Is there any real value of 'a' for which the equation
Question: Is there any real value of ' $a$ ' for which the equation $x^{2}+2 x+\left(a^{2}+1\right)=0$ has real roots? Solution: Let quadratic equation $x^{2}+2 x+\left(a^{2}+1\right)=0$ has real roots. Here, $a=1, b=2$ and,$c=\left(a^{2}+1\right)$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=1, b=2$ and, $c=\left(a^{2}+1\right)$, we get $D=(2)^{2}-4 \times 1 \times\left(a^{2}+1\right)$ $=4-4\left(a^{2}+1\right)$ $=-4 a^{2}$ The given equation will have equal roots, if $D0$ i.e. $-4 a...
Read More →If P (15, r − 1) : P (16, r − 2) = 3 : 4, find r.
Question: IfP(15,r 1) :P(16,r 2) = 3 : 4, findr. Solution: P(15,r 1):P(16,r 2) = 3:4 $\Rightarrow \frac{15 !}{(15-r+1) !} \times \frac{(16-r+2) !}{16 !}=\frac{3}{4}$ $\Rightarrow \frac{15 !}{(16-r) !} \times \frac{(18-r) !}{16 \times 15 !}=\frac{3}{4}$ $\Rightarrow \frac{(18-r)(17-r)(16-r) !}{(16-r) !(16)}=\frac{3}{4}$ $\Rightarrow(18-r)(17-r)=12$ $\Rightarrow(18-r)(17-r)=4 \times 3$ On comparing the LHS and the RHS in above expression, we get: $\Rightarrow 18-r=14$ $\Rightarrow r=14$...
Read More →If (x + 5) is a factor of p(x)
Question: If $(x+5)$ is a factor of $p(x)=x^{3}-20 x+5 k$, then $k=?$ (a) $-5$ (b) 5 (c) 3 (d) $-3$ Solution: (b) 5 $(x+5)$ is a factor of $p(x)=x^{3}-20 x+5 k$. $\therefore p(-5)=0$ $\Rightarrow(-5)^{3}-20 \times(-5)+5 k=0$ $\Rightarrow-125+100+5 k=0$ $\Rightarrow 5 k=25$ $\Rightarrow k=5$...
Read More →If P (15, r − 1) : P (16, r − 2) = 3 : 4, find r.
Question: IfP(15,r 1) :P(16,r 2) = 3 : 4, findr. Solution: P(15,r 1):P(16,r 2) = 3:4 $\Rightarrow \frac{15 !}{(15-r+1) !} \times \frac{(16-r+2) !}{16 !}=\frac{3}{4}$ $\Rightarrow \frac{15 !}{(16-r) !} \times \frac{(18-r) !}{16 \times 15 !}=\frac{3}{4}$ $\Rightarrow \frac{(18-r)(17-r)(16-r) !}{(16-r) !(16)}=\frac{3}{4}$ $\Rightarrow(18-r)(17-r)=12$ $\Rightarrow(18-r)(17-r)=4 \times 3$ On comparing the LHS and the RHS in above expression, we get: $\Rightarrow 18-r=14$ $\Rightarrow r=14$...
Read More →Is there any real value of 'a' for which the equation x2 + 2x + (a2 + 1) = 0
Question: Is there any real value of ' $a$ ' for which the equation $x^{2}+2 x+\left(a^{2}+1\right)=0$ has real roots? Solution: Let quadratic equation $x^{2}+2 x+\left(a^{2}+1\right)=0$ has real roots. Here, $a=1, b=2$ and,$c=\left(a^{2}+1\right)$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=1, b=2$ and, $c=\left(a^{2}+1\right)$, we get $D=(2)^{2}-4 \times 1 \times\left(a^{2}+1\right)$ $=4-4\left(a^{2}+1\right)$ $=-4 a^{2}$ The given equation will have equal roots, if $D0$ i.e. $-4 a...
Read More →Prove that: 1 . P (1, 1) + 2 . P (2, 2) + 3 . P (3, 3) + ... + n . P (n, n) = P (n + 1, n + 1) − 1.
Question: Prove that:1 .P(1, 1) + 2 .P(2, 2) + 3 .P(3, 3) + ... +n.P(n,n) =P(n+ 1,n+ 1) 1. Solution: 1.P(1, 1) + 2.P(2, 2) + 3.P(3, 3) + ... +n.P(n,n) =P(n+ 1,n+ 1) 1 P (n,n) =n! 1.1! + 2.2! + 3.3! ......+n.n! = (n+1)! 1 LHS = 1.1! + 2.2! + 3.3! ......+n.n! $=\sum_{r=1}^{n} r . r !$ $=\sum_{r=1}^{n}[(r+1)-1] r !$ $=\sum_{r=1}^{n}[(r+1) r !-r !]$ $=\sum_{r=1}^{n}\{(r+1) !-r !\}$ $=(2 !-1 !)+(3 !-2 !)+\ldots[(n+1) !-n !]$ $=[(n+1) !-1 !]$ $=[(n+1) !-1]=\operatorname{RHS}$ Hence proved....
Read More →One of the factors of
Question: One of the factors of $\left(25 x^{2}-1\right)+(1+5 x)^{2}$ is (a) $5+x$ (b) $5-x$ (c) $5 x-1$ (d) $10 x$ Solution: $\left(25 x^{2}-1\right)+(1+5 x)^{2}$ $=(5 x-1)(5 x+1)+(1+5 x)^{2}$ $=(5 x+1)[(5 x-1)+(1+5 x)]$ $=(5 x+1)(10 x)$ So, the factors of $\left(25 x^{2}-1\right)+(1+5 x)^{2}$ are $(5 x+1)$ and $10 x$ Hence, the correct answer is option (d)....
Read More →Prove that: 1 . P (1, 1) + 2 . P (2, 2) + 3 . P (3, 3) + ... + n . P (n, n) = P (n + 1, n + 1) − 1.
Question: Prove that:1 .P(1, 1) + 2 .P(2, 2) + 3 .P(3, 3) + ... +n.P(n,n) =P(n+ 1,n+ 1) 1. Solution: 1.P(1, 1) + 2.P(2, 2) + 3.P(3, 3) + ... +n.P(n,n) =P(n+ 1,n+ 1) 1 P (n,n) =n! 1.1! + 2.2! + 3.3! ......+n.n! = (n+1)! 1 LHS = 1.1! + 2.2! + 3.3! ......+n.n! $=\sum_{r=1}^{n} r . r !$ $=\sum_{r=1}^{n}[(r+1)-1] r !$ $=\sum_{r=1}^{n}[(r+1) r !-r !]$ $=\sum_{r=1}^{n}\{(r+1) !-r !\}$ $=(2 !-1 !)+(3 !-2 !)+\ldots[(n+1) !-n !]$ $=[(n+1) !-1 !]$ $=[(n+1) !-1]=\operatorname{RHS}$ Hence proved....
Read More →Write the set of value of 'a' for which the equation
Question: Write the set of value of ' $a$ ' for which the equation $x^{2}+a x-1=0$ has real roots. Solution: The given quadric equation is $x^{2}+a x-1=0$ Then find the value ofa. 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 roots, if $D0$. $a^{2}+40$ $\Rightarrow a^{2}-4$ which is true for all real values of $a$. Therefore, for all real values of $a$, the given...
Read More →Which of the following is a factor of
Question: Which of the following is a factor of $(x+y)^{3}-\left(x^{3}+y^{3}\right)$ ? (a) $x^{2}+y^{2}+2 x y$ (b) $x^{2}+y^{2}-x y$ (c) $x y^{2}$ (d) $3 x y$ Solution: $(x+y)^{3}-\left(x^{3}+y^{3}\right)$ $=x^{3}+y^{3}+3 x y(x+y)-\left(x^{3}+y^{3}\right)$ $=3 x y(x+y)$ Thus, the factors of $(x+y)^{3}-\left(x^{3}+y^{3}\right)$ are $3 x y$ and $(x+y)$. Hence, the correct answer is option (d)....
Read More →If P (n, 5) : P (n, 3) = 2 : 1, find n.
Question: IfP(n, 5) :P(n, 3) = 2 : 1, findn. Solution: We have,P(n, 5):P(n, 3) = 2:1 $\Rightarrow \frac{n !}{(n-5) !} \times \frac{(n-3) !}{n !}=\frac{2}{1}$ $\Rightarrow \frac{n !}{(n-5) !} \times \frac{(n-3)(n-4)(n-5) !}{n !}=\frac{2}{1}$ $\Rightarrow(n-3)(n-4)=2$ $\Rightarrow(n-3)(n-4)=2 \times 1$ Thus, on comparing the LHS and the RHS in above expression, we get, $n-3=2$ $\Rightarrow n=5$...
Read More →If P (n, 5) : P (n, 3) = 2 : 1, find n.
Question: IfP(n, 5) :P(n, 3) = 2 : 1, findn. Solution: We have,P(n, 5):P(n, 3) = 2:1 $\Rightarrow \frac{n !}{(n-5) !} \times \frac{(n-3) !}{n !}=\frac{2}{1}$ $\Rightarrow \frac{n !}{(n-5) !} \times \frac{(n-3)(n-4)(n-5) !}{n !}=\frac{2}{1}$ $\Rightarrow(n-3)(n-4)=2$ $\Rightarrow(n-3)(n-4)=2 \times 1$ Thus, on comparing the LHS and the RHS in above expression, we get, $n-3=2$ $\Rightarrow n=5$...
Read More →Write the sum of real roots of the equation x2 + |x| − 6 = 0.
Question: Write the sum of real roots of the equation $x^{2}+|x|-6=0$. Solution: The given quadric equation is $x^{2}+|x|-6=0$ Here, $a=1, b=\pm 1$ and, $c=-6$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=1, b=\pm 1$ and, $c=-6$ $=(\pm 1)^{2}-4 \times 1 \times-6$ $=1+24$ $=25$ Since, $D \geq 0$ Therefore, root of the given equation are real and distinct. Thus, sum of the roots be $=0$...
Read More →The coefficient of x in the expansion of
Question: The coefficient of $x$ in the expansion of $(x+3)^{3}$ is (a) 1 (b) 9 (c) 18 (d) 27 Solution: $(x+3)^{3}$ $=x^{3}+3^{3}+9 x(x+3)$ $=x^{3}+27+9 x^{2}+27 x$ So, the coefficient of $x$ in $(x+3)^{3}$ is 27 . Hence, the correct answer is option (d)....
Read More →Write the number of real roots of the equation
Question: Write the number of real roots of the equation $x^{2}+3|x|+2=0$. Solution: The given quadric equation is $x^{2}+3|x|+2=0$ $x^{2}+3|x|+2=0$ Here, $a=1, b=\pm 3$ and, $c=2$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=1, b=\pm 3$ and, $c=2$ $=(\pm 3)^{2}-4 \times 1 \times 2$ $=9-8$ $=1$ Since, $D \geq 0$ Therefore, roots of the given equation are real and distinct. $\therefore$ The number of real roots of the given equation is 4 ....
Read More →If P (2n − 1, n) : P (2n + 1, n − 1) = 22 : 7 find n.
Question: IfP(2n 1,n) :P(2n+ 1,n 1) = 22 : 7 findn. Solution: P(2n 1,n):P(2n+ 1,n 1) = 22:7 $\Rightarrow \frac{(2 n-1) !}{(2 n-1-n) !} \times \frac{(2 n+1-n+1) !}{(2 n+1) !}=\frac{22}{7}$ $\Rightarrow \frac{(2 n-1) !}{(n-1) !} \times \frac{(n+2) !}{(2 n+1) !}=\frac{22}{7}$ $\Rightarrow \frac{(2 n-1) !}{(n-1) !} \times \frac{(n+2)(n+1)(n)(n-1) !}{(2 n+1)(2 n)(2 n-1) !}=\frac{22}{7}$ $\Rightarrow \frac{(n+2)(n+1)(n)}{(2 n+1)(2 n)}=\frac{22}{7}$ $\Rightarrow \frac{(n+2)(n+1)}{2(2 n+1)}=\frac{22}{7}...
Read More →If P (2n − 1, n) : P (2n + 1, n − 1) = 22 : 7 find n.
Question: IfP(2n 1,n) :P(2n+ 1,n 1) = 22 : 7 findn. Solution: P(2n 1,n):P(2n+ 1,n 1) = 22:7 $\Rightarrow \frac{(2 n-1) !}{(2 n-1-n) !} \times \frac{(2 n+1-n+1) !}{(2 n+1) !}=\frac{22}{7}$ $\Rightarrow \frac{(2 n-1) !}{(n-1) !} \times \frac{(n+2) !}{(2 n+1) !}=\frac{22}{7}$ $\Rightarrow \frac{(2 n-1) !}{(n-1) !} \times \frac{(n+2)(n+1)(n)(n-1) !}{(2 n+1)(2 n)(2 n-1) !}=\frac{22}{7}$ $\Rightarrow \frac{(n+2)(n+1)(n)}{(2 n+1)(2 n)}=\frac{22}{7}$ $\Rightarrow \frac{(n+2)(n+1)}{2(2 n+1)}=\frac{22}{7}...
Read More →then the value of p is
Question: If $\left(3 x+\frac{1}{2}\right)\left(3 x-\frac{1}{2}\right)=9 x^{2}-p$ then the value of $p$ is (a) 0 (b) $-\frac{1}{4}$ (c) $\frac{1}{4}$ (d) $\frac{1}{2}$ Solution: $\left(3 x+\frac{1}{2}\right)\left(3 x-\frac{1}{2}\right)=9 x^{2}-p$ $9 x^{2}-\frac{1}{4}=9 x^{2}-p$ $\left(\because\left(a^{2}-b^{2}\right)=(a+b)(a-b)\right)$ $\Rightarrow p=\frac{1}{4}$ Hence, the correct answer is option (c)....
Read More →If 1+2–√ is a root of a quadratic equation will rational coefficients
Question: If $1+\sqrt{2}$ is a root of a quadratic equation will rational coefficients, write its other root. Solution: Given that $(1+\sqrt{2})$ is a root of the quadratic equation with rational coefficients. Then find the other root. As we know that if $(1+\sqrt{2})$ is a root of the quadratic equation with rational coefficients then other roots be $(1-\sqrt{2})$. Hence, the require root of the quadratic equation be $(1-\sqrt{2})$...
Read More →If P (n − 1, 3) : P (n, 4) = 1 : 9, find n.
Question: IfP(n 1, 3) :P(n, 4) = 1 : 9, findn. Solution: P(n 1, 3):P(n, 4) = 1:9 $\Rightarrow \frac{(n-1) !}{(n-1-3) !} \times \frac{(n-4) !}{(n) !}=\frac{1}{9}$ $\Rightarrow \frac{(n-1) !}{(n-4) !} \times \frac{(n-4) !}{n !}=\frac{1}{9}$ $\Rightarrow \frac{(n-1) !}{n !}=\frac{1}{9}$ $\Rightarrow \frac{(n-1) !}{n(n-1) !}=\frac{1}{9}$ $\Rightarrow \frac{1}{n}=\frac{1}{9}$ $\Rightarrow n=9$...
Read More →What is the nature of roots of the quadratic equation
Question: What is the nature of roots of the quadratic equation $4 x^{2}-12 x-9=0 ?$ Solution: The given quadric equation is $4 x^{2}-12 x-9=0$ Here, $a=4, b=-12$ and, $c=-9$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=4, b=-12$ and, $c=-9$ $=(-12)^{2}-4 \times 4 \times-9$ $=144+144$ $=288$ Since, $D \geq 0$ Therefore, root of the given equation are real and distinct....
Read More →If P (n − 1, 3) : P (n, 4) = 1 : 9, find n.
Question: IfP(n 1, 3) :P(n, 4) = 1 : 9, findn. Solution: P(n 1, 3):P(n, 4) = 1:9 $\Rightarrow \frac{(n-1) !}{(n-1-3) !} \times \frac{(n-4) !}{(n) !}=\frac{1}{9}$ $\Rightarrow \frac{(n-1) !}{(n-4) !} \times \frac{(n-4) !}{n !}=\frac{1}{9}$ $\Rightarrow \frac{(n-1) !}{n !}=\frac{1}{9}$ $\Rightarrow \frac{(n-1) !}{n(n-1) !}=\frac{1}{9}$ $\Rightarrow \frac{1}{n}=\frac{1}{9}$ $\Rightarrow n=9$...
Read More →If a + b + c = 0, then
Question: If $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=?$ (a) 0 (b) $a b c$ (c) $2 a b c$ (d) $3 a b c$ Solution: (d) 3abc $a+b+c=0$ $\Rightarrow a+b=-c$ $\Rightarrow(a+b)^{3}=(-c)^{3}$ $\Rightarrow a^{3}+b^{3}+3 a b(a+b)=-c^{3}$ $\Rightarrow a^{3}+b^{3}+3 a b(-c)=-c^{3}$ $\Rightarrow a^{3}+b^{3}+c^{3}=3 a b c$...
Read More →Write the value of k for which the quadratic equation
Question: Write the value of $k$ for which the quadratic equation $x^{2}-k x+4=0$ has equal roots. Solution: The given quadric equation is $x^{2}-k x+4=0$, and roots are equal. Then find the value ofk. Here, $a=1, b=-k$ and, $c=4$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=1, b=-k$ and, $c=4$ $=(-k)^{2}-4 \times 1 \times 4$ $=k^{2}-16$ The given equation will have equal roots, if $D=0$ $k^{2}-16=0$ $k^{2}=16$ $k=\sqrt{16}$ $=\pm 4$ Therefore, the value of $k=\pm 4$...
Read More →If P (n, 4) = 12 . P (n, 2), find n.
Question: IfP(n, 4) = 12 .P(n, 2), findn. Solution: P(n, 4) = 12 .P(n, 2) $\Rightarrow \frac{n !}{(n-4) !}=12 \times \frac{n !}{(n-2) !}$ $\Rightarrow \frac{(n-2) !}{(n-4) !}=12 \times \frac{n !}{n !}$ $\Rightarrow \frac{(n-2)(n-3)(n-4) !}{(n-4) !}=12$ $\Rightarrow(n-2)(n-3)=12$ $\Rightarrow(n-2)(n-3)=4 \times 3$ On comparing the LHS and the RHS, we get : $n-2=4$ $\Rightarrow n=6$...
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