Set, Relations and FunctionsMTP March 22Question 1990 of 217
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If f(x)=2+x2−x\displaystyle f(x)=\frac{2+x}{2-x}, then f−1(x)\displaystyle f^{-1}(x)

Options

A2(x−1)x+1\displaystyle \frac{2(x-1)}{x+1}
B2(x+1)x−1\displaystyle \frac{2(x+1)}{x-1}
C(x+1)x−1\displaystyle \frac{(x+1)}{x-1}
D(x−1)x+1\displaystyle \frac{(x-1)}{x+1}
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Correct Answer

✅ Option a — 2(x−1)x+1\displaystyle \frac{2(x-1)}{x+1}

All Options:

  • A2(x−1)x+1\displaystyle \frac{2(x-1)}{x+1}
  • B2(x+1)x−1\displaystyle \frac{2(x+1)}{x-1}
  • C(x+1)x−1\displaystyle \frac{(x+1)}{x-1}
  • D(x−1)x+1\displaystyle \frac{(x-1)}{x+1}

Detailed Solution & Explanation

To find the inverse function f−1(x)\displaystyle f^{-1}(x), we set y=f(x)\displaystyle y = f(x) and solve for x\displaystyle x in terms of y\displaystyle y:
y=2+x2−xy = \frac{2 + x}{2 - x}
Multiply both sides by (2−x)\displaystyle (2 - x):
y(2−x)=2+xy(2 - x) = 2 + x
2y−xy=2+x2y - xy = 2 + x
Rearrange the terms to group all x\displaystyle x terms on one side:
2y−2=x+xy2y - 2 = x + xy
Factor out x\displaystyle x from the right side and factor out 2\displaystyle 2 from the left side:
2(y−1)=x(1+y)2(y - 1) = x(1 + y)<br>Divideby\displaystyle <br>Divide by(1 + y):<br><spanclass="katex−display"><spanclass="katex"><spanclass="katex−mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mrow><mi>x</mi><mo>=</mo><mfrac><mrow><mn>2</mn><mostretchy="false">(</mo><mi>y</mi><mo>−</mo><mn>1</mn><mostretchy="false">)</mo></mrow><mrow><mi>y</mi><mo>+</mo><mn>1</mn></mrow></mfrac></mrow><annotationencoding="application/x−tex">x=2(y−1)y+1</annotation></semantics></math></span><spanclass="katex−html"aria−hidden="true"><spanclass="base"><spanclass="strut"style="height:0.4306em;"></span><spanclass="mordmathnormal">x</span><spanclass="mspace"style="margin−right:0.2778em;"></span><spanclass="mrel">=</span><spanclass="mspace"style="margin−right:0.2778em;"></span></span><spanclass="base"><spanclass="strut"style="height:2.3074em;vertical−align:−0.8804em;"></span><spanclass="mord"><spanclass="mopennulldelimiter"></span><spanclass="mfrac"><spanclass="vlist−tvlist−t2"><spanclass="vlist−r"><spanclass="vlist"style="height:1.427em;"><spanstyle="top:−2.314em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="mord"><spanclass="mordmathnormal"style="margin−right:0.03588em;">y</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mbin">+</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mord">1</span></span></span><spanstyle="top:−3.23em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="frac−line"style="border−bottom−width:0.04em;"></span></span><spanstyle="top:−3.677em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="mord"><spanclass="mord">2</span><spanclass="mopen">(</span><spanclass="mordmathnormal"style="margin−right:0.03588em;">y</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mbin">−</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mord">1</span><spanclass="mclose">)</span></span></span></span><spanclass="vlist−s">​</span></span><spanclass="vlist−r"><spanclass="vlist"style="height:0.8804em;"><span></span></span></span></span></span><spanclass="mclosenulldelimiter"></span></span></span></span></span></span><br>Replacing\displaystyle :<br><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>x</mi><mo>=</mo><mfrac><mrow><mn>2</mn><mo stretchy="false">(</mo><mi>y</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo></mrow><mrow><mi>y</mi><mo>+</mo><mn>1</mn></mrow></mfrac></mrow><annotation encoding="application/x-tex">x = \frac{2(y - 1)}{y + 1}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">x</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:2.3074em;vertical-align:-0.8804em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.427em;"><span style="top:-2.314em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord mathnormal" style="margin-right:0.03588em;">y</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">1</span></span></span><span style="top:-3.23em;"><span class="pstrut" style="height:3em;"></span><span class="frac-line" style="border-bottom-width:0.04em;"></span></span><span style="top:-3.677em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">2</span><span class="mopen">(</span><span class="mord mathnormal" style="margin-right:0.03588em;">y</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">−</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">1</span><span class="mclose">)</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.8804em;"><span></span></span></span></span></span><span class="mclose nulldelimiter"></span></span></span></span></span></span><br>Replacingywith\displaystyle withx$ to express the inverse function, we get:
f−1(x)=2(x−1)x+1f^{-1}(x) = \frac{2(x - 1)}{x + 1}

Hence, **Option A** is the correct answer.

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