Quadratic Equation and Inequalities
If the set of all $\mathrm{a} \in \mathbf{R}-\{1\}$, for which the roots of the equation $(1-\mathrm{a}) x^2+2(\mathrm{a}-3) x+9=0$ are positive is $(-\infty,-\alpha] \cup[\beta, \gamma)$, then $2 \alpha+\beta+\gamma$ is equal to $\qquad$ .
Explanation:
$\begin{aligned} & f(x)=(1-a) x^2+2(a-3) x+9, f(0)=9>0 \\ & D \geq 0 \Rightarrow 4(a-3)^2 \geq 4(1-a) \cdot 9 \end{aligned}$
$\Rightarrow a \in(-\infty,-3] \cup[0, \infty)\quad\text{..... (i)}$
$\begin{aligned} & x_1+x_2=\frac{-2(a-3)}{1-a}, x_1 x_2=\frac{9}{1-a} \\ & x_1+x_2>0 \Rightarrow \frac{a-3}{a-1}>0 \Rightarrow a \in(-\infty, 1) \cup(3, \infty) \ldots \text { (ii) } \end{aligned}$
$x_1 x_2>0 \Rightarrow 1-a>0 \Rightarrow a \in(-\infty, 1) \ldots($ iii $)$
$\begin{aligned} &\Rightarrow \text { Interaction of (i), (ii) and (iii) }\\ &\begin{aligned} & a \in(-\infty,-3] \cup[0,1) \\ \Rightarrow \quad & \alpha=3, \beta=0, \gamma=1 \Rightarrow 2 \alpha+\beta+\gamma=7 \end{aligned} \end{aligned}$
If the equation $\mathrm{a}(\mathrm{b}-\mathrm{c}) \mathrm{x}^2+\mathrm{b}(\mathrm{c}-\mathrm{a}) \mathrm{x}+\mathrm{c}(\mathrm{a}-\mathrm{b})=0$ has equal roots, where $\mathrm{a}+\mathrm{c}=15$ and $\mathrm{b}=\frac{36}{5}$, then $a^2+c^2$ is equal to _________
Explanation:
To solve the given problem, we start with the quadratic equation:
$ a(b-c) x^2 + b(c-a) x + c(a-b) = 0 $
Given that the roots are equal (let’s assume both roots are 1), we know the sum of the roots, $\alpha + \beta$, is twice the value of one root, which leads us to:
$ \alpha + \beta = 2 $
Using the formula for the sum of roots for a quadratic equation, $\alpha + \beta = -\frac{b(c-a)}{a(b-c)}$, we set this equal to 2:
$ -\frac{b(c-a)}{a(b-c)} = 2 $
Solving for this:
$ -bc + ab = 2ab - 2ac \\ 2ac = ab + bc \\ 2ac = b(a + c) $
Given that $a + c = 15$ and $b = \frac{36}{5}$, substitute these into the equation:
$ 2ac = 15b \\ 2ac = 15 \times \frac{36}{5} = 108 \\ ac = 54 $
Now, using the equation $a + c = 15$ and $ac = 54$, find $a^2 + c^2$:
$ a^2 + c^2 = (a + c)^2 - 2ac = 15^2 - 2 \times 54 \\ a^2 + c^2 = 225 - 108 = 117 $
Therefore, $a^2 + c^2$ is equal to 117.
The number of distinct real roots of the equation $|x+1||x+3|-4|x+2|+5=0$, is _______
Explanation:
Let's analyze the equation $ |x+1||x+3|-4|x+2|+5=0 $ based on different intervals of $ x $.
(I) If $ x < -3 $ :
$ \begin{aligned} & |x+1| = -(x+1), \\ & |x+3| = -(x+3), \\ & |x+2| = -(x+2), \\ & -(x+1)*(-(x+3)) - 4(-(x+2)) + 5 = 0 \\ & x^2 + 4x + 3 + 4x + 8 + 5 = 0 \\ & x^2 + 8x + 16 = 0 \\ & \Rightarrow x = -4 \quad (\text{one solution}) \end{aligned} $
(II) If $ -3 \leq x < -2 $ :
$ \begin{aligned} & |x+1| = -(x+1), \\ & |x+3| = -(x+3), \\ & |x+2| = -(x+2), \\ & -(x+1)*(-(x+3)) - 4(-(x+2)) + 5 = 0 \\ & x^2 - 10 = 0 \\ & \Rightarrow x = \pm \sqrt{10} \\ & \text{(Do not satisfy } -3 \leq x < -2) \end{aligned} $
(III) If $ -2 \leq x < -1 $ :
$ \begin{aligned} & |x+1|=-(x+1), \\ & |x+3|=x+3, \\ & |x+2|=-(x+2), \\ & -(x+1)(x+3)-4(-(x+2))+5=0 \\ & -x^2-4x-3-4x-8+5=0 \\ & -x^2-8x-6=0 \\ & \Rightarrow x^2+8x+6=0 \\ & x=\frac{-8 \pm 2\sqrt{10}}{2} = -4 \pm \sqrt{10} \end{aligned} $
(IV) If $ x \geq -1 $ :
$ \begin{aligned} & |x+1|=x+1, \\ & |x+3|=x+3, \\ & |x+2|=x+2, \\ & (x+1)(x+3)-4(x+2)+5=0 \\ & x^2+4x+3-4x-8+5=0 \\ & x^2=0 \\ & \Rightarrow x=0 \quad (\text{one solution}) \end{aligned} $
$\Rightarrow$ The number of distinct real roots are two : $ x = -4 $ and $ x = 0 $.
Let $\alpha, \beta$ be roots of $x^2+\sqrt{2} x-8=0$. If $\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$, then $\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}$ is equal to ________.
Explanation:
$\Rightarrow \alpha^2+\sqrt{2} \alpha=8$
$\begin{aligned} \alpha+\beta=\sqrt{2}, \quad \alpha \beta=-8, & \Rightarrow \alpha+\sqrt{2}=\frac{8}{\alpha} \\ & \Rightarrow \beta+\sqrt{2}=\frac{8}{\beta} \end{aligned}$
$\begin{aligned} & \frac{U_{10}+\sqrt{2} U_9}{2 U_8}=\frac{\alpha^{10}+\beta^{10}+\sqrt{2} \alpha^9+\sqrt{2} \beta^9}{2 \alpha^8+2 \beta^8} \\ & =\frac{\alpha^9(\alpha+\sqrt{2})+\beta^9(\beta+\sqrt{2})}{2\left(\alpha^8+\beta^8\right)} \\ & =\frac{\alpha^9 \cdot\left(\frac{8}{\alpha}\right)+\beta^9\left(\frac{8}{\beta}\right)}{2\left(\alpha^8+\beta^8\right)}=\frac{8}{2}=4 \end{aligned}$
Let $x_1, x_2, x_3, x_4$ be the solution of the equation $4 x^4+8 x^3-17 x^2-12 x+9=0$ and $\left(4+x_1^2\right)\left(4+x_2^2\right)\left(4+x_3^2\right)\left(4+x_4^2\right)=\frac{125}{16} m$. Then the value of $m$ is _________.
Explanation:
First, observe that for a degree-4 polynomial
$P(x)=4x^4+8x^3-17x^2-12x+9$
with roots $x_1,\dots,x_4$, we have
$\prod_{i=1}^4(x_i^2+4) =\prod_{i=1}^4\bigl[(x_i-2i)(x_i+2i)\bigr] =\frac{P(2i)\,P(-2i)}{4^2}\,. $
Compute $P(2i)$:
$ P(2i) =4(2i)^4+8(2i)^3-17(2i)^2-12(2i)+9 =64-64i+68-24i+9 =141-88i. $
By conjugation,
$ P(-2i)=141+88i. $
Hence
$ \prod_{i=1}^4(x_i^2+4) =\frac{(141-88i)(141+88i)}{16} =\frac{141^2+88^2}{16} =\frac{19881+7744}{16} =\frac{27625}{16}. $
We are told
$ \prod_{i=1}^4(x_i^2+4)=\frac{125}{16}\,m =\frac{27625}{16}, $
so
$ 125\,m=27625 \quad\Longrightarrow\quad m=221. $
Answer: $ \displaystyle 221.$
The number of real solutions of the equation $x|x+5|+2|x+7|-2=0$ is __________.
Explanation:
$x|x+5|+2|x+7|-2=0$
$\begin{aligned} & \text { (i) } d x \geq-5 \Rightarrow x(x+5)+2(x+7)-2=0 \\ & x^2+7 x+12=0 \Rightarrow x=-3,-4 \end{aligned}$
$\begin{aligned} \text{(ii)} \quad & x \in(-7,-5) \\ & x(-x-5)+2(x+7)-2=0 \\ & -x^2-3 x+12=0 \\ & \Rightarrow x^2+3 x-12=0 \\ & \Rightarrow x=\frac{-3-\sqrt{57}}{2} \text { satisfy } \end{aligned}$
$\begin{aligned} \text{(iii)} \quad & x \leq-7 \\ & \Rightarrow x(-x-5)+2(-x-7)-2=0 \\ & -x^2-7 x-16=0 \Rightarrow x^2+7 x+16=0 \end{aligned}$
No solution
The number of distinct real roots of the equation $|x||x+2|-5|x+1|-1=0$ is __________.
Explanation:
$|x| \quad|x+2|-5|x+1|-1=0$
$\begin{aligned} & \text { (I) if } x<-2 \\ & x^2+2 x+5 x+5-1=0 \\ & x^2+7 x+4=0 \Rightarrow \text { one root satisfying } x<-2 \end{aligned}$
$\begin{aligned} & \text { (II) if }-2 \leq x<-1 \\ & -x^2-2 x+5 x+5-1=0 \\ & x^2-3 x-4=0 \Rightarrow \text { not root satisfying }-2 \leq x<-1 \end{aligned}$
$\begin{aligned} & \text { (III) if }-1 \leq x<0 \\ & -x^2-2 x-5 x-5-1=0 \\ & x^2+7 x+6=0 \\ & x=-1 \text { is only root satisfying }-1 \leq x<0 \end{aligned}$
(IV) if $x \geq 0$
$\begin{aligned} & x^2+2 x-5 x-5-1=0 \\ & x^2-3 x-6=0 \end{aligned}$
one root satisfying $x \geq 0$
$\Rightarrow$ The number of distinct real roots are three.
Let $a, b, c$ be the lengths of three sides of a triangle satistying the condition $\left(a^2+b^2\right) x^2-2 b(a+c) x+\left(b^2+c^2\right)=0$. If the set of all possible values of $x$ is the interval $(\alpha, \beta)$, then $12\left(\alpha^2+\beta^2\right)$ is equal to __________.
Explanation:
$\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0$
$\begin{aligned} & \Rightarrow a^2 x^2-2 a b x+b^2+b^2 x^2-2 b c x+c^2=0 \\ & \Rightarrow(a x-b)^2+(b x-c)^2=0 \\ & \Rightarrow a x-b=0, \quad b x-c=0 \\ & \Rightarrow a+b>c \quad b+c>a \quad c+a>b \end{aligned}$
$\begin{array}{l|l|l} a+a x>b x & a x+b x>a & a x^2+a>a x \\ a+a x>a x^2 & a x+a x^2>a & x^2-x+1>0 \\ x^2-x-1<0 & x^2+x-1>0 & \text { always true } \end{array}$
$\begin{aligned} & \frac{1-\sqrt{5}}{2}< x<\frac{1+\sqrt{5}}{2} \\ & x< \frac{-1-\sqrt{5}}{2}, \text { or } x >\frac{-1+\sqrt{5}}{2} \end{aligned}$
$\begin{aligned} & \Rightarrow \frac{\sqrt{5}-1}{2}< x<\frac{\sqrt{5}+1}{2} \\ & \Rightarrow \alpha=\frac{\sqrt{5}-1}{2}, \beta=\frac{\sqrt{5}+1}{2} \\ & 12\left(\alpha^2+\beta^2\right)=12\left(\frac{(\sqrt{5}-1)^2+(\sqrt{5}+1)^2}{4}\right)=36 \end{aligned}$
The number of real solutions of the equation $x\left(x^2+3|x|+5|x-1|+6|x-2|\right)=0$ is _________.
Explanation:
The given equation is $x(x^2+3|x|+5|x-1|+6|x-2|)=0$, which can be solved by analyzing it in parts. It can be broken down into: $x=0$ and $x^2+3|x|+5|x-1|+6|x-2|=0$.
For $x=0$, it's clear that it is a solution to the equation since it makes the entire expression equal to zero.
Case (I)$ x<0 $
$ \begin{aligned} & x^2-3 x-5(x-1)-6(x-2)=0 \\\\ & x^2-14 x+17=0 \end{aligned} $
$\because$ All roots are positive $\Rightarrow$ no solution
Case (II)
$ \begin{aligned} & 0 < x < 1 \\\\ & x^2+3 x-5(x-1)-6(x-2)=0 \\\\ & x^2-8 x+17=0 \\\\ & \because D < 0 \Rightarrow \text { no solution } \end{aligned} $
Case (III)
$ 1 < x < 2 $
$ x^2+3 x+5(x-1)-6(x-2)=0 $
$ x^2+2 x+7=0 $
$\Rightarrow$ no solution
Case (IV)
$ \begin{aligned} & x > 2 \\\\ & x^2+3 x+5(x-1)+6(x-2)=0 \\\\ & x^2+14 x-19=0 \end{aligned} $
All roots less than 2
$\Rightarrow$ no solution
Here $x=0$ is only solution.
Let $\alpha, \beta \in \mathbf{N}$ be roots of the equation $x^2-70 x+\lambda=0$, where $\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathbf{N}$. If $\lambda$ assumes the minimum possible value, then $\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}$ is equal to :
Explanation:
$\begin{aligned} & x^2-70 x+\lambda=0 \\ & \alpha+\beta=70 \\ & \alpha \beta=\lambda \\ & \therefore \alpha(70-\alpha)=\lambda \end{aligned}$
Since, 2 and 3 does not divide $\lambda$
$\therefore \alpha=5, \beta=65, \lambda=325$
By putting value of $\alpha, \beta, \lambda$ we get the required value $60$.
Let the set $C=\left\{(x, y) \mid x^2-2^y=2023, x, y \in \mathbb{N}\right\}$. Then $\sum_\limits{(x, y) \in C}(x+y)$ is equal to _________.
Explanation:
First, let's consider the equation $x^2 - 2^y = 2023$ where $x$ and $y$ are natural numbers. Our goal is to find all the pairs $(x, y)$ that satisfy this equation and then sum the values of $x+y$ for each pair in set $C$.
Since $2023$ is an odd number, and $x^2$, the square of any natural number, is even when $x$ is even and odd when $x$ is odd, we can determine that for the left-hand side of the equation to be odd (thus equal to $2023$), $x$ must be odd since the right-hand side of the equation ($2^y$) is always even as it represents a power of two.
Also, $2023$ can be factored into prime factors to further analyze the possible solutions:
$2023 = 7 \times 17 \times 17$
Thus, allowing us to rewrite the equation as:
$x^2 - 2^y = 7 \times 17^2$
The next step is to check for potential values of $x$ that would fit the equation, keeping in mind that $x$ must be odd. We can try to express $x^2$ as $7 \times 17^2$ plus a power of $2$, recognizing that we are looking for the decomposition of the form:
$x^2 = 7 \times 17^2 + 2^y$
By examining the powers of $2$ and keeping in mind that they grow very quickly, we can reason that $y$ cannot be very large because $x^2$ must not exceed $2023$ by a large margin.
Let's start by trying the lowest values for $y$ since that would make $2^y$ small and $x$ has a better chance of being a natural number:
- For $y=1$:
$x^2 = 2023 + 2^1 = 2023 + 2 = 2025$
Surprisingly, we find a perfect square since $45^2 = 2025$. Therefore, $(x, y) = (45, 1)$ is one solution.
- For $y=2$ or higher:
$2^y$ becomes at least $4$ and increases exponentially, so $x^2$ must be at least $2027$ or higher in such cases. There's no natural number between $45$ and $46$, and $46^2$ far exceeds the target (2116), making it impossible for $x^2$ to be less than $2116$ for any larger $y$.
Hence, it appears there is only one possible solution: $(x, y) = (45, 1)$.
Therefore, the sum $\sum_\limits{(x, y) \in C}(x+y)$ for this set will consist of only this one pair:
$\sum_\limits{(x, y) \in C}(x+y) = 45 + 1 = 46$
So the answer is $46$.
Let $a=3 \sqrt{2}$ and $b=\frac{1}{5^{1 / 6} \sqrt{6}}$. If $x, y \in \mathbb{R}$ are such that
$ \begin{aligned} & 3 x+2 y=\log _a(18)^{\frac{5}{4}} \quad \text { and } \\ & 2 x-y=\log _b(\sqrt{1080}), \end{aligned} $
then $4 x+5 y$ is equal to __________.
Explanation:
Given:
$ a = 3\sqrt{2} \Rightarrow a^2 = 18 $
Simplification of $\log_a(18)^{\frac{5}{4}}$
Since $a^2 = 18$, we have:
$ \log_a(18) = \log_a(a^2) = 2 $
Therefore:
$ \log_a(18)^{\frac{5}{4}} = \frac{5}{4} \log_a(18) = \frac{5}{4} \cdot 2 = \frac{5}{2} $
So the equation becomes:
$ 3x + 2y = \frac{5}{2} $
Simplification of $\log_b(\sqrt{1080})$
Given:
$ 1080 = 2^3 \cdot 3^3 \cdot 5 = 6^3 \cdot 5 $
$ b = \frac{1}{5^{1/6} \sqrt{6}} $
$ \Rightarrow \frac{1}{b} = 5^{1/6} \sqrt{6} $
$ \Rightarrow 1080^{1/6} = 5^{1/6} \cdot 6^{1/2} = \frac{1}{b} $
Taking the square root of both sides:
$ \sqrt{1080} = \frac{1}{b^3} $
Thus:
$ \log_b(\sqrt{1080}) = \log_b\left(\frac{1}{b^3}\right) = \log_b(b^{-3}) = -3 $
So the second equation becomes:
$ 2x - y = -3 $
Solving the System of Equations
Now we have:
$ 3x + 2y = \frac{5}{2} $
$ 2x - y = -3 $
Multiply the second equation by 2:
$ 4x - 2y = -6 $
Add this to the first equation:
$ 3x + 2y + 4x - 2y = \frac{5}{2} - 6 $
$ 7x = \frac{5}{2} - 6 $
$ 7x = \frac{5}{2} - \frac{12}{2} $
$ 7x = \frac{5 - 12}{2} $
$ 7x = -\frac{7}{2} $
$ x = -\frac{1}{2} $
Substitute $x$ back into $2x - y = -3$:
$ 2\left(-\frac{1}{2}\right) - y = -3 $
$ -1 - y = -3 $
$ -y = -2 $
$ y = 2 $
Finding $4x + 5y$
$ 4x + 5y = 4 \left(-\frac{1}{2}\right) + 5 \cdot 2 $
$ = -2 + 10 $
$ = 8 $
Thus, the value of $4x + 5y$ is $\boxed{8}$.
Let $[\alpha]$ denote the greatest integer $\leq \alpha$. Then $[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots+[\sqrt{120}]$ is equal to __________
Explanation:
The number of points, where the curve $f(x)=\mathrm{e}^{8 x}-\mathrm{e}^{6 x}-3 \mathrm{e}^{4 x}-\mathrm{e}^{2 x}+1, x \in \mathbb{R}$ cuts $x$-axis, is equal to _________.
Explanation:
$f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1$ intersects the x-axis
where $f(x) = 0$. Setting $f(x)$ equal to zero gives us :
$e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0.$
Let $t = e^{2x}$. The equation now becomes :
$t^4 - t^3 - 3t^2 - t + 1 = 0.$
Dividing by $t^2$ and rearranging the equation gives :
$t^2 - t - 3 - \frac{1}{t} + \frac{1}{t^2} = 0.$
If we let $y = t + \frac{1}{t}$, we get :
$y^2 - y - 5 = 0.$
This quadratic equation in y can be solved using the quadratic formula to give two roots :
$y = \frac{1 \pm \sqrt{21}}{2}.$
Since $y = t + \frac{1}{t}$, and $t > 0$ (as $t = e^{2x}$), we must choose the root where $y > 2$. Thus, we take $y = \frac{1 + \sqrt{21}}{2}$.
So, we have :
$t + \frac{1}{t} = \frac{1 + \sqrt{21}}{2}.$
Solving for $t$ gives us :
$t^2 - yt + 1 = 0,$
or
$t = \frac{y \pm \sqrt{y^2 - 4}}{2}.$
Substituting $y = \frac{1 + \sqrt{21}}{2}$ into the formula gives :
$t = \frac{\frac{1 + \sqrt{21}}{2} \pm \sqrt{\left(\frac{1 + \sqrt{21}}{2}\right)^2 - 4}}{2}.$
This quadratic formula for $t$ yields two possible values (t = 2.37 and t = 0.42). Both of them are positive, thus both are acceptable values for $t = e^{2x}$.
Finally, to get $x$, take the natural log of both sides and divide by 2 :
$x = \frac{1}{2}\ln(t).$
This gives us two solutions for $x$, corresponding to the two positive solutions for $t$. So, the number of points where the curve $f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1$ intersects the x-axis is 2.
If $a$ and $b$ are the roots of the equation $x^{2}-7 x-1=0$, then the value of $\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}$ is equal to _____________.
Explanation:
$ \begin{aligned} & x^2-7 x-1=0 \\\\ & \Rightarrow a^2-7 a-1=0 \Rightarrow a^2-1=7 a .........(i) \end{aligned} $
On squaring both sides, we get $a^4+1=51 a^2$
Similarly, $b^4+1=51 b^2$ ...........(ii)
$ \text { Now, } \frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}=\frac{a^{17}\left(a^4+1\right)+b^{17}\left(b^4+1\right)}{a^{19}+b^{19}} $
$ \begin{aligned} & =\frac{a^{17}\left(51 a^2\right)+b^{17}\left(51 b^2\right)}{a^{19}+b^{19}} \quad[\because \text { From Eq. (i) and (ii) }] \\\\ & =\frac{51\left[a^{19}+b^{19}\right]}{a^{19}+b^{19}}=51 \end{aligned} $
Let m and $\mathrm{n}$ be the numbers of real roots of the quadratic equations $x^{2}-12 x+[x]+31=0$ and $x^{2}-5|x+2|-4=0$ respectively, where $[x]$ denotes the greatest integer $\leq x$. Then $\mathrm{m}^{2}+\mathrm{mn}+\mathrm{n}^{2}$ is equal to __________.
Explanation:
$ \begin{aligned} & \Rightarrow\{x\}-x=x^2-12 x+31 \\\\ & \Rightarrow\{x\}=x^2-11 x+31 \end{aligned} $
So, $0 \leq x^2-11 x+31<1$
$ \begin{aligned} & \Rightarrow x^2-11 x+30 \leq 0 \\\\ & \Rightarrow(x-5)(x-6)<0 \\\\ & \Rightarrow x \in(5,6) \\\\ & \therefore[x]=5 \end{aligned} $
$ \begin{aligned} & \therefore x^2-12 x+5+31=0 \\\\ & \Rightarrow x^2-12 x+36=0 \\\\ & \Rightarrow(x-6)^2=0 \Rightarrow x=6 \end{aligned} $
Hence, $x \in \phi$
$ (\because x \in(5,6)) $
$ \therefore m=0 $
Another equation is $x^2-5[x+2]-4=0$
Case I : $x \geq-2$
$ x^2-5 x-14=0 \Rightarrow x=7,-2 $
Case II : $x<-2$
$ \begin{aligned} & x^2+5 x+6=0 \Rightarrow x=-3-2 \\\\ & \therefore x \in\{-3,-2,7\} \end{aligned} $
$ \therefore n=3 $
Hence, $m^2+m x+n^2=0+0+9=9$
have a common real root is $\frac{3}{\sqrt{2 \beta}}$ then $\beta$ is equal to ___________.
Explanation:
${x^2} - 5\alpha x + 1 = 0$ ..... (1)
${x^2} - \alpha x - 5 = 0$ ...... (2)
have a common root.
Subtracting (1) with (2) we'll get $x = {6 \over {4\alpha }}$
Substituting in (1)
${{36} \over {16{\alpha ^2}}} - {{30} \over 4} + 1 = 0$
$ \Rightarrow {\alpha ^2} = {9 \over {26}}$
$\alpha = {3 \over {\sqrt {2 \times 13} }}$
$\therefore$ $\beta = 13$
Let $\alpha_1,\alpha_2,....,\alpha_7$ be the roots of the equation ${x^7} + 3{x^5} - 13{x^3} - 15x = 0$ and $|{\alpha _1}| \ge |{\alpha _2}| \ge \,...\, \ge \,|{\alpha _7}|$. Then $\alpha_1\alpha_2-\alpha_3\alpha_4+\alpha_5\alpha_6$ is equal to _________.
Explanation:
${x^7} + 3{x^5} - 13{x^3} - 15x = 0$
$x({x^6} + 3{x^4} - 13{x^2} - 15) = 0$
$x = 0 = {\alpha _7}$
Let ${x^2} = t$
${t^3} + 3{t^2} - 13t - 15 = 0$
$(t + 1)(t + 5)(t - 3) = 0$
$t = {x^2} = - 1, - 5,3$
$x\, = \, \pm \,i, \pm \,\sqrt 5 i, \pm \,\sqrt 3 $
${\alpha _1},{\alpha _2} = \pm \,\sqrt 5 i,{\alpha _3},{\alpha _4} = \pm \,\sqrt 3 ,{\alpha _5},{\alpha _6} = \pm \,i$
${\alpha _1}{\alpha _2} - {\alpha _3}{\alpha _4} + {\alpha _5}{\alpha _6} = 5 + 3 + 1 = 9$
Let $\alpha \in\mathbb{R}$ and let $\alpha,\beta$ be the roots of the equation ${x^2} + {60^{{1 \over 4}}}x + a = 0$. If ${\alpha ^4} + {\beta ^4} = - 30$, then the product of all possible values of $a$ is ____________.
Explanation:
$ \therefore \alpha+\beta=-60^{\frac{1}{4}}, \alpha \beta=a $
Now $\alpha^{4}+\beta^{4}=-30$
$\Rightarrow\left(\alpha^{2}+\beta^{2}\right)^{2}-2 a^{2}=-30$
$\Rightarrow\left[(\alpha+\beta)^{2}-2 a\right]^{2}-2 a^{2}=-30$
$\Rightarrow\left(60^{\frac{1}{2}}-2 a\right)^{2}-2 a^{2}=-30$
$\Rightarrow 60+4 a^{2}-4 \cdot 60^{\frac{1}{2}} a-2 a^{2}+30=0$
$\Rightarrow 2 a^{2}-8 \sqrt{15} a+90=0$
Product of value of $a=45$
Let $\lambda \in \mathbb{R}$ and let the equation E be $|x{|^2} - 2|x| + |\lambda - 3| = 0$. Then the largest element in the set S = {$x+\lambda:x$ is an integer solution of E} is ______
Explanation:
$|\lambda-3| \leq 1$
$-1 \leq \lambda-3 \leq 1$
$2 \leq \lambda \leq 4$
$|x|=\frac{2 \pm \sqrt{4-4|\lambda-3|}}{2}$
$=1 \pm \sqrt{1-|\lambda-3|}$
$x_{\text {largest }}=1+1=2$, when $\lambda=3$
Largest element of $S=2+3=5$
Let $\alpha, \beta(\alpha>\beta)$ be the roots of the quadratic equation $x^{2}-x-4=0 .$ If $P_{n}=\alpha^{n}-\beta^{n}$, $n \in \mathrm{N}$, then $\frac{P_{15} P_{16}-P_{14} P_{16}-P_{15}^{2}+P_{14} P_{15}}{P_{13} P_{14}}$ is equal to __________.
Explanation:
$\alpha$ and $\beta$ are the roots of the quadratic equation ${x^2} - x - 4 = 0$.
$\therefore$ $\alpha$ and $\beta$ are satisfy the given equation.
${\alpha ^2} - \alpha - 4 = 0$
$ \Rightarrow {\alpha ^{n + 1}} - {\alpha ^n} - 4{\alpha ^{n - 1}} = 0$ ...... (1)
and ${\beta ^2} - \beta - 4 = 0$
$ \Rightarrow {\beta ^{n + 1}} - {\beta ^n} - 4{\beta ^{n - 1}} = 0$ ...... (2)
Subtracting (2) from (1), we get,
$({\alpha ^{n + 1}} - {\beta ^{n + 1}}) - ({\alpha ^n} - {\beta ^n}) - 4({\alpha ^{n - 1}} - {\beta ^{n - 1}}) = 0$
$ \Rightarrow {P_{n + 1}} - {P_n} - 4{P_{n - 1}} = 0$
$ \Rightarrow {P_{n + 1}} = {P_n} + 4{P_{n - 1}}$
$ \Rightarrow {P_{n + 1}} - {P_n} = 4{P_{n - 1}}$
For $n = 14$, ${P_{15}} - {P_{14}} = 4{P_{13}}$
For $n = 15$, ${P_{16}} - {P_{15}} = 4{P_{14}}$
Now, ${{{P_{15}}{P_{16}} - {P_{14}}{P_{16}} - P_{15}^2 + {P_{14}}{P_{15}}} \over {{P_{13}}{P_{14}}}}$
$ = {{{P_{16}}({P_{15}} - {P_{14}}) - {P_{15}}({P_{15}} - {P_{14}})} \over {{P_{13}}{P_{14}}}}$
$ = {{({P_{15}} - {P_{14}})({P_{16}} - {P_{15}})} \over {{P_{13}}{P_{14}}}}$
$ = {{(4{P_{13}})(4{P_{14}})} \over {{P_{13}}{P_{14}}}}$
$ = 16$
The sum of all real values of $x$ for which $\frac{3 x^{2}-9 x+17}{x^{2}+3 x+10}=\frac{5 x^{2}-7 x+19}{3 x^{2}+5 x+12}$ is equal to __________.
Explanation:
${{3{x^2} - 9x + 17} \over {{x^2} + 3x + 10}} = {{5{x^2} - 7x + 19} \over {3{x^2} + 5x + 12}}$
$ \Rightarrow {{3{x^2} - 9x + 17} \over {5{x^2} - 7x + 19}} = {{{x^2} + 3x + 10} \over {3{x^2} + 5x + 12}}$
${{ - 2{x^2} - 2x - 2} \over {5{x^2} - 7x + 19}} = {{ - 2{x^2} - 2x - 2} \over {3{x^2} + 5x + 12}}$
Either ${x^2} + x + 1 = 0$ or No real roots
$ \Rightarrow 5{x^2} - 7x + 19 = 3{x^2} + 5x + 12$
$2{x^2} - 12x + 7 = 0$
sum of roots = 6
If for some $\mathrm{p}, \mathrm{q}, \mathrm{r} \in \mathbf{R}$, not all have same sign, one of the roots of the equation $\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right) x^{2}-2 \mathrm{q}(\mathrm{p}+\mathrm{r}) x+\mathrm{q}^{2}+\mathrm{r}^{2}=0$ is also a root of the equation $x^{2}+2 x-8=0$, then $\frac{\mathrm{q}^{2}+\mathrm{r}^{2}}{\mathrm{p}^{2}}$ is equal to ____________,
Explanation:
Let roots of
$\therefore$ $\alpha$ + $\beta$ > 0 and $\alpha$$\beta$ > 0
Also, it has a common root with ${x^2} + 2x - 8 = 0$
$\therefore$ The common root between above two equations is 4.
$ \Rightarrow 16({p^2} + {q^2}) - 8q(p + r) + {q^2} + {r^2} = 0$
$ \Rightarrow (16{p^2} - 8pq + {q^2}) + (16{q^2} - 8qr + {r^2}) = 0$
$ \Rightarrow {(4p - q)^2} + {(4q - r)^2} = 0$
$ \Rightarrow q = 4p$ and $r = 16p$
$\therefore$ ${{{q^2} + {r^2}} \over {{p^2}}} = {{16{p^2} + 256{p^2}} \over {{p^2}}} = 272$
The number of distinct real roots of the equation $x^{5}\left(x^{3}-x^{2}-x+1\right)+x\left(3 x^{3}-4 x^{2}-2 x+4\right)-1=0$ is ______________.
Explanation:
${x^8} - {x^7} - {x^6} + {x^5} + 3{x^4} - 4{x^3} - 2{x^2} + 4x - 1 = 0$
$ \Rightarrow {x^7}(x - 1) - {x^5}(x - 1) + 3{x^3}(x - 1) - x({x^2} - 1) + 2x(1 - x) + (x - 1) = 0$
$ \Rightarrow (x - 1)({x^7} - {x^5} + 3{x^3} - x(x + 1) - 2x + 1) = 0$
$ \Rightarrow (x - 1)({x^7} - {x^5} + 3{x^3} - {x^2} - 3x + 1) = 0$
$ \Rightarrow (x - 1)({x^5}({x^2} - 1) + 3x({x^2} - 1) - 1({x^2} - 1)) = 0$
$ \Rightarrow (x - 1)({x^2} - 1)({x^5} + 3x - 1) = 0$
$\therefore$ $x = \, \pm \,1$ are roots of above equation and ${x^5} + 3x - 1$ is a monotonic term hence vanishs at exactly one value of x other than 1 or $-$1.
$\therefore$ 3 real roots.
The number of real solutions of the equation ${e^{4x}} + 4{e^{3x}} - 58{e^{2x}} + 4{e^x} + 1 = 0$ is ___________.
Explanation:
Dividing by e2x
${e^{2x}} + 4{e^x} - 58 + 4{e^{ - x}} + {e^{ - 2x}} = 0$
$ \Rightarrow {({e^x} + {e^{ - x}})^2} + 4({e^x} + {e^{ - x}}) - 60 = 0$
Let ${e^x} + {e^{ - x}} = t \in [2,\infty )$
$ \Rightarrow {t^2} + 4t - 60 = 0$
$ \Rightarrow t = 6$ is only possible solution
${e^x} + {e^{ - x}} = 6 \Rightarrow {e^{2x}} - 6{e^x} + 1 = 0$
Let ${e^x} = p$,
${p^2} - 6p + 1 = 0$
$ \Rightarrow p = {{3 + \sqrt 5 } \over 2}$ or ${{3 - \sqrt 5 } \over 2}$
So $x = \ln \left( {{{3 + \sqrt 5 } \over 2}} \right)$ or $\ln \left( {{{3 - \sqrt 5 } \over 2}} \right)$
Let $\alpha$, $\beta$ be the roots of the equation ${x^2} - 4\lambda x + 5 = 0$ and $\alpha$, $\gamma$ be the roots of the equation ${x^2} - \left( {3\sqrt 2 + 2\sqrt 3 } \right)x + 7 + 3\lambda \sqrt 3 = 0$, $\lambda$ > 0. If $\beta + \gamma = 3\sqrt 2 $, then ${(\alpha + 2\beta + \gamma )^2}$ is equal to __________.
Explanation:
$\because$ $\alpha$, $\beta$ are roots of x2 $-$ 4$\lambda$x + 5 = 0
$\therefore$ $\alpha$ + $\beta$ = 4$\lambda$ and $\alpha$$\beta$ = 5
Also, $\alpha$, $\gamma$ are roots of
${x^2} - (3\sqrt 2 + 2\sqrt 3 )x + 7 + 3\sqrt 3 \lambda = 0,\,\lambda > 0$
$\therefore$ $\alpha + \gamma = 3\sqrt 2 + 2\sqrt 3 $, $\alpha \gamma = 7 + 3\sqrt 3 \lambda $
$\because$ $\alpha$ is common root
$\therefore$ ${\alpha ^2} - 4\lambda \,\,\alpha + 5 = 0$ ....... (i)
and ${\alpha ^2} - (3\sqrt 2 + 2\sqrt 3 )\alpha + 7 + 3\sqrt 3 \lambda = 0$ ...... (ii)
From (i) - (ii) : we get $\alpha = {{2 + 3\sqrt 3 \lambda } \over {3\sqrt 2 + 2\sqrt 3 - 4\lambda }}$
$\because$ $\beta + \gamma = 3\sqrt 2 $
$\therefore$ $4\lambda + 3\sqrt 2 + 2\sqrt 3 - 2\alpha = 3\sqrt 2 $
$ \Rightarrow 3\sqrt 2 = 4\lambda + 3\sqrt 2 + 2\sqrt 3 - {{4 + 6\sqrt 3 \lambda } \over {3\sqrt 2 + 2\sqrt 3 - 4\lambda }}$
$ \Rightarrow 8{\lambda ^2} + 3(\sqrt 3 + 2\sqrt 2 )\lambda - 4 - 3\sqrt 6 = 0$
$\therefore$ $\lambda = {{6\sqrt 2 - 3\sqrt 2 \pm \sqrt {9(11 - 4\sqrt 6 ) + 32(4 + 3\sqrt 6 )} } \over {16}}$
$\therefore$ $\lambda = \sqrt 2 $
$\therefore$ ${(\alpha + 2\beta + \gamma )^2} = {(\alpha + \beta + \beta + \gamma )^2}$
$ = {(4\sqrt 2 + 3\sqrt 2 )^2}$
$ = {(7\sqrt 2 )^2} = 98$
If the sum of all the roots of the equation
${e^{2x}} - 11{e^x} - 45{e^{ - x}} + {{81} \over 2} = 0$ is ${\log _e}p$, then p is equal to ____________.
Explanation:
$e^{2 x}-11 e^x-45 e^{-x}+\frac{81}{2}=0 $
$\Rightarrow 2 e^{3 x}-22 e^{2 x}-90+81 e^x=0 $
$\Rightarrow 2\left(e^x\right)^3-22\left(e^x\right)^2+81 e^x-90=0$
Let $ e^x=y$
$ \Rightarrow 2 y^3-22 y^2+81 y-90=0 $
Product of roots $\left(y_1, y_2, y_3\right)$
$ y_1 \cdot y_2 \cdot y_3=\frac{-(-90)}{2}=45 $
Let $x_1, x_2$, and $x_3$ be roots of given equation
$\Rightarrow e^{x_1} \cdot e^{x_2} \cdot e^{x_3} = 45 $
$\Rightarrow e^{x_1+x_2+x_3} =45$
$\Rightarrow x_1+x_2+x_3 =\log _e 45=\log _e p $
$\Rightarrow p = 45$
Let p and q be two real numbers such that p + q = 3 and p4 + q4 = 369. Then ${\left( {{1 \over p} + {1 \over q}} \right)^{ - 2}}$ is equal to _________.
Explanation:
$\because$ $p + q = 3$ ...... (i)
and ${p^4} + {q^4} = 369$ ...... (ii)
${\{ {(p + q)^2} - 2pq\} ^2} - 2{p^2}{q^2} = 369$
or ${(9 - 2pq)^2} - 2{(pq)^2} = 369$
or ${(pq)^2} - 18pq - 144 = 0$
$\therefore$ $pq = - 6$ or 24
But $pq = 24$ is not possible
$\therefore$ $pq = - 6$
Hence, ${\left( {{1 \over p} + {1 \over q}} \right)^{ - 2}} = {\left( {{{pq} \over {p + q}}} \right)^2} = {( - 2)^2} = 4$
The sum of the cubes of all the roots of the equation
${x^4} - 3{x^3} - 2{x^2} + 3x + 1 = 0$ is _________.
Explanation:
${x^4} - 3{x^3} - {x^2} - {x^2} + 3x + 1 = 0$
$({x^2} - 1)({x^2} - 3x - 1) = 0$
Let the root of ${x^2} - 3x - 1 = 0$ be $\alpha$ and $\beta$ and other two roots of given equation are 1 and $-$1
So sum of cubes of roots
$ = {1^3} + {( - 1)^3} + {\alpha ^3} + {\beta ^3}$
$ = {(\alpha + \beta )^3} - 3\alpha \beta (\alpha + \beta )$
$ = {(3)^3} - 3( - 1)(3)$
$ = 36$
$ x^{\left(16\left(\log _{5} x\right)^{3}-68 \log _{5} x\right)}=5^{-16} $
is __________.
Explanation:
$\left(16\left(\log _{5} x\right)^{3}-68\left(\log _{5} x\right)\right)\left(\log _{5} x\right)=-16$
Let $\left(\log _{5} x\right)=t$
$16 t^{4}-68 t^{2}+16=0$
$ \Rightarrow $ $4 t^{4}-16 t^{2}-t^{2}+4=0$
$ \Rightarrow $ $\left(4 t^{2}-1\right)\left(t^{2}-4\right)=0$
$ \Rightarrow $ $t=\pm \frac{1}{2}, \pm 2$
So $\log _{5} x=\pm \frac{1}{2}$ or $\pm 2$
$\Rightarrow x=5^{\frac{1}{2}}, 5^{\frac{-1}{2}}, 5^{2}, 5^{-2}$
$f(k) = - {2 \over k}$ for k = 2, 3, 4, 5. Then the value of 52 $-$ 10f(10) is equal to :
Explanation:
put x = 0
we get $\lambda = {1 \over {60}}$
Now, put $\lambda$ in equation (1)
$ \Rightarrow kf(k) + 2 = {1 \over {60}}(x - 2)(x - 3)(x - 4)(x - 5)$
Put x = 10
$ \Rightarrow 10f(10) + 2 = {1 \over {60}}(8)(7)(6)(5)$
$ \Rightarrow 52 - 10f(10) = 52 - 26 = 26$
Explanation:
$\alpha$2 $-$ $\alpha$ + 2$\lambda$ = 0 ...... (2)
(1) $-$ 3(2) gives
$-$7$\alpha$ + 21$\lambda$ = 0 $\Rightarrow$ $\alpha$ = 3$\lambda$
Put $\alpha$ = 3$\lambda$ in equation (1) we get
9$\lambda$2 $-$ 3$\lambda$ + 2$\lambda$ $-$ 0
9$\lambda$2 = $\lambda$ $\Rightarrow$ $\lambda$ = ${1 \over 9}$ as $\lambda$ $\ne$ 0
Now, $\alpha$ = 3$\lambda$ $\Rightarrow$ $\lambda$ = ${1 \over 3}$
$\alpha$ + $\beta$ = 1 $\Rightarrow$ $\beta$ = 2/3
$\alpha$ + $\gamma$ = ${10 \over 3}$ $\Rightarrow$ $\gamma$ = 3
${{\beta \gamma } \over \lambda } = {{{2 \over 3} \times 3} \over {{1 \over 9}}} = 18$
Explanation:
$x \in R - \{ 1,2\} $
$ \Rightarrow k(2x - 4 - x + 1) = 2({x^2} - 3x + 2)$
$ \Rightarrow k(x - 3) = 2({x^2} - 3x + 2)$
for x $\ne$ 3, $k = 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right)$
$x - 3 + {2 \over {x - 3}} \ge 2\sqrt 2 ,\forall x > 3$
& $x - 3 + {2 \over {x - 3}} \le - 2\sqrt 2 ,\forall x < - 3$
$ \Rightarrow 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right) \in \left( { - \infty ,6 - 4\sqrt 2 } \right] \cup \left[ {6 + 4\sqrt 2 ,\infty } \right)$
for no real roots
$k \in (6 - 4\sqrt 2 ,6 + 4\sqrt 2 ) - \{ 0\} $
Integral k$\in${1, 2 ..... 11}
Sum of k = 66
Explanation:
$ \Rightarrow {t^2} - t - 4 - {1 \over t} + {1 \over {{t^2}}} = 0$
$ \Rightarrow {\alpha ^2} - \alpha - 6 = 0,\alpha = t + {1 \over t} \ge 2$
$ \Rightarrow \alpha = 3, - 2$ (reject)
$ \Rightarrow t + {1 \over t} = 3$
$\Rightarrow$ The number of real roots = 2
Explanation:
$ \Rightarrow $ a2 + b2 + c2 + 2(ab + bc + ca) = 1
$ \Rightarrow $ a2 + b2 + c2 = – 3 ….(i)
$ \Rightarrow $ ab + bc + ca = 2 ….(ii)
Squaring of equation (ii),
$ \Rightarrow $ a2b2 + b2c2 + c2a2 + 2(ab2c + bc2a + ca2b) = 4
$ \Rightarrow $ a2b2 + b2c2 + c2a2 + 2abc(a + b + c) = 4
$ \Rightarrow $ a2b2 + b2c2 + c2a2 + 6 = 4
$ \Rightarrow $ a2b2 + b2c2 + c2a2 = – 2 ….(iii)
Squaring of equation (i),
$ \Rightarrow $ a4 + b4 + c4 + 2(a2b2 + b2c2 + c2a2) = 9
$ \Rightarrow $ a4 + b4 + c4 – 4 = 9
$ \Rightarrow $ a4 + b4 + c4 = 13
Explanation:
& ${P_n} = {\alpha ^n} - {\beta ^n}$ (Given)
Now, ${{{{P_{17}}{P_{20}} + 5\sqrt 2 {P_{17}}{P_{19}}} \over {{P_{18}}{P_{19}} + 5\sqrt 2 P_{18}^2}}}$ = ${{{{P_{17}}({P_{20}} + 5\sqrt 2 {P_{19}})} \over {{P_{18}}({P_{19}} + 5\sqrt 2 P_{18}^{})}}}$
${{{P_{17}}({\alpha ^{20}} - {\beta ^{20}} + 5\sqrt 2 ({\alpha ^{19}} - {\beta ^{19}}))} \over {{P_{18}}({\alpha ^{19}} - {\beta ^{19}} + 5\sqrt 2 ({\alpha ^{18}} - {\beta ^{18}}))}}$
${{{P_{17}}({\alpha ^{19}}(\alpha + 5\sqrt 2 ) - {\beta ^{19}}(\beta + 5\sqrt 2 ))} \over {{P_{18}}({\alpha ^{18}}(\alpha + 5\sqrt 2 ) - {\beta ^{18}}(\beta + 5\sqrt 2 ))}}$
Since, $\alpha + 5\sqrt 2 = - 10/\alpha $ ..... (1)
& $\beta + 5\sqrt 2 = - 10/\beta $ ....... (2)
Now, put there values in above expression $ = - {{10{P_{17}}{P_{18}}} \over { - 10{P_{18}}{P_{17}}}} = 1$
Explanation:
$ \therefore $ Quadratic equation with roots $\alpha$, $\beta$ is x2 $-$ x $-$ 1 = 0
$ \Rightarrow $ $\alpha$2 = $\alpha$ + 1
Multiplying both sides by $\alpha$n$-$1
$\alpha$n$+$1 = $\alpha$n + $\alpha$n$-$1 ......(1)
Similarly,
$\beta$n + 1 = $\beta$n + $\beta$n + 1 ..... (2)
Adding (1) & (2)
${\alpha ^{n + 1}} + {\beta ^{n + 1}} = ({\alpha ^n} + {\beta ^n}) + ({\alpha ^{n - 1}} + {\beta ^{n - 1}})$
$ \Rightarrow $ Pn+1 = Pn + Pn$-$1
$ \Rightarrow $ 29 = Pn + 11 (Given, Pn + 1 = 29, Pn $-$ 1 = 11)
$ \Rightarrow $ Pn = 18
$ \therefore $ $P_n^2$ = 182 = 324
Explanation:
x = 1 satisfying the equation
$ \therefore $ x $-$ 1 is factor of
x3 $-$ 2x2 + 2x $-$ 1
= (x $-$ 1) (x2 $-$ x + 1) = 0
x = 1, ${{1 + i\sqrt 3 } \over 2},{{1 - i\sqrt 3 } \over 2}$
x = 1, $-$ $\omega$2, $-$$\omega$
Sum of 162th power of roots
= (1)162 + ($-$$\omega$2)162 + ($-$$\omega$)162
= 1 + ($\omega$)324 + ($\omega$)162
= 1 + 1 + 1 = 3
Explanation:
${(x + 1)^2} + (x - 5) = {{27} \over 4}$
$ \Rightarrow {x^2} + 3x - 4 = {{27} \over 4}$
$ \Rightarrow {x^2} + 3x - {{43} \over 4} = 0$
$ \Rightarrow 4{x^2} + 12x - 43 = 0$
$x = {{ - 12 \pm \sqrt {144 + 688} } \over 8}$
$x = {{ - 12 \pm \sqrt {832} } \over 8} = {{ - 12 \pm 28.8} \over 8}$
$ = {{ - 3 + 7.2} \over 2}$
$ = {{ - 3 + 7.2} \over 2},{{ - 3 - 7.2} \over 2}$
= 2.1, -5.1 [ both are rejected as x should be > 5 ]
(Therefore no solution)
For $x \le 5$
${(x + 1)^2} - (x - 5) = {{27} \over 4}$
${x^2} + x + 6 - {{27} \over 4} = 0$
$4{x^2} + 4x - 3 = 0$
$x = {{ - 4 \pm \sqrt {16 + 48} } \over 8}$
$x = {{ - 4 \pm 8} \over 8} \Rightarrow x = - {{12} \over 8},{4 \over 8}$
$ \therefore $ So, the equation have two real roots.
Explanation:
$3{x^2} - 4\left| {{x^2} - 1} \right| + x - 1 = 0$ .... (i)
For $-$1 $\le$ x $\le$ 1 i.e., x$\in$[$-$1, 1]
From Eq. (i), we get
$3{x^2} - 4( - {x^2} + 1) + x - 1 = 0$
$ \Rightarrow 3{x^2} + 4{x^2} - 4 + x - 1 = 0$
$ \Rightarrow 7{x^2} + x - 5 = 0$
$ \Rightarrow x = {{ - 1 \pm \sqrt {1 + 140} } \over {(2 \times 7)}}$
Here, both values of x are acceptable.
For | x | > | i.e. x $\in$($-$ $\infty$, $-$1) $\cup$ (1, $\infty$)
From Eq. (i), we get
$3{x^2} - 4({x^2} - 1) + x - 1 = 0$
$ \Rightarrow {x^2} - x - 3 = 0$
$ \Rightarrow x = {{1 \pm \sqrt {1 + 12} } \over 2}$
Again here, both values of x are acceptable.
Hence, total number of solutions is 4.
2x2 + (a – 10)x + ${{33} \over 2}$ = 2a has real roots is
Explanation:
(a – 10)2 – 4$\left( {{{33} \over 2} - 2a} \right).2$ $ \ge $ 0
$ \Rightarrow $ a2 + 100 – 20a – 132 + 16a $ \ge $ 0
$ \Rightarrow $ a 2 – 4a – 32 $ \ge $ 0
$ \Rightarrow $ (a – 8) (a + 4) $ \ge $ 0
$ \Rightarrow $ a $ \le $ -4 $ \cup $ a $ \ge $ 8
$ \therefore $ least positive a = 8
Explanation:
$\sqrt 3 a\cos x + 2b\sin x = c$
$ \therefore $ $\sqrt 3 a\cos \alpha + 2b\sin \alpha = c$ ... (i)
and $\sqrt 3 a\cos \beta + 2b\sin \beta = c$ ... (ii)
On subtracting Eq. (ii) from Eq. (i), we get
$\sqrt 3 a(\cos \alpha - \cos \beta ) + 2b(\sin \alpha - \sin \beta ) = 0$
$ \Rightarrow \sqrt 3 a\left( { - 2\sin \left( {{{\alpha + \beta } \over 2}} \right)} \right)\sin \left( {{{\alpha - \beta } \over 2}} \right) + 2b\left( {2\cos \left( {{{\alpha + \beta } \over 2}} \right)} \right)\sin \left( {{{\alpha - \beta } \over 2}} \right) = 0$
$ \Rightarrow \sqrt 3 a\sin \left( {{{\alpha + \beta } \over 2}} \right) = 2b\cos \left( {{{\alpha + \beta } \over 2}} \right)$
$ \Rightarrow \tan \left( {{{\alpha + \beta } \over 2}} \right) = {{2b} \over {\sqrt 3 a}}$
$ \Rightarrow \tan \left( {{\pi \over 6}} \right) = {{2b} \over {\sqrt 3 a}}$ [$ \because $ $\alpha $ + $\beta $ = ${\pi \over 3}$, given]
$ \Rightarrow {1 \over {\sqrt 3 }} = {{2b} \over {\sqrt 3 a}} \Rightarrow {b \over a} = {1 \over 2}$
$ \Rightarrow {b \over a} = 0.5$
The value of $6 + {\log _{3/2}}\left( {{1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}...} } } } \right)$ is __________.
Explanation:
$6 + {\log _{{3 \over 2}}}\left( {{1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}...} } } } \right)$
Let $\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}} \sqrt {...} } = y$
$\therefore$ $y = \sqrt {4 - {1 \over {3\sqrt 2 }}y} $
$ \Rightarrow {y^2} + {1 \over {3\sqrt 2 }}y - 4 = 0$
$ \Rightarrow 3\sqrt 2 {y^2} + y - 12\sqrt 2 = 0$
$\therefore$ $y = {{ - 1 \pm 17} \over {6\sqrt 2 }}$ or $y = {8 \over {3\sqrt 2 }}$
Now,
$6 + {\log _{{3 \over 2}}}\left( {{1 \over {3\sqrt 2 }}.y} \right) = 6 + {\log _{{3 \over 2}}}\left( {{1 \over {3\sqrt 2 }}.{8 \over {3\sqrt 2 }}} \right)$
$ = 6 + {\log _{{3 \over 2}}}\left( {{4 \over 9}} \right) = 6 + {\log _{{3 \over 2}}}{\left( {{3 \over 2}} \right)^{ - 2}}$
$ = 6 - 2.{\log _{{3 \over 2}}}\left( {{3 \over 2}} \right) = 4$
Explanation:
We have ${{{a^{ - 5}} + {a^{ - 4}} + {a^{ - 3}} + {a^{ - 3}} + {a^{ - 3}} + {a^8} + {a^{10}} + 1} \over 8} \ge 1$
Therefore, the minimum value is 8.
Explanation:
Let $f(x) = {x^4} - 4{x^3} + 12{x^2} + x - 1 = 0$
$f'(x) = 4{x^3} - 12{x^2} + 24x + 1 = 4({x^3} - 3{x^2} + 6x) + 1$
$f''(x) = 12{x^2} - 24x + 24 = 12({x^2} - 2x + 2)$
f''(x) has 0 real roots.
f(x) has maximum two distinct real roots as f(0) = $-$1.
Explanation:
We have ${x^2} - 8kx + 16({k^2} - k + 1) = 0$
$D > 0 \Rightarrow k > 1$ ..... (1)
${{ - b} \over {2a}} > 4 \Rightarrow {{8k} \over 2} > 4$
$ \Rightarrow k > 1$ ..... (2)
Now, $f(4) \ge 0 \Rightarrow 16 - 32k + 16({k^2} - k + 1) \ge 0$
${k^2} - 3k + 2 \ge 0$
$k \le 1 \cup k \ge 2$ ..... (3)
Using Eqs. (1), (2) and (3), we get ${k_{\min }} = 2$.
If roots of the equation $x^2-10 c x-11 d=0$ are $a, b$ and those of $x^2-10 a x-11 b=0$ are $c, d$, then the value of $a+b+c+d$ is $(a, b, c$ and $d$ are distinct numbers)
Explanation:
( $a, b$ ) roots of $x^2-10 c x-11 d=0$
( $c, \mathrm{~d}$ ) roots of $x^2-10 a x-11 b=0$
We know sum of roots
$ \quad \begin{aligned} a+b=10 c \text { and } c+d & =10 a \\ \Rightarrow \quad a+b+c+d & =10(c+a) \end{aligned} $
$ \Rightarrow \quad(b+d)=9(a+c) $
Now, we calculate $(c+a)$
Here product of root
$ \begin{aligned} & & a b=-11 d \text { and } c d & =-11 b \\ \Rightarrow & & a b c d & =121 b d \\ \Rightarrow & & a c & =121 \end{aligned} $
$ \begin{aligned} & \text { Also } \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad a^2-10 c a-11 d=0 \\ &\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, c^2-10 a c-11 b=0 \\ & \Rightarrow \quad \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, c^2+a^2-20 a c-11(b+\mathrm{d})=0 \\ & \Rightarrow\,\,\, c^2+a^2+2 a c-22 a c-11 \times 9(a+c)=0 \\ & \Rightarrow \quad(a+c)^2-22 \times 121-11 \times 9(a+c)=0 \\ & \Rightarrow\,\,\,\,\,\,\,\,\, \quad(a+c)^2-99(a+c)-22 \times 121=0 \end{aligned} $
$ \begin{aligned} a+c & =\frac{99 \pm \sqrt{9801+88 \times 121}}{2} \\ & =\frac{99 \pm \sqrt{9801+10648}}{2} \\ & =\frac{99 \pm \sqrt{20449}}{2}=\frac{99 \pm 143}{2} \\ & =\frac{242}{2}, \frac{-44}{2}=121,-22 \end{aligned} $
Discard the value of $a+c=-22$
$ \begin{aligned}Hence,\,\, a+b+c+d & =121 \times 10 \\ & =1210 \end{aligned} $