If 2 is a root of the equation x2 + bx + 12 = 0 and the
Question: If 2 is a root of the equation $x^{2}+b x+12=0$ and the equation $x^{2}+b x+q=0$ has equal roots, then $q=$ (a) 8(b) 8(c) 16(d) 16 Solution: 2 is the common roots given quadric equation are $x^{2}+b x+12=0$, and $x^{2}+b x+q=0$ Then find the value ofq. Here, $x^{2}+b x+12=0$.......(1) $x^{2}+b x+q=0 \ldots \ldots(2)$ Putting the value of $x=2$ in equation (1) we get $2^{2}+b \times 2+12=0$ $4+2 b+12=0$ $2 b=-16$ $b=-8$ Now, putting the value of $b=-8$ in equation (2) we get $x^{2}-8 x+...
Read More →If (n + 1)! = 90 [(n − 1)!], find n.
Question: If (n + 1)! = 90 [(n 1)!], findn. Solution: (n+ 1)! = 90 [(n 1)!] $\Rightarrow(n+1) \times(n) \times(n-1) !=90[(n-1) !]$ $\Rightarrow(n+1) \times(n)=90$ $\Rightarrow(n+1) \times(n)=10 \times 9$ On comparing, we get: n= 9...
Read More →If (n + 1)! = 90 [(n − 1)!], find n.
Question: If (n + 1)! = 90 [(n 1)!], findn. Solution: (n+ 1)! = 90 [(n 1)!] $\Rightarrow(n+1) \times(n) \times(n-1) !=90[(n-1) !]$ $\Rightarrow(n+1) \times(n)=90$ $\Rightarrow(n+1) \times(n)=10 \times 9$ On comparing, we get: n= 9...
Read More →If (n + 2)! = 60 [(n − 1)!], find n.
Question: If (n+ 2)! = 60 [(n 1)!], findn. Solution: (n+ 2)! = 60 [(n 1)!] $\Rightarrow(n+2) \times(n+1) \times(n) \times(n-1) !=60[(n-1) !]$ $\Rightarrow(n+2) \times(n+1) \times(n)=60$ $\Rightarrow(n+2) \times(n+1) \times(n)=5 \times 4 \times 3$ n= 3...
Read More →If (n + 2)! = 60 [(n − 1)!], find n.
Question: If (n+ 2)! = 60 [(n 1)!], findn. Solution: (n+ 2)! = 60 [(n 1)!] $\Rightarrow(n+2) \times(n+1) \times(n) \times(n-1) !=60[(n-1) !]$ $\Rightarrow(n+2) \times(n+1) \times(n)=60$ $\Rightarrow(n+2) \times(n+1) \times(n)=5 \times 4 \times 3$ n= 3...
Read More →The value of
Question: The value of $\sqrt{6+\sqrt{6+\sqrt{6+}} \ldots .}$ is (a) 4(b) 3(c) 2(d) 3.5 Solution: Let $x=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+} \ldots . .}}}$ Squaring both sides we get $x^{2}=6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\ldots . . .}}}}$ $x^{2}=6+x$ $x^{2}-x-6=0$ $x^{2}-3 x+2 x-6=0$ $x(x-3)+2(x-3)=0$ $(x-3)(x+2)=0$ The value ofxcannot be negative. Thus, the value ofx= 3 Therefore, the correct answer is $(b)$...
Read More →Prove that: n! (n + 2) = n! + (n + 1)!
Question: Prove that:n! (n+ 2) =n! + (n+ 1)! Solution: RHS =n! + (n+ 1)! = n! + (n+ 1)(n!) =n!( 1+n+ 1) =n!(n+2) = LHS Hence, proved....
Read More →Which of the following are true:
Question: Which of the following are true: (i) (2 +3)! = 2! + 3! (ii) (2 3)! = 2! 3! Solution: (i) LHS = (2 +3)! = 5! = 120 RHS = 2! + 3! = 2 + 6 = 8 Since LHS RHS, Thus, (i) is false. (ii) LHS = (2 3)! = 6! = 720 RHS = 2! 3! = 2 6 = 12 LHS RHS Thus, (ii) is false....
Read More →The positive value of k for which the equation x2 + kx + 64 = 0 and x2 − 8x + k = 0
Question: The positive value of $k$ for which the equation $x^{2}+k x+64=0$ and $x^{2}-8 x+k=0$ will both have real roots, is (a) 4 (b) 8 (c) 12 (d) 16 Solution: The given quadric equation are $x^{2}+k x+64=0$, and $x^{2}-8 x+k=0$ roots are real. Then find the value ofa. Here, $x^{2}+k x+64=0$.......(1) $x^{2}-8 x+k=0 \cdots(2)$ $a_{1}=1, b_{1}=k$ and, $c_{1}=64$ $a_{2}=1, b_{2}=-8$ and,$c_{2}=k$ As we know that $D_{1}=b^{2}-4 a c$ Putting the value of $a_{1}=1, b_{1}=k$ and, $c_{1}=64$ $=(k)^{2...
Read More →Convert the following products into factorials:
Question: Convert the following products into factorials: (i) 5 6 7 8 9 10 (ii) 3 6 9 12 15 18 (iii) (n+ 1) (n+ 2) (n+ 3) ... (2n) (iv) 1 3 5 7 9 ... (2n 1) Solution: (i) $5 \times 6 \times 7 \times 8 \times 9 \times 10=\frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10}{1 \times 2 \times 3 \times 4}$ $=\frac{10 !}{4 !}$ (ii) $3 \times 6 \times 9 \times 12 \times 15 \times 18=(3 \times 1) \times(3 \times 2) \times(3 \times 3) \times(3 \times 4) \times(3 \ti...
Read More →Find x in each of the following:
Question: Findxin each of the following: (i) $\frac{1}{4 !}+\frac{1}{5 !}=\frac{x}{6 !}$ (ii) $\frac{x}{10 !}=\frac{1}{8 !}+\frac{1}{9 !}$ (iii) $\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}$ Solution: (i) $\frac{1}{4 !}+\frac{1}{5 !}=\frac{x}{6 !}$ $\Rightarrow \frac{1}{4 !}+\frac{1}{5(4 !)}=\frac{x}{6 !}$ $\Rightarrow \frac{5+1}{5(4 !)}=\frac{x}{6 !}$ $\Rightarrow \frac{6}{5 !}=\frac{x}{6 !}$ $\Rightarrow \frac{6}{5 !}=\frac{x}{6 \times 5 !}$ $\Rightarrow x=36$ (ii) $\frac{x}{10 !}=\frac{1}{8 !}+...
Read More →Find x in each of the following:
Question: Findxin each of the following: (i) $\frac{1}{4 !}+\frac{1}{5 !}=\frac{x}{6 !}$ (ii) $\frac{x}{10 !}=\frac{1}{8 !}+\frac{1}{9 !}$ (iii) $\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}$ Solution: (i) $\frac{1}{4 !}+\frac{1}{5 !}=\frac{x}{6 !}$ $\Rightarrow \frac{1}{4 !}+\frac{1}{5(4 !)}=\frac{x}{6 !}$ $\Rightarrow \frac{5+1}{5(4 !)}=\frac{x}{6 !}$ $\Rightarrow \frac{6}{5 !}=\frac{x}{6 !}$ $\Rightarrow \frac{6}{5 !}=\frac{x}{6 \times 5 !}$ $\Rightarrow x=36$ (ii) $\frac{x}{10 !}=\frac{1}{8 !}+...
Read More →If the equation ax2 + 2x + a = 0 has two distinct roots, if
Question: If the equation $a x^{2}+2 x+a=0$ has two distinct roots, if (a) $a=\pm 1$ (b) $a=0$ (c) $a=0,1$ (d) $a=-1,0$ Solution: The given quadric equation is $a x^{2}+2 x+a=0$, and roots are distinct. Then find the value ofa. Here, $a=a, b=2$ and, $c=a$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=a, b=2$ and, $c=a$ $=(2)^{2}-4 \times a \times a$ $=4-4 a^{2}$ The given equation will have real and distinct roots, if $D0$ $4-4 a^{2}=0$ $4 a^{2}=4$ $a^{2}=\frac{4}{4}$ $a=\sqrt{1}$ $=\p...
Read More →Express each of the following equations in the form ax
Question: Express each of the following equations in the formax + by + c= 0 and indicate the values ofa, b, cin each case. (i) $3 x+5 y=7.5$ (ii) $2 x-\frac{y}{5}+6=0$ (iii) $3 y-2 x=6$ (iv) $4 x=5 y$ (v) $\frac{x}{5}-\frac{y}{6}=1$ (vi) $\sqrt{2} x+\sqrt{3} y=5$ Solution: (i) 3x+ 5y= 7.5 This can be expressed in the form $a x+b y+c=0$ as $3 x+5 y+(-7.5)=0$. (ii) $2 x-\frac{y}{5}+6=0$ This can be expressed in the form $a x+b y+c=0$ as $2 x+\left(-\frac{1}{5}\right) y+6=0$ (iii) $3 y-2 x=6$ This ...
Read More →Solve the following
Question: Prove that $\frac{1}{9 !}+\frac{1}{10 !}+\frac{1}{11 !}=\frac{122}{11 !}$ Solution: $\mathrm{LHS}=\frac{1}{9 !}+\frac{1}{10 !}+\frac{1}{11 !}$ $=\frac{1}{9 !}+\frac{1}{10 \times 9 !}+\frac{1}{11 \times 10 \times 9 !}$ $=\frac{110+11+1}{11 \times 10 \times 9 !}$ $=\frac{122}{11 !}=\mathrm{RHS}$ Hence proved....
Read More →Compute:
Question: Compute: (i) $\frac{30 !}{28 !}$ (ii) $\frac{11 !-10 !}{9 !}$ (iii) L.C.M. $(6 !, 7 !, 8 !)$ Solution: (i) $\frac{30 !}{28 !}=\frac{30 \times 29 \times 28 !}{28 !} \quad[\because n !=n(n-1) !]$ $=30 \times 29$ $=870$ (ii) $\frac{11 !-10 !}{9 !}=\frac{11 \times 10 \times 9 !-10 \times 9 !}{9 !} \quad[\because n !=n(n-1) !]$ $=\frac{9 !(110-10)}{9 !}$ $=100$ (iii) LCM of (6!,7! and 8!): $n !=n(n-1) !$ Therefore, (6!,7! and 8!) can be rewritten as: $8 !=8 \times 7 \times 6 !$ $7 !=7 \time...
Read More →Compute:
Question: Compute: (i) $\frac{30 !}{28 !}$ (ii) $\frac{11 !-10 !}{9 !}$ (iii) L.C.M. $(6 !, 7 !, 8 !)$ Solution: (i) $\frac{30 !}{28 !}=\frac{30 \times 29 \times 28 !}{28 !} \quad[\because n !=n(n-1) !]$ $=30 \times 29$ $=870$ (ii) $\frac{11 !-10 !}{9 !}=\frac{11 \times 10 \times 9 !-10 \times 9 !}{9 !} \quad[\because n !=n(n-1) !]$ $=\frac{9 !(110-10)}{9 !}$ $=100$ (iii) LCM of (6!,7! and 8!): $n !=n(n-1) !$ Therefore, (6!,7! and 8!) can be rewritten as: $8 !=8 \times 7 \times 6 !$ $7 !=7 \time...
Read More →If ax2 + bx + c = 0 has equal roots, then c =
Question: If $a x^{2}+b x+c=0$ has equal roots, then $c=$ (a) $\frac{-b}{2 a}$ (b) $\frac{b}{2 a}$ (c) $\frac{-b^{2}}{4 a}$ (d) $\frac{b^{2}}{4 a}$ Solution: The given quadric equation is $a x^{2}+b x+c=0$, and roots are equal Then find the value ofc. Let $\alpha$ and $\beta$ be two roots of given equation $\alpha=\beta$ Then, as we know that sum of the roots $\alpha+\beta=\frac{-b}{a}$ $\alpha+\alpha=\frac{-b}{a}$ $2 \alpha=\frac{-b}{a}$ $a=\frac{-b}{2 a}$ And the product of the roots $\alpha \...
Read More →If the equation 9x2 + 6kx + 4 = 0 has equal roots,
Question: If the equation $9 x^{2}+6 k x+4=0$ has equal roots, then the roots are both equal to (a) $\pm \frac{2}{3}$ (b) $\pm \frac{3}{2}$ (C) 0 (d) $\pm 3$ Solution: The given quadric equation is $9 x^{2}+6 k x+4=0$, and roots are equal. Then find roots of given equation. Here, $a=9, b=6 k$ and,$c=4$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=9, b=6 k$ and, $c=4$ $=(6 k)^{2}-4 \times 9 \times 4$ $=36 k^{2}-144$ The given equation will have equal roots, if $D=0$ $36 k^{2}-144=0$ $3...
Read More →If the equation x2 − ax + 1 = 0 has two distinct roots, then
Question: If the equation $x^{2}-a x+1=0$ has two distinct roots, then (a) $|a|=2$ (b) $|a|2$ (c) $|a|2$ (d) None of these Solution: The given quadric equation is $x^{2}-a x+1=0$, and roots are distinct. Then find the value of $a$. Here, $a=1, b=a$ and,$c=1$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=1, b=a$ and, $c=1$ $=(a)^{2}-4 \times 1 \times 1$ $=a^{2}-4$ The given equation will have real and distinct roots, if $D0$ $a^{2}-40$ $a^{2}4$ $a\sqrt{4}$ $\pm 2$ Therefore, the value o...
Read More →How many three letter words can be made using the letters of the word 'ORIENTAL'?
Question: How many three letter words can be made using the letters of the word 'ORIENTAL'? Solution: The word ORIENTAL consists of 8 letters. In order to make three letter words, we need to permute these 8 letters, taken three at a time. $\Rightarrow{ }^{8} P_{3}=8 \times 7 \times 6=336$...
Read More →Find the value
Question: If $\frac{a}{b}+\frac{b}{a}=-1$ then $\left(a^{3}-b^{3}\right)=?$ (a) 3(b) 2(c) 1(d) 0 Solution: $\frac{a}{b}+\frac{b}{a}=-1$ $\Rightarrow \frac{a^{2}+b^{2}}{a b}=-1$ $\Rightarrow a^{2}+b^{2}=-a b$ $\Rightarrow a^{2}+b^{2}+a b=0$ Thus, we have: $\left(a^{3}-b^{3}\right)=(a-b)\left(a^{2}+b^{2}+a b\right)$ $=(a-b) \times 0$ $=0$...
Read More →How many words (with or without dictionary meaning) can be made from the letters in the word MONDAY,
Question: How many words (with or without dictionary meaning) can be made from the letters in the word MONDAY, assuming that no letter is repeated, if (i) 4 letters are used at a time? (ii) all letters are used at a time. (iii) all letters are used but first is vowel. Solution: (i) The word MONDAY consists of 6 letters. Number of words formed using 4 letters = Number of arrangements of 6 letters, taken 4 at a time $={ }^{6} P_{4}=\frac{6 !}{2 !}=6 \times 5 \times 4 \times 3=360$ (ii) Number of w...
Read More →If the equation x2 + 4x + k = 0 has real and distinct roots, then
Question: If the equation $x^{2}+4 x+k=0$ has real and distinct roots, then (a) $k4$ (b) $k4$ (c) $k \geq 4$ (d) $k \leq 4$ Solution: The given quadric equation is $x^{2}+4 x+k=0$, and roots are real and distinct. Then find the value ofk. Here, $a=1, b=4$ and, $c=k$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=1, b=4$ and,$c=k$ $=(4)^{2}-4 \times 1 \times k$ $=16-4 k$ The given equation will have real and distinct roots, if $D0$ $16-4 k0$ $4 k16$ $k\frac{16}{4}$ $4$ Therefore, the val...
Read More →m men and n women are to be seated in a row so that no two women sit together.
Question: mmen andnwomen are to be seated in a row so that no two women sit together. ifmnthen show that the number of ways in which they can be seated as $\frac{m !(m+1) !}{(m-n+1) !}$ Solution: 'm' men can be seated in a row inm! ways. ' $m$ ' men will generate $(m+1)$ gaps that are to be filled by ' $n$ ' women $=$ Number of arrangements of $(m+1)$ gaps, taken ' $n$ ' at a time $=m+1 P_{n}=$ $\frac{(m+1) !}{(m+1-n) !}$ $\therefore$ By fundamental principle of counting, total number of ways in...
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