\sqrt{12}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{For the equation }|x|^2+|x|-6=0,\text{ the sum of the real roots is}
\displaystyle \text{(a) }1\qquad \text{(b) }0\qquad \text{(c) }2\qquad \text{(d) none of these}
\displaystyle \text{Answer: (b) }0
\displaystyle \text{Let }y=|x|,\text{ where }y\geq0.
\displaystyle y^2+y-6=0.
\displaystyle (y+3)(y-2)=0.
\displaystyle y=-3\text{ or }y=2.
\displaystyle \text{Since }y=|x|\geq0,\text{ we reject }y=-3.
\displaystyle |x|=2.
\displaystyle \therefore x=2,-2.
\displaystyle \text{Sum of the real roots}=2+(-2)=0.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }a,b\text{ are the roots of the equation }x^2+x+1=0,\text{ then }a^2+b^2=
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad\text{(c) }-1\qquad\text{(d) }3
\displaystyle \text{Answer: (c) }-1
\displaystyle a+b=-1,\qquad ab=1.
\displaystyle a^2+b^2=(a+b)^2-2ab.
\displaystyle =(-1)^2-2(1)=1-2=-1.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }\alpha,\beta\text{ are roots of the equation }4x^2+3x+7=0,
\displaystyle \text{then }\frac{1}{\alpha}+\frac{1}{\beta}\text{ is equal to}
\displaystyle \text{(a) }\frac{7}{3}\qquad\text{(b) }-\frac{7}{3}\qquad \text{(c) }\frac{3}{7}\qquad\text{(d) }-\frac{3}{7}
\displaystyle \text{Answer: (d) }-\frac{3}{7}
\displaystyle \alpha+\beta=-\frac{3}{4},\qquad \alpha\beta=\frac{7}{4}.
\displaystyle \frac{1}{\alpha}+\frac{1}{\beta} =\frac{\alpha+\beta}{\alpha\beta}.
\displaystyle =\frac{-\frac{3}{4}}{\frac{7}{4}} =-\frac{3}{7}.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The values of }x\text{ satisfying }\log_3(x^2+4x+12)=2\text{ are}
\displaystyle \text{(a) }2,-4\qquad\text{(b) }1,-3\qquad \text{(c) }-1,3\qquad\text{(d) }-1,-3
\displaystyle \text{Answer: (d) }-1,-3
\displaystyle \log_3(x^2+4x+12)=2.
\displaystyle x^2+4x+12=3^2=9.
\displaystyle x^2+4x+3=0.
\displaystyle (x+1)(x+3)=0.
\displaystyle \therefore x=-1,-3.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The number of real roots of the equation}
\displaystyle (x^2+2x)^2-(x+1)^2-55=0\text{ is}
\displaystyle \text{(a) }2\qquad\text{(b) }1\qquad \text{(c) }4\qquad\text{(d) none of these}
\displaystyle \text{Answer: (a) }2
\displaystyle \text{Let }y=(x+1)^2.
\displaystyle x^2+2x=(x+1)^2-1=y-1.
\displaystyle (y-1)^2-y-55=0.
\displaystyle y^2-3y-54=0.
\displaystyle (y-9)(y+6)=0.
\displaystyle y=9\text{ or }y=-6.
\displaystyle \text{Since }y=(x+1)^2\geq0,\text{ }y=-6\text{ is rejected.}
\displaystyle (x+1)^2=9.
\displaystyle x+1=\pm3.
\displaystyle x=2,-4.
\displaystyle \therefore \text{The equation has }2\text{ real roots.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }\alpha,\beta\text{ are the roots of the equation }ax^2+bx+c=0,\text{ then}
\displaystyle \frac{1}{a\alpha+b}+\frac{1}{a\beta+b}=
\displaystyle \text{(a) }\frac{c}{ab}\qquad\text{(b) }\frac{a}{bc}\qquad \text{(c) }\frac{b}{ac}\qquad\text{(d) none of these}
\displaystyle \text{Answer: (c) }\frac{b}{ac}
\displaystyle \text{Since }\alpha\text{ is a root, }a\alpha^2+b\alpha+c=0.
\displaystyle \alpha(a\alpha+b)=-c.
\displaystyle \therefore \frac{1}{a\alpha+b}=-\frac{\alpha}{c}.
\displaystyle \text{Similarly, }\frac{1}{a\beta+b}=-\frac{\beta}{c}.
\displaystyle \therefore \frac{1}{a\alpha+b}+\frac{1}{a\beta+b} =-\frac{\alpha+\beta}{c}.
\displaystyle \alpha+\beta=-\frac{b}{a}.
\displaystyle \therefore \frac{1}{a\alpha+b}+\frac{1}{a\beta+b} =\frac{b}{ac}.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }\alpha,\beta\text{ are the roots of the equation }x^2+px+1=0;
\displaystyle \gamma,\delta\text{ the roots of the equation }x^2+qx+1=0,\text{ then}
\displaystyle (\alpha-\gamma)(\alpha+\delta)(\beta-\gamma)(\beta+\delta)=
\displaystyle \text{(a) }q^2-p^2\qquad\text{(b) }p^2-q^2\qquad \text{(c) }p^2+q^2\qquad\text{(d) none of these}
\displaystyle \text{Answer: (a) }q^2-p^2
\displaystyle \alpha+\beta=-p,\qquad\alpha\beta=1,
\displaystyle \gamma+\delta=-q,\qquad\gamma\delta=1.
\displaystyle (\alpha-\gamma)(\beta-\gamma) =\alpha\beta-\gamma(\alpha+\beta)+\gamma^2.
\displaystyle =1+p\gamma+\gamma^2.
\displaystyle \text{Since }\gamma^2+q\gamma+1=0,\quad 1+\gamma^2=-q\gamma.
\displaystyle \therefore(\alpha-\gamma)(\beta-\gamma) =(p-q)\gamma.
\displaystyle (\alpha+\delta)(\beta+\delta) =\alpha\beta+\delta(\alpha+\beta)+\delta^2.
\displaystyle =1-p\delta+\delta^2.
\displaystyle \text{Since }\delta^2+q\delta+1=0,\quad 1+\delta^2=-q\delta.
\displaystyle \therefore(\alpha+\delta)(\beta+\delta) =-(p+q)\delta.
\displaystyle \therefore(\alpha-\gamma)(\alpha+\delta) (\beta-\gamma)(\beta+\delta)
\displaystyle =-(p-q)(p+q)\gamma\delta.
\displaystyle =-(p^2-q^2)(1)=q^2-p^2.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The number of real solutions of }|2x-x^2-3|=1\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }2\qquad \text{(c) }3\qquad\text{(d) }4
\displaystyle \text{Answer: (b) }2
\displaystyle |2x-x^2-3|=1.
\displaystyle 2x-x^2-3=1\quad\text{or}\quad 2x-x^2-3=-1.
\displaystyle \text{For }2x-x^2-3=1,
\displaystyle x^2-2x+4=0.
\displaystyle D=(-2)^2-4(1)(4)=-12<0.
\displaystyle \therefore \text{There are no real roots in this case.}
\displaystyle \text{For }2x-x^2-3=-1,
\displaystyle x^2-2x+2=0.
\displaystyle D=(-2)^2-4(1)(2)=-4<0.
\displaystyle \therefore \text{There are no real roots in this case.}
\displaystyle \therefore \text{The equation has }0\text{ real solutions.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The number of solutions of }x^2+|x-1|=1\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad \text{(c) }2\qquad\text{(d) }3
\displaystyle \text{Answer: (c) }2
\displaystyle \text{Case I: }x\geq1.
\displaystyle |x-1|=x-1.
\displaystyle x^2+x-1=1.
\displaystyle x^2+x-2=0.
\displaystyle (x-1)(x+2)=0.
\displaystyle x=1\text{ or }x=-2.
\displaystyle \text{Since }x\geq1,\text{ only }x=1\text{ is valid.}
\displaystyle \text{Case II: }x<1.
\displaystyle |x-1|=1-x.
\displaystyle x^2+1-x=1.
\displaystyle x^2-x=0.
\displaystyle x(x-1)=0.
\displaystyle x=0\text{ or }x=1.
\displaystyle \text{Since }x<1,\text{ only }x=0\text{ is valid.}
\displaystyle \therefore \text{The solutions are }x=0,1.
\displaystyle \therefore \text{The number of solutions is }2.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }x\text{ is real and }k= \frac{x^2-x+1}{x^2+x+1},\text{ then}
\displaystyle \text{(a) }k\in\left[\frac{1}{3},3\right]\qquad \text{(b) }k\geq3\qquad\text{(c) }k\leq\frac{1}{3}\qquad \text{(d) none of these}
\displaystyle \text{Answer: (a) }k\in\left[\frac{1}{3},3\right]
\displaystyle k(x^2+x+1)=x^2-x+1.
\displaystyle (k-1)x^2+(k+1)x+(k-1)=0.
\displaystyle \text{Since }x\text{ is real, the discriminant must be non-negative.}
\displaystyle (k+1)^2-4(k-1)^2\geq0.
\displaystyle k^2+2k+1-4(k^2-2k+1)\geq0.
\displaystyle -3k^2+10k-3\geq0.
\displaystyle 3k^2-10k+3\leq0.
\displaystyle (3k-1)(k-3)\leq0.
\displaystyle \therefore \frac{1}{3}\leq k\leq3.
\displaystyle \therefore k\in\left[\frac{1}{3},3\right].
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If the roots of }x^2-bx+c=0 \text{ are two consecutive integers, then }b^2-4c\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad \text{(c) }2\qquad\text{(d) none of these}
\displaystyle \text{Answer: (b) }1
\displaystyle \text{Let the two consecutive integer roots be }n\text{ and }n+1.
\displaystyle b=n+(n+1)=2n+1.
\displaystyle c=n(n+1).
\displaystyle b^2-4c=(2n+1)^2-4n(n+1).
\displaystyle =4n^2+4n+1-4n^2-4n=1.
\displaystyle \therefore b^2-4c=1.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The value of }a\text{ such that }x^2-11x+a=0\text{ and}
\displaystyle x^2-14x+2a=0\text{ may have a common root is}
\displaystyle \text{(a) }0\qquad\text{(b) }12\qquad \text{(c) }24\qquad\text{(d) }32
\displaystyle \text{Answer: (a) }0\text{ and (c) }24
\displaystyle \text{Let the common root be }\alpha.
\displaystyle \alpha^2-11\alpha+a=0\qquad ...(1)
\displaystyle \alpha^2-14\alpha+2a=0\qquad ...(2)
\displaystyle \text{Subtracting (1) from (2),}
\displaystyle -3\alpha+a=0.
\displaystyle \therefore a=3\alpha.
\displaystyle \text{Substituting }a=3\alpha\text{ in (1),}
\displaystyle \alpha^2-11\alpha+3\alpha=0.
\displaystyle \alpha^2-8\alpha=0.
\displaystyle \alpha(\alpha-8)=0.
\displaystyle \therefore \alpha=0\text{ or }8.
\displaystyle \text{Hence }a=3\alpha=0\text{ or }24.
\displaystyle \therefore a=0\text{ or }24.
\displaystyle \text{Thus, both options (a) and (c) satisfy the given question.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The values of }k\text{ for which the quadratic equation}
\displaystyle kx^2+1=kx+3x-11x^2\text{ has real and equal roots are}
\displaystyle \text{(a) }-11,-3\qquad\text{(b) }5,7\qquad \text{(c) }5,-7\qquad\text{(d) none of these}
\displaystyle \text{Answer: (c) }5,-7
\displaystyle kx^2+1=kx+3x-11x^2.
\displaystyle (k+11)x^2-(k+3)x+1=0.
\displaystyle \text{For real and equal roots, }D=0.
\displaystyle [-(k+3)]^2-4(k+11)(1)=0.
\displaystyle (k+3)^2-4(k+11)=0.
\displaystyle k^2+6k+9-4k-44=0.
\displaystyle k^2+2k-35=0.
\displaystyle (k+7)(k-5)=0.
\displaystyle \therefore k=5,-7.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If the equations }x^2+2x+3\lambda=0\text{ and}
\displaystyle 2x^2+3x+5\lambda=0\text{ have a non-zero common roots, then }\lambda=
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad \text{(c) }3\qquad\text{(d) none of these}
\displaystyle \text{Answer: (b) }-1
\displaystyle \text{Let the non-zero common root be }\alpha.
\displaystyle \alpha^2+2\alpha+3\lambda=0.\qquad ...(1)
\displaystyle 2\alpha^2+3\alpha+5\lambda=0.\qquad ...(2)
\displaystyle 2\times(1)-(2)\text{ gives}
\displaystyle \alpha+\lambda=0.
\displaystyle \therefore \lambda=-\alpha.
\displaystyle \text{Substituting }\lambda=-\alpha\text{ in (1),}
\displaystyle \alpha^2+2\alpha-3\alpha=0.
\displaystyle \alpha^2-\alpha=0.
\displaystyle \alpha(\alpha-1)=0.
\displaystyle \text{Since the common root is non-zero, }\alpha=1.
\displaystyle \therefore \lambda=-1.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If one root of the equation }x^2+px+12=0\text{ is }4,
\displaystyle \text{while the equation }x^2+px+q=0\text{ has equal roots, the value of }q\text{ is}
\displaystyle \text{(a) }\frac{49}{4}\qquad\text{(b) }\frac{4}{49}\qquad \text{(c) }4\qquad\text{(d) none of these}
\displaystyle \text{Answer: (a) }\frac{49}{4}
\displaystyle \text{Since }4\text{ is a root of }x^2+px+12=0,
\displaystyle 4^2+4p+12=0.
\displaystyle 16+4p+12=0.
\displaystyle 4p=-28.
\displaystyle p=-7.
\displaystyle \text{For }x^2+px+q=0\text{ to have equal roots, }D=0.
\displaystyle p^2-4q=0.
\displaystyle (-7)^2-4q=0.
\displaystyle 49-4q=0.
\displaystyle \therefore q=\frac{49}{4}.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The value of }p\text{ and }q\ (p\ne0,\ q\ne0)\text{ for which }p,q
\displaystyle \text{are the roots of the equation }x^2+px+qab=0\text{ are}
\displaystyle \text{(a) }p=1,\ q=-2\qquad \text{(b) }p=-1,\ q=-2
\displaystyle \text{(c) }p=-1,\ q=2\qquad \text{(d) }p=1,\ q=2
\displaystyle \text{Answer: (c) }p=-1,\ q=2
\displaystyle \text{Since }p,q\text{ are the roots,}
\displaystyle p+q=-p,\qquad pq=qab.
\displaystyle p+q=-p\Rightarrow q=-2p.
\displaystyle pq=qab.
\displaystyle \text{Since }q\ne0,\quad p=ab.
\displaystyle \text{For }ab=-1,\quad p=-1.
\displaystyle q=-2p=2.
\displaystyle \therefore p=-1,\qquad q=2.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The set of all values of }m\text{ for which both the roots of the equation}
\displaystyle x^2-(m+1)x+m+4=0\text{ are real and negative, is}
\displaystyle \text{(a) }(-\infty,-3]\cup[5,\infty)\qquad \text{(b) }[-3,5]
\displaystyle \text{(c) }(-4,-3]\qquad \text{(d) }(-3,-1]
\displaystyle \text{Answer: (c) }(-4,-3]
\displaystyle \text{Let the roots be }\alpha,\beta.
\displaystyle \alpha+\beta=m+1,\qquad\alpha\beta=m+4.
\displaystyle \text{For both roots to be real, }D\geq0.
\displaystyle (m+1)^2-4(m+4)\geq0.
\displaystyle m^2-2m-15\geq0.
\displaystyle (m-5)(m+3)\geq0.
\displaystyle \therefore m\leq-3\text{ or }m\geq5.
\displaystyle \text{For both roots to be negative, }\alpha+\beta<0.
\displaystyle m+1<0\Rightarrow m<-1.
\displaystyle \text{Also, }\alpha\beta>0.
\displaystyle m+4>0\Rightarrow m>-4.
\displaystyle \text{Combining all the conditions,}
\displaystyle -4<m\leq-3.
\displaystyle \therefore m\in(-4,-3].
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The number of roots of the equation}
\displaystyle \frac{(x+2)(x-5)}{(x-3)(x+6)} =\frac{x-2}{x+4}\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad \text{(c) }2\qquad\text{(d) }3
\displaystyle \text{Answer: (b) }1
\displaystyle \text{Here }x\ne3,-6,-4.
\displaystyle (x+2)(x-5)(x+4)=(x-2)(x-3)(x+6).
\displaystyle (x^2-3x-10)(x+4)=(x^2-5x+6)(x+6).
\displaystyle x^3+x^2-22x-40=x^3+x^2-24x+36.
\displaystyle 2x=76.
\displaystyle x=38.
\displaystyle \text{Since }x=38\text{ satisfies the domain restrictions, it is a valid root.}
\displaystyle \therefore \text{The equation has }1\text{ root.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }\alpha\text{ and }\beta\text{ are the roots of }4x^2+3x+7=0,
\displaystyle \text{then the value of }\frac{1}{\alpha}+\frac{1}{\beta}\text{ is}
\displaystyle \text{(a) }\frac{4}{7}\qquad \text{(b) }-\frac{3}{7}\qquad \text{(c) }\frac{3}{7}\qquad \text{(d) }-\frac{3}{4}
\displaystyle \text{Answer: (b) }-\frac{3}{7}
\displaystyle \alpha+\beta=-\frac{3}{4},\qquad \alpha\beta=\frac{7}{4}.
\displaystyle \frac{1}{\alpha}+\frac{1}{\beta} =\frac{\alpha+\beta}{\alpha\beta}.
\displaystyle =\frac{-\frac{3}{4}}{\frac{7}{4}} =-\frac{3}{7}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }\alpha,\beta\text{ are the roots of the equation }x^2+px+q=0,\text{ then}
\displaystyle -\frac{1}{\alpha},-\frac{1}{\beta}\text{ are the roots of the equation}
\displaystyle \text{(a) }x^2-px+q=0\qquad \text{(b) }x^2+px+q=0
\displaystyle \text{(c) }qx^2+px+1=0\qquad \text{(d) }qx^2-px+1=0
\displaystyle \text{Answer: (d) }qx^2-px+1=0
\displaystyle \alpha+\beta=-p,\qquad\alpha\beta=q.
\displaystyle \text{Sum of the new roots} =-\frac{1}{\alpha}-\frac{1}{\beta}.
\displaystyle =-\frac{\alpha+\beta}{\alpha\beta} =-\frac{-p}{q}=\frac{p}{q}.
\displaystyle \text{Product of the new roots} =\left(-\frac{1}{\alpha}\right)\left(-\frac{1}{\beta}\right) =\frac{1}{q}.
\displaystyle \therefore x^2-\frac{p}{q}x+\frac{1}{q}=0.
\displaystyle \therefore qx^2-px+1=0.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If the difference of the roots of }x^2-px+q=0\text{ is unity, then}
\displaystyle \text{(a) }p^2+4q=1\qquad \text{(b) }p^2-4q=1
\displaystyle \text{(c) }p^2+4q^2=(1+2q)^2\qquad \text{(d) }4p^2+q^2=(1+2p)^2
\displaystyle \text{Answer: (b) }p^2-4q=1
\displaystyle \text{Let the roots be }\alpha,\beta.
\displaystyle \alpha+\beta=p,\qquad\alpha\beta=q.
\displaystyle \text{Given, }\alpha-\beta=1.
\displaystyle (\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta.
\displaystyle 1^2=p^2-4q.
\displaystyle \therefore p^2-4q=1.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }\alpha,\beta\text{ are the roots of the equation}
\displaystyle x^2-p(x+1)-c=0,\text{ then }(\alpha+1)(\beta+1)=
\displaystyle \text{(a) }c\qquad\text{(b) }c-1\qquad \text{(c) }1-c\qquad\text{(d) none of these}
\displaystyle \text{Answer: (c) }1-c
\displaystyle x^2-px-p-c=0.
\displaystyle \alpha+\beta=p,\qquad\alpha\beta=-(p+c).
\displaystyle (\alpha+1)(\beta+1) =\alpha\beta+\alpha+\beta+1.
\displaystyle =-(p+c)+p+1.
\displaystyle =1-c.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The least value of }k\text{ which makes the roots of the equation}
\displaystyle x^2+5x+k=0\text{ imaginary is}
\displaystyle \text{(a) }4\qquad\text{(b) }5\qquad \text{(c) }6\qquad\text{(d) }7
\displaystyle \text{Answer: (d) }7
\displaystyle \text{For imaginary roots, }D<0.
\displaystyle 5^2-4(1)(k)<0.
\displaystyle 25-4k<0.
\displaystyle 4k>25.
\displaystyle k>\frac{25}{4}=6.25.
\displaystyle \text{Hence, from the given options, the least value of }k\text{ is }7.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{The equation of the smallest degree with real coefficients having}
\displaystyle 1+i\text{ as one of the roots is}
\displaystyle \text{(a) }x^2+x+1=0\qquad \text{(b) }x^2-2x+2=0
\displaystyle \text{(c) }x^2+2x+2=0\qquad \text{(d) }x^2+2x-2=0
\displaystyle \text{Answer: (b) }x^2-2x+2=0
\displaystyle \text{Since the coefficients are real and }1+i\text{ is a root,}
\displaystyle 1-i\text{ is also a root.}
\displaystyle \text{Required equation}=[x-(1+i)][x-(1-i)]=0.
\displaystyle =[(x-1)-i][(x-1)+i]=0.
\displaystyle (x-1)^2+1=0.
\displaystyle x^2-2x+2=0.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{Write the number of real roots of the equation}
\displaystyle (x-1)^2+(x-2)^2+(x-3)^2=0.
\displaystyle \text{Answer:}
\displaystyle (x-1)^2\geq 0,\quad (x-2)^2\geq 0,\quad (x-3)^2\geq 0.
\displaystyle \text{For their sum to be }0,\text{ each term must be }0.
\displaystyle \therefore x=1,\quad x=2,\quad x=3\text{ simultaneously, which is impossible.}
\displaystyle \therefore \text{The equation has no real roots.}
\displaystyle \therefore \text{Number of real roots}=0.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }a\text{ and }b\text{ are roots of the equation}
\displaystyle x^2-px+q=0,\text{ then write the value of }\frac{1}{a}+\frac{1}{b}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }a\text{ and }b\text{ are the roots,}
\displaystyle a+b=p,\qquad ab=q.
\displaystyle \therefore \frac{1}{a}+\frac{1}{b} =\frac{a+b}{ab}=\frac{p}{q}.
\displaystyle \therefore \frac{1}{a}+\frac{1}{b}=\frac{p}{q}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If roots }\alpha,\beta\text{ of the equation}
\displaystyle x^2-px+16=0\text{ satisfy the relation }\alpha^2+\beta^2=9,
\displaystyle \text{then write the value of }p.
\displaystyle \text{Answer:}
\displaystyle \alpha+\beta=p,\qquad \alpha\beta=16.
\displaystyle \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta.
\displaystyle 9=p^2-2(16).
\displaystyle p^2=41.
\displaystyle \therefore p=\pm\sqrt{41}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }2+\sqrt{3}\text{ is a root of the equation}
\displaystyle x^2+px+q=0,\text{ then write the values of }p\text{ and }q.
\displaystyle \text{Answer:}
\displaystyle \text{Since the coefficients are rational, the other root is }2-\sqrt{3}.
\displaystyle \text{Sum of roots}=(2+\sqrt{3})+(2-\sqrt{3})=4.
\displaystyle \therefore -p=4\quad\Rightarrow\quad p=-4.
\displaystyle \text{Product of roots}=(2+\sqrt{3})(2-\sqrt{3})=4-3=1.
\displaystyle \therefore q=1.
\displaystyle \therefore p=-4,\qquad q=1.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If the difference between the roots of the equation}
\displaystyle x^2+ax+8=0\text{ is }2,\text{ write the values of }a.
\displaystyle \text{Answer:}
\displaystyle \text{Let the roots be }\alpha\text{ and }\beta.
\displaystyle \alpha+\beta=-a,\qquad \alpha\beta=8.
\displaystyle (\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta.
\displaystyle 2^2=(-a)^2-4(8).
\displaystyle 4=a^2-32\quad\Rightarrow\quad a^2=36.
\displaystyle \therefore a=\pm6.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Write the roots of the equation}
\displaystyle (a-b)x^2+(b-c)x+(c-a)=0.
\displaystyle \text{Answer:}
\displaystyle \text{Putting }x=1,
\displaystyle (a-b)+(b-c)+(c-a)=0.
\displaystyle \therefore x=1\text{ is one root.}
\displaystyle \text{Product of roots}=\frac{c-a}{a-b}.
\displaystyle \therefore 1\times x_2=\frac{c-a}{a-b}.
\displaystyle \therefore \text{The roots are }1\text{ and }\frac{c-a}{a-b}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }a\text{ and }b\text{ are roots of the equation}
\displaystyle x^2-x+1=0,\text{ then write the value of }a^2+b^2.
\displaystyle \text{Answer:}
\displaystyle a+b=1,\qquad ab=1.
\displaystyle a^2+b^2=(a+b)^2-2ab.
\displaystyle =1^2-2(1)=-1.
\displaystyle \therefore a^2+b^2=-1.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Write the number of quadratic equations, with real roots,}
\displaystyle \text{which do not change by squaring their roots.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the roots be }\alpha,\beta.
\displaystyle \text{The roots remain unchanged on squaring, so the possible real roots are }0,1.
\displaystyle \text{Hence, the possible pairs of roots are }(0,0),(0,1),(1,1).
\displaystyle \text{The corresponding equations are}
\displaystyle x^2=0,\qquad x^2-x=0,\qquad x^2-2x+1=0.
\displaystyle \therefore \text{Number of quadratic equations}=3.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }\alpha,\beta\text{ are roots of the equation}
\displaystyle x^2+lx+m=0,\text{ write an equation whose roots are }-\frac1\alpha\text{ and }-\frac1\beta.
\displaystyle \text{Answer:}
\displaystyle \alpha+\beta=-l,\qquad \alpha\beta=m.
\displaystyle \text{Sum of new roots}=-\frac1\alpha-\frac1\beta =-\frac{\alpha+\beta}{\alpha\beta}=\frac{l}{m}.
\displaystyle \text{Product of new roots} =\left(-\frac1\alpha\right)\left(-\frac1\beta\right)=\frac1m.
\displaystyle \therefore x^2-\frac{l}{m}x+\frac1m=0.
\displaystyle \therefore mx^2-lx+1=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }\alpha,\beta\text{ are roots of the equation}
\displaystyle x^2-a(x+1)-c=0,\text{ then write the value of }(1+\alpha)(1+\beta).
\displaystyle \text{Answer:}
\displaystyle x^2-ax-(a+c)=0.
\displaystyle \therefore \alpha+\beta=a,\qquad \alpha\beta=-(a+c).
\displaystyle (1+\alpha)(1+\beta)=1+\alpha+\beta+\alpha\beta.
\displaystyle =1+a-(a+c)=1-c.
\displaystyle \therefore (1+\alpha)(1+\beta)=1-c.
\displaystyle \\

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