If <math> <mrow> <mi>g</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mi>g</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math> and <math> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mrow> <mo>[</mo> <mrow> <mi>g</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>+</mo> <mi>g</mi> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mo>−</mo> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mo>]</mo> </mrow> </mrow> </math> and <math> <mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math> &nbsp;then

f"(x) = 0 has at least two roots in (0, 2)

f"(x) = 0 has exactly one root in (0, 1)

f"(x) = 0 has no root in (0, 1)

f"(x) = 0 has 3 roots in (0, 2)

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Ncert Solutions Maths class 12th Maths Class 12th

If $g' (\frac{3}{2}) = g' (\frac{1}{2})$ and $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ and $f' (\frac{3}{2}) = f' (\frac{1}{2})$ then

Option 1 -

f"(x) = 0 has exactly one root in (0, 1)

Option 2 -

f"(x) = 0 has no root in (0, 1)

Option 3 -

f"(x) = 0 has at least two roots in (0, 2)

Option 4 -

f"(x) = 0 has 3 roots in (0, 2)

2 Views | Posted a week ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a week ago

Correct Option - 3

Detailed Solution:

$f' (x) = \frac{g' (x) - g' (2 - x)}{2}, f' (\frac{3}{2}) = \frac{g' (\frac{3}{2}) - g' (\frac{1}{2})}{2} = 0$

Also $f' (\frac{1}{2}) = \frac{g' (\frac{1}{2}) - g' (\frac{3}{2})}{2} = 0$ , f' (1) = 0

-> $f' (\frac{3}{2}) = f' (\frac{1}{2}) = 0$

->roots in $(\frac{1}{2}, 1)$ and $(1, \frac{3}{2})$

->f" (x) is zero at least twice in $(\frac{1}{2}, \frac{3}{2})$

<math> <mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mi>g</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>−</mo> <mi>g</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mn>2</mn> <mo>−</mo> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>, </mo> <mtext> </mtext> <mi>f</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mi>g</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>−</mo> <mi>g</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> </mrow> </math>  Also  <math> <mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mi>g</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>−</mo> <mi>g</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> </mrow> </math> , f' (1) = 0-> <math> <mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mi>f</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>0</mn> </mrow> </math>  ->roots in <math> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>, </mo> <mtext> </mtext> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math> and <math> <mrow> <mrow> <mo> (</mo> <mrow> <mn>1</mn> <mo>, </mo> <mtext> </mtext> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math>  ->f" (x) is zero at least twice in <math> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>, </mo> <mtext> </mtext> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1030" DrawAspect="Content" ObjectID="_1820650922"> </o:OLEObject></xml><! [endif]->

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If $g' (\frac{3}{2}) = g' (\frac{1}{2})$ and $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ and $f' (\frac{3}{2}) = f' (\frac{1}{2})$ then

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If $g' (\frac{3}{2}) = g' (\frac{1}{2})$ and $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ and $f' (\frac{3}{2}) = f' (\frac{1}{2})$ then