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Standard XII

Mathematics

Definition of Functions

If a function...

Question

If a function $f (x) = x + \frac{1}{x}$ is shifted along $y - a x i s$ as $g (x) = f (x) + 2$ . Then calculate the number of solution for $g (x) = 3$ .

4

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Solution

The correct option is D $0$
$g (x) = 3 \Rightarrow x + \frac{1}{x} + 2 = 3 \Rightarrow x^{2} - x + 1 = 0$

Now $b^{2} - 4 a c = (- 1)^{2} - 4 \times 1 \times 1 = 1 - 4 = - 3$

Hence as discriminant is negative there is no real solution to the equation but 1 complex pair as the solutions .

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$y = f (x)$ is differentiable function and $g (x) = f (x - x^{2})$ If $y = g (x)$ has local maxima at x= $\frac{1}{2}$ but the absolute maximum exists at some other points, then a minimum number of a solution of $g (x) = 0$ is

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If a function f(x)=x+1x is shifted along y−axis as g(x)=f(x)+2. Then calculate the number of solution for g(x)=3.

The correct option is D 0g(x)=3⇒x+1x+2=3⇒x2−x+1=0 Now b2−4ac=(−1)2−4×1×1=1−4=−3Hence as discriminant is negative there is no real solution to the equation but 1 complex pair as the solutions .

If a function $f (x) = x + \frac{1}{x}$ is shifted along $y - a x i s$ as $g (x) = f (x) + 2$ . Then calculate the number of solution for $g (x) = 3$ .

The correct option is D $0$
$g (x) = 3 \Rightarrow x + \frac{1}{x} + 2 = 3 \Rightarrow x^{2} - x + 1 = 0$

Now $b^{2} - 4 a c = (- 1)^{2} - 4 \times 1 \times 1 = 1 - 4 = - 3$

Hence as discriminant is negative there is no real solution to the equation but 1 complex pair as the solutions .