SATHEE: Functions 1 Question 4

Functions 1 Question 4

4. Range of the function $f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1}; x \in R$ is

(a) $(1, \infty)$

(b) $(1, 11 / 7)$

(d) $(1, 7 / 5)$

$(2003, 2 M)$

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Answer:

Correct Answer: 4. $(d)$

Solution:

Let $y = f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1}, x \in R$

$\begin{array}{ll} ∴ & y = \frac{x^{2} + x + 2}{x^{2} + x + 1} \\ y = 1 + \frac{1}{x^{2} + x + 1} [i.e. y > 1] \\ \Rightarrow & y x^{2} + y x + y = x^{2} + x + 2 \\ \Rightarrow & x^{2} (y - 1) + x (y - 1) + (y - 2) = 0, \forall x \in R \\ Since, x is real, D \geq 0 \\ \Rightarrow & (y - 1)^{2} - 4 (y - 1) (y - 2) \geq 0 \\ \Rightarrow & (y - 1) (y - 1) - 4 (y - 2) \geq 0 \end{array}$

$\begin{array}{lr} \Rightarrow & (y - 1) (- 3 y + 7) \geq 0 \\ \Rightarrow & 1 \leq y \leq \frac{7}{3} \end{array}$

From Eqs. (i) and (ii), Range $\in 1, \frac{7}{3}$

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Functions 1 Question 4

4. Range of the function f(x)=x2+x+2x2+x+1;x∈R is

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4. Range of the function $f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1}; x \in R$ is