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Quadratic Equation and Inequalities question

2023 · 13 Apr · Shift 1 · Q29
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  5. /2023 · 13 Apr · Shift 1 · Q29

Quadratic Equation and Inequalities question

2023 · 13 Apr · Shift 1 · Q29

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
The set of all a∈Ra \in \mathbb{R}a∈R for which the equation x∣x−1∣+∣x+2∣+a=0x|x-1|+|x+2|+a=0x∣x−1∣+∣x+2∣+a=0 has exactly one real root, is :
  1. A
    (−∞,∞)(-\infty, \infty)(−∞,∞)
  2. B
    (−6,∞)(-6, \infty)(−6,∞)
  3. C
    (−∞,−3)(-\infty,-3)(−∞,−3)
  4. D
    (−6,−3)(-6,-3)(−6,−3)
View written solutionFree

Correct answer: A

Let f(x)=x∣x−1∣+∣x+2∣.f(x)=x|x-1|+|x+2|.f(x)=x∣x−1∣+∣x+2∣. We need the values of a∈Ra\in\mathbb Ra∈R such that x∣x−1∣+∣x+2∣+a=0x|x-1|+|x+2|+a=0x∣x−1∣+∣x+2∣+a=0 has exactly one real root, i.e. f(x)=−af(x)=-af(x)=−a has exactly one real solution.

So we study the graph/range/monotonicity of f(x)f(x)f(x).

1. Break into intervals using the moduli

Critical points are x=−2x=-2x=−2 and x=1x=1x=1.

Case 1: x≤−2x\le -2x≤−2

Then ∣x−1∣=1−x,∣x+2∣=−(x+2)=−x−2.|x-1|=1-x,\qquad |x+2|=-(x+2)=-x-2.∣x−1∣=1−x,∣x+2∣=−(x+2)=−x−2. So f(x)=x(1−x)+(−x−2)=x−x2−x−2=−x2−2.f(x)=x(1-x)+(-x-2)=x-x^2-x-2=-x^2-2.f(x)=x(1−x)+(−x−2)=x−x2−x−2=−x2−2.

Hence for x≤−2x\le -2x≤−2, f(x)=−x2−2.f(x)=-x^2-2.f(x)=−x2−2. This is strictly decreasing on this interval since f′(x)=−2x>0 for x<0?f'(x)=-2x>0 \text{ for }x<0?f′(x)=−2x>0 for x<0? Let us be careful: for x≤−2x\le -2x≤−2, −2x>0-2x>0−2x>0, so fff is actually increasing on (−∞,−2](-\infty,-2](−∞,−2]. Also, f(−2)=−4−2=−6,f(-2)=-4-2=-6,f(−2)=−4−2=−6, and as x→−∞x\to-\inftyx→−∞, f(x)→−∞.f(x)\to -\infty.f(x)→−∞. So on (−∞,−2](-\infty,-2](−∞,−2], range is (−∞,−6].(-\infty,-6].(−∞,−6].

Case 2: −2≤x≤1-2\le x\le 1−2≤x≤1

Then ∣x−1∣=1−x,∣x+2∣=x+2.|x-1|=1-x,\qquad |x+2|=x+2.∣x−1∣=1−x,∣x+2∣=x+2. So f(x)=x(1−x)+(x+2)=x−x2+x+2=−x2+2x+2.f(x)=x(1-x)+(x+2)=x-x^2+x+2=-x^2+2x+2.f(x)=x(1−x)+(x+2)=x−x2+x+2=−x2+2x+2. This is a downward parabola. Its derivative is f′(x)=−2x+2=2(1−x),f'(x)=-2x+2=2(1-x),f′(x)=−2x+2=2(1−x), which is positive for x<1x<1x<1. Thus on [−2,1][-2,1][−2,1], fff is strictly increasing.

Values at endpoints:

\qquad f(1)=-1+2+2=3.$$ So range on this interval is $$[-6,3].$$ --- ### Case 3: $x\ge 1$ Then $$|x-1|=x-1,\qquad |x+2|=x+2.$$ So $$f(x)=x(x-1)+(x+2)=x^2-x+x+2=x^2+2.$$ This is strictly increasing for $x\ge1$. At $x=1$, $$f(1)=3,$$ and as $x\to\infty$, $$f(x)\to\infty.$$ So range on $[1,\infty)$ is $$[3,\infty).$$ --- ## 2. Combine the behavior Thus: - on $(-\infty,-2]$, $f$ is increasing from $-\infty$ to $-6$, - on $[-2,1]$, $f$ is increasing from $-6$ to $3$, - on $[1,\infty)$, $f$ is increasing from $3$ to $\infty$. So overall, $f(x)$ is increasing on the entire real line. Indeed, the pieces join continuously: $$f(-2)=-6,\qquad f(1)=3.$$ Hence $f:\mathbb R\to\mathbb R$ is continuous and strictly increasing, with $$\lim_{x\to-\infty}f(x)=-\infty, \qquad \lim_{x\to\infty}f(x)=\infty.$$ Therefore, for **every** real number $c$, the equation $$f(x)=c$$ has exactly one real solution. Taking $c=-a$, for every $a\in\mathbb R$, the equation $$f(x)+a=0$$ has exactly one real root. --- ## 3. Check options Thus the required set of all $a$ is $$(-\infty,\infty).$$ So the correct option is: $$\boxed{\text{A}}$$ --- ## 4. Compare with stored correct answer Stored correct answer: A Our derived answer: A They agree.
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