\displaystyle \textbf{Question 1. }\text{The complete set of values of }k,\text{ for which the quadratic equation } \\ 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-\sqrt2
\displaystyle \text{Answer:}
\displaystyle \text{For equal roots, discriminant }=0
\displaystyle (-k)^2-4(1)(k+2)=0
\displaystyle k^2-4k-8=0
\displaystyle k=\frac{4\pm\sqrt{16+32}}{2}
\displaystyle =\frac{4\pm4\sqrt3}{2}=2\pm2\sqrt3
\displaystyle =2\pm\sqrt{12}
\displaystyle \therefore \text{Correct option is (b).}
\\

\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:}
\displaystyle \text{Let }|x|=y,\text{ where }y\geq0
\displaystyle y^2+y-6=0
\displaystyle (y+3)(y-2)=0
\displaystyle \therefore y=2
\displaystyle \therefore |x|=2
\displaystyle \therefore x=\pm2
\displaystyle \text{Sum of real roots }=2+(-2)=0
\displaystyle \therefore \text{Correct option is (b).}
\\

\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:}
\displaystyle a+b=-1,\qquad ab=1
\displaystyle a^2+b^2=(a+b)^2-2ab
\displaystyle =(-1)^2-2(1)=-1
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 4. }\text{If }\alpha,\beta\text{ are roots of the equation }4x^2+3x+7=0,\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:}
\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{Correct option is (d).}
\\

\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:}
\displaystyle \log_3(x^2+4x+12)=2
\displaystyle \therefore x^2+4x+12=3^2=9
\displaystyle \therefore x^2+4x+3=0
\displaystyle \therefore (x+1)(x+3)=0
\displaystyle \therefore x=-1,\ -3
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 6. }\text{The number of real roots of the equation } \\ (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:}
\displaystyle x^2+2x=(x+1)^2-1
\displaystyle \text{Let }y=x+1
\displaystyle \therefore x^2+2x=y^2-1
\displaystyle (y^2-1)^2-y^2-55=0
\displaystyle y^4-3y^2-54=0
\displaystyle (y^2-9)(y^2+6)=0
\displaystyle \therefore y^2=9
\displaystyle \therefore y=\pm3
\displaystyle \therefore x+1=3\text{ or }x+1=-3
\displaystyle \therefore x=2\text{ or }x=-4
\displaystyle \therefore \text{Number of real roots }=2
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 7. }\text{If }\alpha,\beta\text{ are the roots of the equation }ax^2+bx+c=0,\text{ then } \\ \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:}
\displaystyle \alpha+\beta=-\frac{b}{a},\qquad \alpha\beta=\frac{c}{a}
\displaystyle \frac{1}{a\alpha+b}+\frac{1}{a\beta+b}
\displaystyle =\frac{a(\alpha+\beta)+2b}{(a\alpha+b)(a\beta+b)}
\displaystyle =\frac{-b+2b}{a^2\alpha\beta+ab(\alpha+\beta)+b^2}
\displaystyle =\frac{b}{ac-b^2+b^2}=\frac{b}{ac}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 8. }\text{If }\alpha,\beta\text{ are the roots of } \\ x^2+px+1=0;\ \gamma,\delta\text{ the roots of }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:}
\displaystyle \alpha+\beta=-p,\quad \alpha\beta=1,\quad \gamma+\delta=-q,\quad \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,\text{ we get }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,\text{ we get }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)\gamma\cdot[-(p+q)\delta]
\displaystyle =-(p^2-q^2)\gamma\delta=q^2-p^2
\displaystyle \therefore \text{Correct option is (a).}
\\

\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:}
\displaystyle |2x-x^2-3|=1
\displaystyle \therefore 2x-x^2-3=1\qquad \text{or}\qquad 2x-x^2-3=-1
\displaystyle x^2-2x+4=0\qquad \text{or}\qquad x^2-2x+2=0
\displaystyle \text{For }x^2-2x+4=0,
\displaystyle D=(-2)^2-4(1)(4)=4-16=-12<0
\displaystyle \therefore \text{No real root}
\displaystyle \text{For }x^2-2x+2=0,
\displaystyle D=(-2)^2-4(1)(2)=4-8=-4<0
\displaystyle \therefore \text{No real root}
\displaystyle \therefore \text{Number of real solutions}=0
\displaystyle \therefore\text{Correct option is (a).}
\\

\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:}
\displaystyle x^2+|x-1|=1
\displaystyle \text{Case 1: }x\geq1
\displaystyle |x-1|=x-1
\displaystyle x^2+x-1=1
\displaystyle x^2+x-2=0
\displaystyle (x+2)(x-1)=0
\displaystyle x=-2,\ 1
\displaystyle \text{Since }x\geq1,\ \text{valid solution is }x=1
\displaystyle \text{Case 2: }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,\ 1
\displaystyle \text{Since }x<1,\ \text{valid solution is }x=0
\displaystyle \therefore \text{Solutions are }x=0,\ 1
\displaystyle \therefore \text{Number of solutions}=2
\displaystyle \therefore\text{Correct option is (c).}
\\

\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:}
\displaystyle k=\frac{x^2-x+1}{x^2+x+1}
\displaystyle \therefore k(x^2+x+1)=x^2-x+1
\displaystyle \therefore (k-1)x^2+(k+1)x+(k-1)=0
\displaystyle \text{Since }x\text{ is real, discriminant }\geq0
\displaystyle (k+1)^2-4(k-1)^2\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 \text{Correct option is (a).}
\\

\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:}
\displaystyle \text{Let the roots be }n\text{ and }n+1.
\displaystyle \text{Difference between roots }=1
\displaystyle \therefore \sqrt{b^2-4c}=1
\displaystyle \therefore b^2-4c=1
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 13. }\text{The value of }a\text{ such that }x^2-11x+a=0\text{ and } \\ 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:}
\displaystyle \text{Let the common root be }\alpha.
\displaystyle \alpha^2-11\alpha+a=0\qquad \text{and}\qquad \alpha^2-14\alpha+2a=0
\displaystyle \text{Subtracting, }3\alpha-a=0
\displaystyle \therefore a=3\alpha
\displaystyle \alpha^2-11\alpha+3\alpha=0
\displaystyle \therefore \alpha^2-8\alpha=0
\displaystyle \therefore \alpha=0\text{ or }8
\displaystyle \therefore a=0\text{ or }24
\displaystyle \therefore \text{Both options (a) and (c) are possible.}
\\

\displaystyle \textbf{Question 14. }\text{The values of }k\text{ for which }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:}
\displaystyle kx^2+1=kx+3x-11x^2
\displaystyle \therefore (k+11)x^2-(k+3)x+1=0
\displaystyle \text{For equal roots, discriminant }=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{Correct option is (c).}
\\

\displaystyle \textbf{Question 15. }\text{If }x^2+2x+3\lambda=0\text{ and }2x^2+3x+5\lambda=0\text{ have a non-zero common root, then }\lambda=
\displaystyle \text{(a) }1\qquad \text{(b) }-1\qquad \text{(c) }3\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the common root be }\alpha.
\displaystyle \alpha^2+2\alpha+3\lambda=0
\displaystyle 2\alpha^2+3\alpha+5\lambda=0
\displaystyle \text{Twice the first equation gives }2\alpha^2+4\alpha+6\lambda=0
\displaystyle \text{Subtracting the second equation, }\alpha+\lambda=0
\displaystyle \therefore \alpha=-\lambda
\displaystyle \lambda^2-2\lambda+3\lambda=0
\displaystyle \therefore \lambda^2+\lambda=0
\displaystyle \therefore \lambda(\lambda+1)=0
\displaystyle \text{Since the common root is non-zero, }\lambda\neq0
\displaystyle \therefore \lambda=-1
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 16. }\text{If one root of }x^2+px+12=0\text{ is }4,\text{ while } \\ 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:}
\displaystyle 4^2+4p+12=0
\displaystyle \therefore 16+4p+12=0
\displaystyle \therefore 4p=-28
\displaystyle \therefore p=-7
\displaystyle x^2+px+q=0\text{ has equal roots.}
\displaystyle \therefore p^2-4q=0
\displaystyle \therefore 49-4q=0
\displaystyle \therefore q=\frac{49}{4}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 17. }\text{The value of }p\text{ and }q\ (p\neq0,\ q\neq0) \\ \text{ for which }p,q\text{ are the roots of the equation }x^2+px+q=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:}
\displaystyle \text{Since }p,q\text{ are roots of }x^2+px+q=0,
\displaystyle p+q=-p
\displaystyle \therefore q=-2p
\displaystyle pq=q
\displaystyle q(p-1)=0
\displaystyle \text{Given }q\neq0,\ \therefore p-1=0
\displaystyle \therefore p=1
\displaystyle q=-2p=-2
\displaystyle \therefore p=1,\ q=-2
\displaystyle \therefore\text{Correct option is (a).}
\\

\displaystyle \textbf{Question 18. }\text{The set of all values of }m\text{ for which both the roots of the equation } \\ 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:}
\displaystyle x^2-(m+1)x+m+4=0
\displaystyle \text{For both roots to be real and negative:}
\displaystyle \text{(i) Discriminant }\geq0
\displaystyle \text{(ii) Sum of roots }<0
\displaystyle \text{(iii) Product of roots }>0
\displaystyle D=(m+1)^2-4(m+4)
\displaystyle =m^2+2m+1-4m-16
\displaystyle =m^2-2m-15
\displaystyle =(m-5)(m+3)\geq0
\displaystyle \therefore m\leq-3\qquad \text{or}\qquad m\geq5
\displaystyle \text{Sum of roots}=m+1<0
\displaystyle \therefore m<-1
\displaystyle \text{Product of roots}=m+4>0
\displaystyle \therefore m>-4
\displaystyle \text{Combining all conditions,}
\displaystyle -4<m\leq-3
\displaystyle \therefore \text{Required set is }(-4,-3]
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 19. }\text{The number of roots of }\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:}
\displaystyle \frac{(x+2)(x-5)}{(x-3)(x+6)}=\frac{x-2}{x+4}
\displaystyle (x^2-3x-10)(x+4)=(x-2)(x^2+3x-18)
\displaystyle x^3+x^2-22x-40=x^3+x^2-24x+36
\displaystyle 2x=76
\displaystyle x=38
\displaystyle \therefore \text{Number of roots }=1
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 20. }\text{If }\alpha,\beta\text{ are the roots of }4x^2+3x+7=0,\text{ then }\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:}
\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{Correct option is (b).}
\\

\displaystyle \textbf{Question 21. }\text{If }\alpha,\beta\text{ are the roots of }x^2+px+q=0,\text{ then }-\frac{1}{\alpha},-\frac{1}{\beta}\text{ are roots of}
\displaystyle \text{(a) }x^2-px+q=0\qquad \text{(b) }x^2+px+q=0\qquad \\ \text{(c) }qx^2+px+1=0\qquad \text{(d) }qx^2-px+1=0
\displaystyle \text{Answer:}
\displaystyle \alpha+\beta=-p,\qquad \alpha\beta=q
\displaystyle \text{New sum }=-\frac{1}{\alpha}-\frac{1}{\beta}=-\frac{\alpha+\beta}{\alpha\beta}=\frac{p}{q}
\displaystyle \text{New product }=\frac{1}{\alpha\beta}=\frac{1}{q}
\displaystyle \therefore \text{Required equation is }x^2-\frac{p}{q}x+\frac{1}{q}=0
\displaystyle \therefore qx^2-px+1=0
\displaystyle \therefore \text{Correct option is (d).}
\\

\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\qquad \text{(c) }p^2+4q^2=(1+2q)^2\qquad \text{(d) }4p^2+q^2=(1+2p)^2
\displaystyle \text{Answer:}
\displaystyle \text{Let the roots be }\alpha,\beta.
\displaystyle \alpha+\beta=p,\qquad \alpha\beta=q
\displaystyle \alpha-\beta=1
\displaystyle (\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta
\displaystyle 1=p^2-4q
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 23. }\text{If }\alpha,\beta\text{ are the roots of }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:}
\displaystyle x^2-p(x+1)-c=0
\displaystyle \Rightarrow 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=1-c
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 24. }\text{The least value of }k\text{ which makes the roots of } \\ 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:}
\displaystyle \text{For imaginary roots, discriminant }<0
\displaystyle 5^2-4k<0
\displaystyle 25<4k
\displaystyle k>\frac{25}{4}
\displaystyle \therefore \text{Least integral value of }k=7
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 25. }\text{The equation of the smallest degree with real coefficients having } \\ 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\qquad \text{(c) }x^2+2x+2=0\qquad \text{(d) }x^2+2x-2=0
\displaystyle \text{Answer:}
\displaystyle \text{Since coefficients are real, the other root is }1-i
\displaystyle \text{Sum of roots }=(1+i)+(1-i)=2
\displaystyle \text{Product of roots }=(1+i)(1-i)=2
\displaystyle \therefore \text{Required equation is }x^2-2x+2=0
\displaystyle \therefore \text{Correct option is (b).}
\\

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