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)

Ask & Answer Question

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 month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month 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]->

0 Upvotes 0 Downvotes

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
682k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

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

1 Answer

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

If g ' ( 3 2 ) = g ' ( 1 2 ) and f ( x ) = 1 2 [ g ( x ) + g ( 2 − x ) ] and f ' ( 3 2 ) = f ' ( 1 2 ) then

1 Answer

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

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