The domain of fx-1-x2-14x-49 is

Given : f(x)=√(1 √(x^2 14x+49)) ⇒ f(x )=√(1 |x 7|) f(x) is defined when 1 |x 7|≥0 ⇒|x 7|≤1 ⇒ 1≤ x 7≤1 ⇒ 6≤ x≤ 8 nbsp; ...

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Standard XII

Mathematics

Logarithmic Inequalities

The domain of...

Question

The domain of $f (x) = \sqrt{1 - \sqrt{x^{2} - 14 x + 49}}$ is

[7, 8]

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[6, 7]

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[6, 8]

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[0, 8]

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Solution

The correct option is C $[6, 8]$
Given : $f (x) = \sqrt{1 - \sqrt{x^{2} - 14 x + 49}}$
$\Rightarrow f (x) = \sqrt{1 - | x - 7 |}$
$f (x)$ is defined when
$1 - | x - 7 | \geq 0 \Rightarrow | x - 7 | \leq 1 \Rightarrow - 1 \leq x - 7 \leq 1 \Rightarrow 6 \leq x \leq 8$

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Similar questions

Q. The domain of

f (x) = \sqrt{x - \sqrt{1 - x^{2}}}

Let f(x) = $x^{2}$ -1, 0<x<2
and 2x+3, 2 $\leq$ x<3, The quadratic equation whose roots are ${lim}_{x \to 2^{-}} f (x), a n d {lim}_{x \to 2^{+}} f (x)$

Q. Factorise :

x^{2} + 14 x + 49

Q. Solve

x^{2} + 49 + 14 x

Q. Simplify:

x^{2} + 14 x + 49

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Logarithmic Inequalities

Standard XII Mathematics

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The domain of f(x)=√1−√x2−14x+49 is

The correct option is C [6,8]Given : f(x)=√1−√x2−14x+49 ⇒f(x)=√1−|x−7| f(x) is defined when 1−|x−7|≥0⇒|x−7|≤1⇒−1≤x−7≤1⇒6≤x≤8

The domain of $f (x) = \sqrt{1 - \sqrt{x^{2} - 14 x + 49}}$ is

The correct option is C $[6, 8]$
Given : $f (x) = \sqrt{1 - \sqrt{x^{2} - 14 x + 49}}$
$\Rightarrow f (x) = \sqrt{1 - | x - 7 |}$
$f (x)$ is defined when
$1 - | x - 7 | \geq 0 \Rightarrow | x - 7 | \leq 1 \Rightarrow - 1 \leq x - 7 \leq 1 \Rightarrow 6 \leq x \leq 8$