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Let \alpha and \beta   be the roots of equation

x^{2}-6x-2=0.\; if a_{n}=\alpha ^{n}-\beta ^{n},\; for\: n\geq 1,\; then\; the \: value\; of\; \frac{a_{10}-2a_{8}}{2a_{9}}

is equal to:

  • Option 1)

    6

  • Option 2)

    -6

  • Option 3)

    3

  • Option 4)

    -3

3

  1.  -5
  2. 6
  3. 1
  4. -3
Volume of parallelopiped,  Here  and,  Given that V is 546, put it in the above equation. Option (4) is correct

\int_{0}^{\pi /2} \frac{1}{1+\sqrt{tanx}} dx

  1. \pi

  2. \pi/2

  3. \pi/4

  4. 3\pi/2

...............................(1) We also know that,  Using this property in equation (1) ,  .....................(2) Adding equation (1) and (2),  Option (3) is correct
  1. x2+y2-3x+1=0
  2. x2+y2-x+5=0

  3. x2+y2-8x+6y=8

  4. x2+y2-4x+8y=7

General equation of circle ,  Now find the points of intersection of the two circles. At the intersection,              Putting the value of x1 in the equation of any of the two circle to find y1. Now we have 3 points through which the circle is passing,     ,       ,     Now putting these points in general equation we will have 3...
  1. \frac{x}{\sqrt{1+x^{2}}}
  2. \frac{1}{\sqrt{1+x^{2}}}
  3. \frac{\sqrt{1+x^{2}}}{x}
  4. \sqrt{1+x^{2}}
Let  Using Pythagoras theorem,  Option (2) is correct

(1+\omega^{2}-\omega)(1-\omega^{2}+\omega) is

  1. 4

  2. \omega

  3. 2

  4. Zero

With the help of 2 properties of cube roots of unity, we can solve this question. Property 1,  Property 2,  Now we have to find the value of (1+-)(1-+) Using property 2,  Now use property 1, Option (1) is correct  
Engineering
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Let f_{k}\left ( x \right )= \frac{1}{k}\left ( \sin ^{k}x+\cos ^{k}x \right )where x\epsilon R\: \: and\: \: k\geqslant 1      Then f_{4}(x)-f_{6}(x)  equals :

  • Option 1)

    \frac{1}{4}

  • Option 2)

    \frac{1}{12}

  • Option 3)

    \frac{1}{6}

  • Option 4)

    \frac{1}{3}

 

Option 2

Engineering
53 Views   |  

 Let ABC be a triangle whose circumcentre is at P.  If the position vectors of A, B, C

and P are\vec{a},\vec{b},\vec{c} \: \: and \: \: \frac{\vec{a}+\vec{b}+\vec{c}}{4}

respectively, then the position vector of the orthocentre of this triangle, is :

  • Option 1)

    \vec{a}+\vec{b}+\vec{c}

  • Option 2)

    -\left ( \frac{\vec{a}+\vec{b}+\vec{c}}{2} \right )

  • Option 3)

    \vec{0}

  • Option 4)

    \left ( \frac{\vec{a}+\vec{b}+\vec{c}}{2} \right )

 

https://www.youtube.com/watch?v=y30g2lfwv_c

Engineering
56 Views   |  

If (10)9 + 2(11)1   (10)8 + 3(11)2  (10)7 +......  +10 (11)9 = k (10)9, then k is equal to :

  • Option 1)

    100

  • Option 2)

    110

  • Option 3)

    \frac{121}{10}

  • Option 4)

    \frac{441}{100}

 

Use

Sum of n terms of a GP -

S_{n}= \left\{\begin{matrix} a\frac{\left ( r^{n}-1 \right )}{r-1}, &if \: r\neq 1 \\ n\, a, & if \, r= 1 \end{matrix}\right.

 

- wherein

a\rightarrow first term

r\rightarrow common ratio

n\rightarrow number of terms    

 

 and

 

(10)9 + 2(11)1 (10)8 + 3(11)2 (10)7 +......  +10 (11)9 = k(10)9

Take common 109

10^{9}\left [ 1+2\times \frac{11}{10}+3\times \left ( \frac{11}{10} \right )^{2}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot 10\times \left ( \frac{11}{10} \right ) ^{9}\right ]= k\left ( 10 \right )^{9}

\therefore k= 1+2x+3x^{2}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot 10x^{9} \: where \:\:x=\frac{11}{10}

   kx= x+2x^{2}+3x^{3}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot 10x^{10}

Subtract

k-kx= 1+x+x^{2}+x^{3}+x^{4}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (-10x^{10})

k(1-x)= \frac{1(x^{10}-1)}{x-1}-10x^{10}

k\left ( 1-\frac{11}{10} \right )= \frac{\left ( \frac{11}{10} \right )^{10}-1}{\frac{11}{10}-1}-10\times \left ( \frac{11}{10} \right )^{10}

-\frac{k}{10}= \frac{\left ( \frac{11}{10} \right )^{10}}{\frac{1}{10}}-10-\frac{11^{10}}{10^{9}}

 

\therefore k=100


Option 1)

100

Option 2)

110

Option 3)

\frac{121}{10}

Option 4)

\frac{441}{100}

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Engineering
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The number of common tangents to the circles x^{2}+y^{2}-4x-6y-12=0\, and\, x^{2}+y^{2}+6x+18y+26=0,is:

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    3

  • Option 4)

    4

 

 

Common tangents of two circles -

When two circles touch  each other externally, there are three common tangents, two of them are direct.

 

- wherein

 

 C_1=(2,3)

C_2=(-3,-9)

C_1C_2=\sqrt{25+144}

r_1=\sqrt{4+9+12}= 5

r_2=\sqrt{9+81-26}= 8

\therefore r_1+r_2=C_1C_2

 

3 Tangents for circle touching externally


Option 1)

1

Option 2)

2

Option 3)

3

Option 4)

4

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Engineering
102 Views   |  

The sum of all real values of x satisfying the equation

\left ( x^{2} \right-5x+5 )^{x^{2}+4x-60}=1

is

 

  • Option 1)

    3

  • Option 2)

    -4

  • Option 3)

    6

  • Option 4)

    5

 
As we have learned Roots of Quadratic Equation with real Coefficients - are roots if is satisfied by   - wherein       Also there is a case  (x-3)(x-2)= 0  = x = 2,3  For x = 3 :  So sum 4+1+6+(-10)+2= 3              Option 1) 3 Option 2) -4 Option 3) 6 Option 4) 5
Engineering
319 Views   |  

If the coefficients of x3 and x4 in the expansion of (1 + ax + bx2) (1 - 2x)18 in powers of x are both zero, then (a, b) is equal to :

  • Option 1)

    \left ( 14,\; \frac{272}{3} \right )

  • Option 2)

    \left ( 16,\; \frac{272}{3} \right )

  • Option 3)

    \left ( 16,\; \frac{251}{3} \right )

  • Option 4)

    \left ( 14,\; \frac{251}{3} \right )

 
As we have learned Expression of Binomial Theorem -   - wherein for n  +ve integral .     coeff of   coeff of    and  and   a = 16  b = 272/3        Option 1) Option 2) Option 3) Option 4)
Engineering
159 Views   |  

 If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is :

 

  • Option 1)

     46th

     

  • Option 2)

     59th

     

  • Option 3)

    52nd

  • Option 4)

     58th

 
As we have learned Rank of any Word - We arrange the words according to dictionary. Eq. for SUNIL        Rank is 95 - wherein                                                                               Tarting with A     4! / 2! = 12 words  Starting with L   ---->  4! = 24 words  Starting with m ----->  4! / 2! = 12 words  Starting with S ----> A ----> 3! / 2! = 3 words                     ...
Engineering
134 Views   |  

If the function.

g(x)=   \left \{ \right. 

is differentiable, then the value of k + m is :

  • Option 1)

    2

  • Option 2)

    \frac{16}{5}

  • Option 3)

    \frac{10}{3}

  • Option 4)

    4

 
As we have learned Continuity at a point - A function f(x)  is said to be continuous at  x = a in its domain if  1.  f(a) is defined  : at  x = a. of  f(x) at  x = a exists from left and right. -     Condition for differentiability - A function  f(x) is said to be differentiable at    if      both exist and are equal otherwise non differentiable -    FOr continuity at x= 3 ,    and for...
Engineering
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At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P w.r.t. additional number of workers x is given by \frac{dp}{dx}=100-12\sqrt{x}. If the firm employs 25 more workers,then the new level of production of items is :

 

  • Option 1)

    4500

  • Option 2)

    2500

  • Option 3)

    3000

  • Option 4)

    3500

 
As we have learned Variable Separation Method - integrating, we get -   and    We have              Option 1) 4500 Option 2) 2500 Option 3) 3000 Option 4) 3500
Engineering
108 Views   |  

 If fand g are differentiable functions in [0, 1] satisfying f (0) = 2 = g (1), g (0) = 0 andf(1)=6, then for some c\epsilon]0, 1[ : 

  • Option 1)

    f'\left ( c \right )=g'\left ( c \right )

  • Option 2)

    f'\left ( c \right )=2g'\left ( c \right )

  • Option 3)

    2f'\left ( c \right )=g'\left ( c \right )

  • Option 4)

    2f'\left ( c \right )=3g'\left ( c \right )

 
As we have learned cauchy's Theorem -   for some  - wherein     We have  For some    Option 1) Option 2) Option 3) Option 4)
Engineering
282 Views   |  

The sum of first 20 terms of the sequence 0.7 ,0.77,0.777,.........,is :

  • Option 1)

    \frac{7}{9}\left ( 99+10^{-20} \right )

  • Option 2)

    \frac{7}{81}\left ( 179-10^{-20} \right )

  • Option 3)

    \frac{7}{9}\left ( 99-10^{-20} \right )

  • Option 4)

    \frac{7}{81}\left ( 179+10^{-20} \right )

 

As we learnt

 

Sum of infinite terms of a GP -

a+ar+ar^{2}+- - - - -= \frac{a}{1-r}\\here \left | r \right |<1

- wherein

a\rightarrow first term

r\rightarrow common ratio

 

 S=0.7+0.77+0.777...upto\: \: 20\: \: terms

S=\frac{7}{9}(0.9+0.99+0.999...)

S=\frac{7}{9}(1-0.1+1-0.01+1-0.001...)

S=\frac{7}{9}(20-(\frac{1}{10}+\frac{1}{100}+...upto\: \: 20\: \: terms))

S=\frac{7}{9}(20-\frac{\frac{1}{10}(1-\frac{1}{10^{20}})}{(1-\frac{1}{10})})

S=\frac{7}{9}(20-\frac{1-10^{-20}}{9})

S=\frac{7}{81}(179+10^{-20})

 


Option 1)

\frac{7}{9}\left ( 99+10^{-20} \right )

Option 2)

\frac{7}{81}\left ( 179-10^{-20} \right )

Option 3)

\frac{7}{9}\left ( 99-10^{-20} \right )

Option 4)

\frac{7}{81}\left ( 179+10^{-20} \right )

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Engineering
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Let T_{n} be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If T_{n+1}-T_n=10 then the value of n is :

 

 

  • Option 1)

    8

  • Option 2)

    7

  • Option 3)

    5

  • Option 4)

    10

 
As we have learned Theorem of Combination - Each of the different groups or selection which can be made by taking r things from n things is called a combination. - wherein Where       no. of selection of 3 vertices out of n vertices                Option 1) 8 Option 2) 7 Option 3) 5 Option 4) 10
Engineering
134 Views   |  

The graph of the function  y=f\left ( x \right )  is symmetrical about the line x=2 ,then

  • Option 1)

    f\left ( x \right )= f\left ( -x \right )

  • Option 2)

    f\left (2+ x \right )= f\left ( 2-x \right )

  • Option 3)

    f\left ( x +2\right )= f\left ( x-2 \right )

  • Option 4)

    f\left ( x \right )=- f\left ( -x \right )

 

As we learnt in

Even Function -

f(-x)= f(x)

- wherein

Symmetric about Y - axis

 

 Since a graph symmetric about y-axis

means  x = 0 then it is even function and f(-x) = f(x)

\therefore    f(0 - x) = f(0 + x)     (b < z it is symmetric about v = 0 )

But in question it is symmetric about x = 2

then f(x - 2) = f(x + 2) 

Correct option is 3.

 


Option 1)

f\left ( x \right )= f\left ( -x \right )

Option 2)

f\left (2+ x \right )= f\left ( 2-x \right )

Option 3)

f\left ( x +2\right )= f\left ( x-2 \right )

Option 4)

f\left ( x \right )=- f\left ( -x \right )

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Engineering
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if   f(x)+2f\left ( \frac{1}{x} \right )=3x,x\neq 0,  and  S=\left \{ x\, \epsilon \, R : f(x)=f(-x)\right \};then S:

  • Option 1)

    is an empty set.

  • Option 2)

    contains exactly one element.

  • Option 3)

    contains exactly two elements.

  • Option 4)

    contains more than two elements.

 

As we learnt in

FUNCTIONS -

A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B.

-

 

 f(x)+2f \left(\frac{1}{x} \right )=3x

Put \frac{1}{x} at the place of 

f\left(\frac{1}{x} \right )+2f(x)=\frac{3}{x}                                                    (i)

2f\left(\frac{1}{x} \right )+f(x)=3x                                                (ii)

Multiplying (i) by 2

2f\left(\frac{1}{x} \right )+4f(x)=\frac{6}{x}

\underline{2f\left(\frac{1}{x} \right )+f(x)=3x}

                      3f(x)=\frac{6}{x}-3x

                    f(x)=\frac{2}{x}-3x

and             f(-x)=\frac{2}{-x}+x

\therefore\ \; \frac{2}{x}-x=-\frac{2}{x}+x

\Rightarrow\ \; \frac{4}{x}-2x=0

\Rightarrow\ \; \frac{4-2x^{2}}{x}=0

\Rightarrow\ \; 4=2x^{2}

\Rightarrow\ \; x^{2}=2

x=\pm \sqrt{2}, \; x \neq 0

Correct option is 3.

 

 

 


Option 1)

is an empty set.

Option 2)

contains exactly one element.

Option 3)

contains exactly two elements.

Option 4)

contains more than two elements.

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