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Functions question

2023 · 31 Jan · Shift 2 · Q24
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Functions question

2023 · 31 Jan · Shift 2 · Q24

JEE MainMathematicsFunctionsMCQ+4 / −1
Let f:R−{2,6}→Rf: \mathbb{R}-\{2,6\} \rightarrow \mathbb{R}f:R−{2,6}→R be real valued function defined as f(x)=x2+2x+1x2−8x+12f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}f(x)=x2−8x+12x2+2x+1​. Then range of fff is
  1. A
    (−∞,−214]∪[1,∞)\left(-\infty,-\frac{21}{4}\right] \cup[1, \infty)(−∞,−421​]∪[1,∞)
  2. B
    (−∞,−214)∪(0,∞)\left(-\infty,-\frac{21}{4}\right) \cup(0, \infty)(−∞,−421​)∪(0,∞)
  3. C
    (−∞,−214]∪[0,∞)\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)(−∞,−421​]∪[0,∞)
  4. D
    (−∞,−214]∪[214,∞)\left(-\infty,-\frac{21}{4}\right] \cup\left[\frac{21}{4}, \infty\right)(−∞,−421​]∪[421​,∞)
View written solutionFree

Correct answer: C

  1. Given function
f(x)=x2+2x+1x2−8x+12=(x+1)2(x−2)(x−6),x∈R∖{2,6} f(x)=\frac{x^2+2x+1}{x^2-8x+12} = \frac{(x+1)^2}{(x-2)(x-6)}, \qquad x\in \mathbb R\setminus\{2,6\}f(x)=x2−8x+12x2+2x+1​=(x−2)(x−6)(x+1)2​,x∈R∖{2,6}

We need the range of fff.

  1. Let y=f(x)y=f(x)y=f(x) and solve for xxx

Set

y=x2+2x+1x2−8x+12y=\frac{x^2+2x+1}{x^2-8x+12}y=x2−8x+12x2+2x+1​

Cross-multiplying,

y(x2−8x+12)=x2+2x+1y(x^2-8x+12)=x^2+2x+1y(x2−8x+12)=x2+2x+1 (y−1)x2+(−8y−2)x+(12y−1)=0(y-1)x^2+(-8y-2)x+(12y-1)=0(y−1)x2+(−8y−2)x+(12y−1)=0

For a real value of yyy to belong to the range, this quadratic in xxx must have at least one real solution.

So its discriminant must satisfy

Δ≥0\Delta \ge 0Δ≥0
  1. Compute the discriminant

Here,

a=y−1,b=−8y−2,c=12y−1a=y-1,\quad b=-8y-2,\quad c=12y-1a=y−1,b=−8y−2,c=12y−1

Thus,

Δ=(−8y−2)2−4(y−1)(12y−1)\Delta = (-8y-2)^2-4(y-1)(12y-1)Δ=(−8y−2)2−4(y−1)(12y−1)

First,

(−8y−2)2=64y2+32y+4(-8y-2)^2=64y^2+32y+4(−8y−2)2=64y2+32y+4

and

(y−1)(12y−1)=12y2−13y+1(y-1)(12y-1)=12y^2-13y+1(y−1)(12y−1)=12y2−13y+1

So,

4(y−1)(12y−1)=48y2−52y+44(y-1)(12y-1)=48y^2-52y+44(y−1)(12y−1)=48y2−52y+4

Therefore,

Δ=(64y2+32y+4)−(48y2−52y+4)\Delta=(64y^2+32y+4)-(48y^2-52y+4)Δ=(64y2+32y+4)−(48y2−52y+4) \Delta=16y^2+84y n$$

\Delta=4y(4y+21)

For real $x$,

4y(4y+21)\ge 0

y(4y+21)\ge 0

Thisgives This gives Thisgives

y\le -\frac{21}{4} \quad \text{or} \quad y\ge 0

Sothepossiblerangeis So the possible range is Sothepossiblerangeis

(-\infty,-\tfrac{21}{4}]\cup[0,\infty)

--- 4. **Check boundary values are actually attained** ### (i) $y=0$

\frac{(x+1)^2}{(x-2)(x-6)}=0

Thisrequires This requires Thisrequires

(x+1)^2=0 \Rightarrow x=-1

Since $-1\ne 2,6$, $y=0$ is attained. ### (ii) $y=-\frac{21}{4}$ At the discriminant boundary, the quadratic has a repeated real root, so this value is attained as well. Let us verify:

-\frac{21}{4}=\frac{x^2+2x+1}{x^2-8x+12}

Cross−multiplying: Cross-multiplying: Cross−multiplying:

-21(x^2-8x+12)=4(x^2+2x+1)

-21x^2+168x-252=4x^2+8x+4

25x^2-160x+256=0

(5x-16)^2=0

So So So

x=\frac{16}{5}

which is allowed. Hence $y=-\frac{21}{4}$ is also attained. --- 5. **Check whether excluded points $x=2,6$ create any issue** If the only solutions for some $y$ were $x=2$ or $x=6$, that $y$ would need to be removed from the range. So check whether $x=2$ or $x=6$ satisfy the quadratic obtained after cross-multiplication. Substitute $x=2$:

(y-1)(4)+(-8y-2)(2)+(12y-1)

=4y-4-16y-4+12y-1=-9\ne 0

Substitute $x=6$:

(y-1)(36)+(-8y-2)(6)+(12y-1)

=36y-36-48y-12+12y-1=-49\ne 0

So $x=2,6$ are never solutions of this equation for any $y$. Hence no additional values need to be excluded. --- 6. **Final range** Therefore,

\boxed{(-\infty,-\tfrac{21}{4}]\cup[0,\infty)}

Thismatches∗∗OptionC∗∗. This matches **Option C**.Thismatches∗∗OptionC∗∗.
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