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4. Study the diagram. The line l is perpendicular to line m

            (a) Is CE = EG?

            (b) Does PE bisect CG?

            (c) Identify any two line segments for which PE is the perpendicular bisector.

            (d) Are these true?

                    (i) AC > FG

                    (ii) CD = GH

                    (iii) BC < EH.

                

(a) CE = 5 - 3 = 2 units

     EG = 7 - 5 = 2 units    

     Therefore CE = EG.

(b) CE = EG therefore PE bisects CG.

(c) PE is the perpendicular bisector for line segments DF and BH

(d) (i) AC = 3 - 1 = 2 units

         FG = 7 - 6 = 1 unit

           Therefore AC > FG

            True

    (ii) CD = 4 - 3 = 1 unit

         GH = 8 - 7 = 1 unit

          Therefore CD = GH

          True    

    (iii) BC = 3 - 2 = 1 unit

          EH = 8 - 5 = 3 units

           Therefore BC < EH

            True 

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Posted by

Sayak

3. There are two set-squares in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?

The angles of the two set quares are

(i) 90o, 60o and 30o

(ii) 90o, 45o and 45o

Yes they have the common angle measure 90o

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Posted by

Sayak

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2.  Let \overline {PQ} be the perpendicular to the line segment \overline {XY} . Let \overline {PQ} and \overline {XY} intersect in the point A. What is the measure of \angle PAY ?

PQ\perp XY

 PQ and XY intersect at A

PA\perp AY

Therefore \angle PAY=90^{o}

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Sayak

1.  Which of the following are models for perpendicular lines :

            (a) The adjacent edges of a table top.

            (b) The lines of a railway track.

            (c) The line segments forming the letter ‘L’.

            (d) The letter V.

 (a) The adjacent edges of a table top are models for perpendicular lines.

(b) The lines of a railway track are not models for perpendicular lines as they are parallel to each other.

(c) The line segments forming the letter ‘L’ are models for perpendicular lines.

(d) The line segments forming the letter ‘V’ are models for perpendicular lines.

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Sayak

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 Q18         If u, v and w are functions of x, then show that                                                                                                                  \frac{d}{dx} ( u,v,w) = \frac{du}{dx} v. w +u . \frac{dv }{dx } v. w+ u . \frac{dv}{dx } . w+u.v \frac{dw}{dx}
in two ways - first by repeated application of product rule, second by logarithmic differentiation.

It is given that u, v and w are the functions of x
Let y = u.v.w
Now, we differentiate using product rule w.r.t x
First, take  y = u.(vw)
Now,
\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{d(v.w)}{dx}.u                      -(i)
Now, again by the product rule 
\frac{d(v.w)}{dx}= \frac{dv}{dx}.w + \frac{dw}{dx}.v
Put this in equation (i)
we get,
\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)
Hence, by product rule we proved it

Now, by taking the log 
Again take y = u.v.w
Now, take log on both sides 
\log y = \log u + \log v + \log w
Now, differentiate w.r.t. x
we get,
\frac{1}{y}.\frac{dy}{dx} = \frac{1}{u}.\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}\\ \frac{dy}{dx}= y. \left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\ \frac{dy}{dx} = (u.v.w)\left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\
\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)
Hence, we proved it by taking the log

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Posted by

Gautam harsolia

 Q17 (3)   Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below:
              (iii)  by logarithmic differentiation.
                 Do they all give the same answer?

Given function is
y=(x^2 - 5x + 8) (x^3 + 7x + 9)
Now, take log on both the sides
\log y = \log (x^2-5x+8)+\log (x^3+7x+9)
Now, differentiate w.r.t. x
we get,
\frac{1}{y}.\frac{dy}{dx} = \frac{1}{x^2-5x+8}.(2x-5) + \frac{1}{x^3+7x+9}.(3x^2+7)\\ \frac{dy}{dx}= y.\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\ \frac{dy}{dx}=(x^2-5x+8)(x^3+7x+9).\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\ \frac{dy}{dx} = (2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)\\ \frac{dy}{dx} = 5x^4-20x^3+45x^2-56x+11
Therefore, the answer is 5x^4-20x^3+45x^2-56x+11
And yes they all give the same answer

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Gautam harsolia

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 Q17 (2)   Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below:
                (ii) by expanding the product to obtain a single polynomial.

Given function is
f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)
Multiply both to  obtain a single higher degree polynomial
f(x) = x^2(x^3+7x+9)-5x(x^3+7x+9)+8(x^3+7x+9)
            = x^5+7x^3+9x^2-5x^4-35x^2-45x+8x^3+56x+72
            = x^5-5x^4+15x^3-26x^2+11x+72
Now, differentiate w.r.t. x
we get,
f^{'}(x)=5x^4-20x^3+45x^2-52x+11
Therefore, the answer is 5x^4-20x^3+45x^2-52x+11
 

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Posted by

Gautam harsolia

 Q17 (1)   Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below:
                (i) by using product rule

Given function  is 
f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)
Now, we need to differentiate using the product rule
f^{'}(x)=\frac{d((x^2 - 5x + 8))}{dx}. (x^3 + 7x + 9)+(x^2 - 5x + 8).\frac{d( (x^3 + 7x + 9))}{dx}\\
             = (2x-5).(x^3+7x+9)+(x^2-5x+8)(3x^2+7)\\ =2x^4+14x^2+18x-5x^3-35x-45+3x^4-15x^3+24x^2+7x^2-35x+56\\ = 5x^4 -20x^3+45x^2-52x+11
Therefore, the answer is 5x^4 -20x^3+45x^2-52x+11

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Gautam harsolia

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 Q16  Find the derivative of the function given by f (x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) and hence find 

           f ' (1)

Given function is
y = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)
Take log on both sides
\log y =\log (1 + x) + \log (1 + x^2) +\log (1 + x^4) +\log (1 + x^8)
NOW, differentiate w.r.t. x
\frac{1}{y}.\frac{dy}{dx} = \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8}\\ \frac{dy}{dx}=y.\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )\\ \frac{dy}{dx}= (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )
Therefore, f^{'}(x)= (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )
Now, the vale of  f^{'}(1)  is
f^{'}(1)= (1 + 1) (1 + 1^2) (1 + 1^4) (1 + 1^8).\left ( \frac{1}{1+1}+ \frac{2(1)}{1+1^2}+ \frac{4(1)^3}{1+1^4}+ \frac{8(1)^7}{1+1^8} \right )\\ f^{'}(1)=16.\frac{15}{2} = 120

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Gautam harsolia

 Q15  Find dy/dx of the functions given in Exercises 12 to 15.  xy = e ^{x-y}

 

Given function is
f(x)\Rightarrow xy = e ^{x-y}
Now, take  take log on both the sides
\log x+\ log y = (x-y)(1) \ \ \ \ \ \ \ \ \ \ \ \ (\because \log e = 1)\\ \log x+\ log y = (x-y)
Now, differentiate w.r.t  x
\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}
By taking similar terms on same side
We get, 
(\frac{1}{y}+1)\frac{dy}{dx}=1-\frac{1}{x}\\ \frac{y+1}{y}.\frac{dy}{dx}= \frac{x-1}{x}\\ \frac{dy}{dx}= \frac{y}{x}.\frac{x-1}{y+1}
Therefore, the answer is  \frac{y}{x}.\frac{x-1}{y+1} 

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Gautam harsolia

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