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The number of distinct real roots of the equation x5 ( x 3 − x 2 − x + 1 ) + x ( 3 x 3 − 4 x 2 − 2 x + 4 ) − 1 = 0  is …………..

0 5 Views | Posted 2 weeks ago
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    Answered by

    Raj Pandey | Contributor-Level 9

    2 weeks ago

    x 8 − x 7 − x 6 + x 5 + 3 x 4 − 4 x 3 − 2 x 2 + 4 x − 1 = 0  

    ⇒ x 7 ( x − 1 ) − x 5 ( x − 1 ) + 3 x 3 ( x − 1 ) − x ( x 2 − 1 ) + 2 x ( 1 − x ) + ( x − 1 ) = 0  

    ⇒ ( x − 1 ) ( x 2 − 1 ) ( x 5 + 3 x − 1 ) = 0 ∴ x = ± 1  are roots of above equation and x5 + 3x – 1 is a monotonic term hence vanishes at exactly one value of x other then 1 or -1.

     3 real roots.

Similar Questions for you

If |z + 1| = αz + β (i + 1) and   z = 1 2 – 2i, find α + β.
A
alok kumar singh

  | 1 2 − 2 i + 1 | = α ( 1 2 − 2 i ) + β ( 1 + i )  

9 4 + 4 = α ( 1 2 − 2 i ) + β ( 1 + i )

5 2 = α ( 1 2 ) + β + i ( − 2 α + β )             

α 2 + β = 5 2      ...(1)

 –2α + β = 0                    …(2)

Solving (1) and (2)

α 2 + 2 α = 5 2

5 2 α = 5 2            

a = 1

b = 2

-> a + b = 3

For a non-zero complex number z satisfying z 2 + i z ¯ = 0 , then value of |z|2 is
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alok kumar singh

z 2 = − i z ¯

| z 2 | = | − i z ¯ |

| z 2 | = | z |

| z 2 | − | z | = 0

| z | ( | z | − 1 ) = 0

|z| = 0 (not acceptable)

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Given x2 – 70x + λ = 0 with positive roots a and b where one of the root is less than 10 and λ 2 and λ 3 are not integers then find value of α − 1 + β − 1 | α − β |  is equal to
A
alok kumar singh

Given : x2 – 70x + l = 0

->Let roots be a and b

->b = 70 – a

->= a (70 – a)

l is not divisible by 2 and 3

->a = 5, b = 65


-> 5 − 1 + 6 5 − 1 | 6 0 | = | 4 + 8 6 0 | = 1 5

Let z1 and z2 be two complex numbers such that z1 + z2 = 5 and , then the value of  is equal to
A
alok kumar singh

z1 + z2 = 5

z 1 3 + z 2 3 = 2 0 + 1 5 i            

  z 1 3 + z 2 3 = ( z 1 + z 2 ) 3 − 3 z 1 z 2 ( z 1 + z 2 )           

z 1 3 + z 2 3 = 1 2 5 − 3 z 1 ⋅ z 2 ( 5 )            

 ⇒ 20 + 15i = 125 – 15z1z2

⇒ 3z1z2 = 25 – 4 – 3i

3z1z2 = 21– 3i

z1⋅z2 = 7 – i

(z1 + z2)2 = 25

z 1 2 + z 2 2 = 2 5 − 2 7 ( 7 − i )     

= 11 + 2i

  ( z ? 1 2 + z 2 2 ) 2         = 121 − 4 + 44i

⇒   z 1 4 + z 2 4 + 2 ( 7 − i ) 2 = 1 1 7 + 4 4 i

⇒   z 1 4 + z 2 4 = 117 + 44i − 2(49 −1−14i )

= 21 + 72i

⇒   | Z 1 4 + Z 2 4 | = 7 5

The integer ‘k’ for which the inequality x2 – 2(3k – 1) x + 8k2 – 7 > 0 is valid for every x in R, is:
V
Vishal Baghel

x 2 − 2 ( 3 k − 1 ) x + 8 k 2 − 7 > 0 ⇒ a > 0 , & D < 0 ,

a = 1 > 0 and D < 0

4 (3k – 1)2 – 4 (8k2 – 7) < 0

  ( 9 k 2 + 1 − 6 k − 8 k 2 + 7 < 0 )

k ( k − 4 ) − 2 ( k − 4 ) < 0

k ∈ ( 2 , 4 )

K = 3

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