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

\displaystyle \textbf{Question 1: }\text{If }x^n-1\text{ is divisible by }x-\lambda,\text{ then the least}
\displaystyle \text{positive integral value of }\lambda\text{ is}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad\text{(c) }3\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle \text{Since }x^n-1\text{ is divisible by }x-\lambda,
\displaystyle \text{by the Factor Theorem, }\lambda^n-1=0.
\displaystyle \therefore \lambda^n=1.
\displaystyle \therefore \text{the least positive integral value of }\lambda\text{ is }1.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{For all }n\in N,\ 3\times5^{2n+1}+2^{3n+1}\text{ is divisible by}
\displaystyle \text{(a) }19\qquad\text{(b) }17\qquad\text{(c) }23\qquad\text{(d) }25
\displaystyle \text{Answer:}
\displaystyle 3\times5^{2n+1}+2^{3n+1}=15(25)^n+2(8)^n.
\displaystyle \text{Since }25\equiv8\pmod{17},
\displaystyle 15(25)^n+2(8)^n\equiv15(8)^n+2(8)^n\pmod{17}.
\displaystyle \therefore 3\times5^{2n+1}+2^{3n+1}\equiv17(8)^n\equiv0\pmod{17}.
\displaystyle \therefore 3\times5^{2n+1}+2^{3n+1}\text{ is divisible by }17.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }10^n+3\times4^{n+2}+\lambda\text{ is divisible by }9
\displaystyle \text{for all }n\in N,\text{ then the least positive integral value of }\lambda\text{ is}
\displaystyle \text{(a) }5\qquad\text{(b) }3\qquad\text{(c) }7\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle 10^n\equiv1^n\equiv1\pmod9.
\displaystyle \text{Also, }4^3\equiv1\pmod9\text{ and }3\times4^k\equiv3\pmod9.
\displaystyle \therefore 10^n+3\times4^{n+2}+\lambda\equiv1+3+\lambda\pmod9.
\displaystyle \therefore 4+\lambda\equiv0\pmod9.
\displaystyle \therefore \lambda\equiv5\pmod9.
\displaystyle \therefore \text{the least positive integral value of }\lambda\text{ is }5.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }P(n):2^n<(1\times2\times3\times\cdots\times n).
\displaystyle \text{Then the smallest positive integer for which }P(n)\text{ is true is}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad\text{(c) }3\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle P(n):2^n<n!.
\displaystyle n=1:\quad 2^1=2>1!=1.
\displaystyle n=2:\quad 2^2=4>2!=2.
\displaystyle n=3:\quad 2^3=8>3!=6.
\displaystyle n=4:\quad 2^4=16<4!=24.
\displaystyle \therefore \text{the smallest positive integer for which }P(n)\text{ is true is }4.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{A student was asked to prove a statement }P(n)\text{ by induction.}
\displaystyle \text{He proved }P(k+1)\text{ is true whenever }P(k)\text{ is true for all }k>5\in N
\displaystyle \text{and also }P(5)\text{ is true. On the basis of this he could conclude that }P(n)\text{ is true.}
\displaystyle \text{(a) for all }n\in N\qquad\text{(b) for all }n>5
\displaystyle \text{(c) for all }n\ge5\qquad\text{(d) for all }n<5
\displaystyle \text{Answer:}
\displaystyle P(5)\text{ is true, but the induction step is given only for }k>5.
\displaystyle \text{Thus, }P(5)\Rightarrow P(6)\text{ has not been established.}
\displaystyle \therefore \text{strictly, none of the given options follows from the statement.}
\displaystyle \text{If the intended condition is }k\ge5,\text{ then }P(n)\text{ is true for all }n\ge5.
\displaystyle \therefore \text{under the intended condition, Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }P(n):49^n+16n+\lambda\text{ is divisible by }64
\displaystyle \text{for }n\in N\text{ is true, then the least negative integral value of }\lambda\text{ is}
\displaystyle \text{(a) }-3\qquad\text{(b) }-2\qquad\text{(c) }-1\qquad\text{(d) }-4
\displaystyle \text{Answer:}
\displaystyle 49^n=(1+48)^n.
\displaystyle \therefore 49^n\equiv1+48n\pmod{64},
\displaystyle \text{since all the remaining terms contain }48^2\text{ as a factor.}
\displaystyle \therefore 49^n+16n+\lambda\equiv1+48n+16n+\lambda\pmod{64}.
\displaystyle \therefore 49^n+16n+\lambda\equiv1+\lambda\pmod{64}.
\displaystyle \text{For divisibility by }64,\quad1+\lambda\equiv0\pmod{64}.
\displaystyle \therefore \lambda\equiv-1\pmod{64}.
\displaystyle \therefore \text{the least negative integral value of }\lambda\text{ is }-1.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{State the first principle of mathematical induction.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be a statement involving the natural number }n\text{ such that:}
\displaystyle \text{(i) }P(1)\text{ is true, i.e. }P(n)\text{ is true for }n=1,\text{ and}
\displaystyle \text{(ii) }P(m+1)\text{ is true whenever }P(m)\text{ is true.}
\displaystyle \text{Then, }P(n)\text{ is true for all natural numbers }n.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the set of value of }n\text{ for which the statement}
\displaystyle P(n):2n<n!\text{ is true.}
\displaystyle \text{Answer:}
\displaystyle n=1:\quad 2n=2,\quad n!=1,\quad\therefore 2n>n!.
\displaystyle n=2:\quad 2n=4,\quad n!=2,\quad\therefore 2n>n!.
\displaystyle n=3:\quad 2n=6,\quad n!=6,\quad\therefore 2n=n!.
\displaystyle n=4:\quad 2n=8,\quad n!=24,\quad\therefore 2n<n!.
\displaystyle \text{The inequality remains true for every natural number }n\ge4.
\displaystyle \therefore \text{the required set is }\{4,5,6,7,\ldots\}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{State the second principle of mathematical induction.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be a statement involving the natural number }n\text{ such that:}
\displaystyle \text{(i) }P(1)\text{ is true, i.e. }P(n)\text{ is true for }n=1,\text{ and}
\displaystyle \text{(ii) }P(m+1)\text{ is true whenever }P(r)\text{ is true for all }r,
\displaystyle \text{where }1\le r\le m.
\displaystyle \text{Then, }P(n)\text{ is true for all natural numbers }n.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }P(n):2\times4^{2n+1}+3^{3n+1}\text{ is divisible by }\lambda
\displaystyle \text{for all }n\in N\text{ is true, then find the value of }\lambda.
\displaystyle \text{Answer:}
\displaystyle \text{Since the statement is true for all }n\in N,\text{ put }n=1.
\displaystyle 2\times4^{2(1)+1}+3^{3(1)+1}=2\times4^3+3^4.
\displaystyle =2\times64+81=128+81=209.
\displaystyle 209=11\times19.
\displaystyle \therefore \lambda=11.
\displaystyle \\

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