Let m and 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 m^2+mn+n^2 is equal to \underline\hspace2cm .

Quadratic Equation and Inequalities PYQ — JEE Main 2023 · 8 Apr · Shift 2 · Q37 with Solution

View written solutionFree

Correct answer: 9

We need to find:

$m$ = number of real roots of $x^2-12x+[x]+31=0$
$n$ = number of real roots of $x^2-5|x+2|-4=0$

Then compute $m^2+mn+n^2.$

1. Finding $m$

We must solve $x^2-12x+[x]+31=0.$

Since $[x]$ is involved, isolate it: $[x]=-x^2+12x-31.$

Now note that $[x]$ is an integer. Therefore the right-hand side must be an integer.

Let $f(x)=-x^2+12x-31.$ Complete the square: $f(x)=-(x^2-12x+36)+5=-(x-6)^2+5.$

So the equation becomes $[x]=5-(x-6)^2.$

Because $[x]$ is an integer, we need $5-(x-6)^2$ to be an integer.

Let $x=n+t,$ where $n=[x]$ and $0\le t<1$ .

Then the equation is $x^2-12x+n+31=0.$ Substitute $x=n+t$ : $(n+t)^2-12(n+t)+n+31=0$ $n^2+2nt+t^2-11n-12t+31=0.$ This is messy, so instead use interval-wise checking.

For $x\in[k,k+1)$ , we have $[x]=k$ . Then the equation becomes $x^2-12x+k+31=0.$ For this root to belong to that interval, solve for each possible $k$ .

But since $[x]=5-(x-6)^2,$ and $[x]=x-t$ , it is more efficient to look at possible integer values of the RHS.

Because $(x-6)^2\ge 0$ , $[x]\le 5.$ Also, since $[x]$ is close to $x$ , roots should lie near where the quadratic is small. Let us directly test intervals $[k,k+1)$ for plausible $k$ .

Interval method

If $x\in[k,k+1)$ , then $[x]=k$ , so $x^2-12x+k+31=0.$ Its roots are

=\frac{12\pm\sqrt{20-4k}}{2} =6\pm\sqrt{5-k}.$$ These are real only if $$5-k\ge 0\quad\Rightarrow\quad k\le 5.$$ Now each root must satisfy $x\in[k,k+1)$. --- ### Check root $x=6+\sqrt{5-k}$ Since $\sqrt{5-k}\ge 0$, this root is at least $6$. But $k\le 5$, so $x\ge 6>k+1$. Hence this root cannot lie in $[k,k+1)$. So only the root $$x=6-\sqrt{5-k}$$ can work. We need $$k\le 6-\sqrt{5-k}<k+1.$$ Let us test integer $k\le 5$ near the expected values. - For $k=1$: $$x=6-\sqrt{4}=4,$$ but $4\notin [1,2)$. Not valid. - For $k=2$: $$x=6-\sqrt{3}\approx 4.268,$$ not in $[2,3)$. Not valid. - For $k=3$: $$x=6-\sqrt{2}\approx 4.586,$$ not in $[3,4)$. Not valid. - For $k=4$: $$x=6-1=5,$$ and $5\notin [4,5)$ since upper end excluded. Not valid. - For $k=5$: $$x=6,$$ and $6\notin [5,6)$. Not valid. Try smaller $k$: - For $k=0$: $$x=6-\sqrt{5}\approx 3.764,$$ not in $[0,1)$. - For negative $k$, the root only gets larger than the interval. So the above direct testing suggests no roots, but let us carefully derive the interval condition properly. For root $x=6-\sqrt{5-k}$ to lie in $[k,k+1)$: ### First inequality: $$6-\sqrt{5-k}\ge k$$ $$6-k\ge \sqrt{5-k}.$$ Since RHS is nonnegative, square: $$(6-k)^2\ge 5-k$$ $$36-12k+k^2\ge 5-k$$ $$k^2-11k+31\ge 0.$$ This is always positive for integer $k$ near the range, so lower bound is okay. ### Second inequality: $$6-\sqrt{5-k}<k+1$$ $$5-k<\sqrt{5-k}.$$ Let $y=\sqrt{5-k}\ge 0$. Then $5-k=y^2$, so condition becomes $$y^2<y$$ $$0<y<1.$$ Hence $$0<\sqrt{5-k}<1$$ $$0<5-k<1.$$ But $k$ is an integer, and there is no integer $k$ satisfying this. Therefore, **no real root** exists. So, $$m=0.$$ --- ## 2. Finding $n$ Solve $$x^2-5|x+2|-4=0.$$ We split into cases. ### Case 1: $x\ge -2$ Then $$|x+2|=x+2.$$ Equation becomes $$x^2-5(x+2)-4=0$$ $$x^2-5x-14=0$$ $$(x-7)(x+2)=0.$$ So roots are $$x=7,\,-2.$$ Both satisfy $x\ge -2$. ### Case 2: $x<-2$ Then $$|x+2|=-(x+2).$$ Equation becomes $$x^2-5(-(x+2))-4=0$$ $$x^2+5x+10-4=0$$ $$x^2+5x+6=0$$ $$(x+2)(x+3)=0.$$ So roots are $$x=-2,\,-3.$$ Among these, only $$x=-3$$ is allowed because this case requires $x<-2$. Thus the real roots are $$x=-3,-2,7.$$ Hence, $$n=3.$$ --- ## 3. Compute the required value We have $$m=0,\quad n=3.$$ Therefore, $$m^2+mn+n^2=0^2+0\cdot 3+3^2=9.$$ --- ## 4. Comparison with stored answer Derived answer: $$9$$ Stored correct answer: $$9$$ They match.

More from Quadratic Equation and Inequalities

1. Finding

m

We must solve

x^2-12x+[x]+31=0.

Since

[x]

is involved, isolate it:

[x]=-x^2+12x-31.

Now note that

[x]

is an integer. Therefore the right-hand side must be an integer.

Let

f(x)=-x^2+12x-31.

Complete the square:

f(x)=-(x^2-12x+36)+5=-(x-6)^2+5.

So the equation becomes

[x]=5-(x-6)^2.

Because

[x]

is an integer, we need

5-(x-6)^2

to be an integer.

Let

x=n+t,

where

n=[x]

and

0\le t<1

Then the equation is

x^2-12x+n+31=0.

Substitute

x=n+t

(n+t)^2-12(n+t)+n+31=0

n^2+2nt+t^2-11n-12t+31=0.

This is messy, so instead use interval-wise checking.

For

x\in[k,k+1)

, we have

[x]=k

. Then the equation becomes

x^2-12x+k+31=0.

For this root to belong to that interval, solve for each possible

k

But since

[x]=5-(x-6)^2,

and

[x]=x-t

, it is more efficient to look at possible integer values of the RHS.

Because

(x-6)^2\ge 0

[x]\le 5.

Also, since

[x]

is close to

x

, roots should lie near where the quadratic is small. Let us directly test intervals

[k,k+1)

for plausible

k

Interval method

x\in[k,k+1)

, then

[x]=k

, so

x^2-12x+k+31=0.

Its roots are

=\frac{12\pm\sqrt{20-4k}}{2} =6\pm\sqrt{5-k}.$$ These are real only if $$5-k\ge 0\quad\Rightarrow\quad k\le 5.$$ Now each root must satisfy $x\in[k,k+1)$. --- ### Check root $x=6+\sqrt{5-k}$ Since $\sqrt{5-k}\ge 0$, this root is at least $6$. But $k\le 5$, so $x\ge 6>k+1$. Hence this root cannot lie in $[k,k+1)$. So only the root $$x=6-\sqrt{5-k}$$ can work. We need $$k\le 6-\sqrt{5-k}<k+1.$$ Let us test integer $k\le 5$ near the expected values. - For $k=1$: $$x=6-\sqrt{4}=4,$$ but $4\notin [1,2)$. Not valid. - For $k=2$: $$x=6-\sqrt{3}\approx 4.268,$$ not in $[2,3)$. Not valid. - For $k=3$: $$x=6-\sqrt{2}\approx 4.586,$$ not in $[3,4)$. Not valid. - For $k=4$: $$x=6-1=5,$$ and $5\notin [4,5)$ since upper end excluded. Not valid. - For $k=5$: $$x=6,$$ and $6\notin [5,6)$. Not valid. Try smaller $k$: - For $k=0$: $$x=6-\sqrt{5}\approx 3.764,$$ not in $[0,1)$. - For negative $k$, the root only gets larger than the interval. So the above direct testing suggests no roots, but let us carefully derive the interval condition properly. For root $x=6-\sqrt{5-k}$ to lie in $[k,k+1)$: ### First inequality: $$6-\sqrt{5-k}\ge k$$ $$6-k\ge \sqrt{5-k}.$$ Since RHS is nonnegative, square: $$(6-k)^2\ge 5-k$$ $$36-12k+k^2\ge 5-k$$ $$k^2-11k+31\ge 0.$$ This is always positive for integer $k$ near the range, so lower bound is okay. ### Second inequality: $$6-\sqrt{5-k}<k+1$$ $$5-k<\sqrt{5-k}.$$ Let $y=\sqrt{5-k}\ge 0$. Then $5-k=y^2$, so condition becomes $$y^2<y$$ $$0<y<1.$$ Hence $$0<\sqrt{5-k}<1$$ $$0<5-k<1.$$ But $k$ is an integer, and there is no integer $k$ satisfying this. Therefore, **no real root** exists. So, $$m=0.$$ --- ## 2. Finding $n$ Solve $$x^2-5|x+2|-4=0.$$ We split into cases. ### Case 1: $x\ge -2$ Then $$|x+2|=x+2.$$ Equation becomes $$x^2-5(x+2)-4=0$$ $$x^2-5x-14=0$$ $$(x-7)(x+2)=0.$$ So roots are $$x=7,\,-2.$$ Both satisfy $x\ge -2$. ### Case 2: $x<-2$ Then $$|x+2|=-(x+2).$$ Equation becomes $$x^2-5(-(x+2))-4=0$$ $$x^2+5x+10-4=0$$ $$x^2+5x+6=0$$ $$(x+2)(x+3)=0.$$ So roots are $$x=-2,\,-3.$$ Among these, only $$x=-3$$ is allowed because this case requires $x<-2$. Thus the real roots are $$x=-3,-2,7.$$ Hence, $$n=3.$$ --- ## 3. Compute the required value We have $$m=0,\quad n=3.$$ Therefore, $$m^2+mn+n^2=0^2+0\cdot 3+3^2=9.$$ --- ## 4. Comparison with stored answer Derived answer: $$9$$ Stored correct answer: $$9$$ They match.

Quadratic Equation and Inequalities question

Quadratic Equation and Inequalities question

1. Finding $m$

Interval method

More from Quadratic Equation and Inequalities

Quadratic Equation and Inequalities question

Quadratic Equation and Inequalities question

1. Finding $m$

Interval method

More from Quadratic Equation and Inequalities

1. Finding mmm

Interval method

More from Quadratic Equation and Inequalities

1. Finding mmm

Interval method

More from Quadratic Equation and Inequalities

1. Finding $m$

1. Finding $m$