Question

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Answer 1

Question :-

p ( x ) = 2x⁴ - 3x³ + 2x² + 2x - 1 is divided by ( x - 2 ) and q ( x ) = 3x³ - 2x² + x - 1 is divided by ( x - 1 ) . So, twice the sum of the remainders is ?

(A) 21 (B) 35 (C) 54 (D) 40

Answer :-

Given :-

p ( x ) = 2x⁴ - 3x³ + 2x² + 2x - 1 is divided by ( x - 2 )

q ( x ) = 3x³ - 2x² + x - 1 is divided by ( x - 1 )

Required to find :-

• Value when the sum of the remainder is twiced ?

Concept used :-

• Remainder theorem

Solution :-

Given information :-

p ( x ) = 2x⁴ - 3x³ + 2x² + 2x - 1 is divided by ( x - 2 )

q ( x ) = 3x³ - 2x² + x - 1 is divided by ( x - 1 )

we need to find the value when the sum of the remainder is twiced !

So,

Consider the given statement 1

p ( x ) = 2x⁴ - 3x³ + 2x² + 2x - 1 is divided by ( x - 2 )

From this above statement we can conclude that ,

When ( x + 2 ) divides p ( x ) it leaves reminder . So factor theorem is not applicable here !

So,

p ( x ) = 2x⁴ - 3x³ + 2x² + 2x - 1

( x - 2 ) when divides p ( x ) leaves remainder

So,

Let,

=> x - 2 = 0

=> x = 2

Substitute this value in place of x in p ( x )

So,

p ( 2 ) = 2 ( 2 )⁴ - 3 ( 2 )³ + 2 ( 2 )² + 2 ( 2 ) - 1

p ( 2 ) = 2 ( 16 ) - 3 ( 8 ) + 2 ( 4 ) + 4 - 1

p ( 2 ) = 32 - 24 + 8 + 4 - 1

p ( 2 ) = 44 - 25

p ( 2 ) = 19

Hence,

When ( x - 2 ) divides p ( x ) the remainder ( R1 ) = 19

Similarly,

Consider statement - 2

q ( x ) = 3x³ - 2x² + x - 1 is divided by ( x - 1 )

From this statement we get the same above conclusion

So,

Let,

=> x - 1 = 0

=> x = 1

This implies,

q ( 1 ) = 3 ( 1 )³ - 2 ( 1 )² + 1 - 1

+ 1 & - 1 get cancelled

q ( 1 ) = 3 ( 1 ) - 2 ( 1 )

q ( 1 ) = 3 - 2

q ( 1 ) = 1

Hence,

when ( x - 1 ) divides q ( x ) the remainder ( R2 ) = 1

This implies ,

Sum of the remainders in 2 cases = R1 + R2

=> 19 + 1

=> 20

Hence,

Sum of the remainders = 20

But ,

we need to find the value when the sum of the remainders is twiced

This can be easily understood in the below way ,

Let, the sum of the remainders be x

Twice the sum of the remainders = 2x

So,

As we know that ,

Sum of the remainders = 20

Hence,

Twice the sum of the remainders = 2 ( 20 ) = 40

Hence,

Twice the sum of the remainders = 40

Therefore,

Option - D is correct !

Answer 2

$\red{\huge{\boxed{\sf{Question:}}}}$

p(x) = 2x⁴ - 3x³ + 2x² + 2x - 1 is divided by (x - 2) and q(x) = 3x³ - 2x² + x - 1 is divided by (x - 1) . So, twice the sum of the remainders is ?

(A) 21         (B) 35           (C) 54           (D) 40

$\red{\huge{\boxed{\sf{Solution:}}}}$

p(x) = 2x⁴ - 3x³ + 2x² + 2x - 1 is divided by (x - 2)

→ x = 2

So,

p(2) = 2(2)⁴ - 3(2)³ + 2(2)² + 2(2) - 1

       = 2(16) - 3(8) + 2(4) + 4 - 1

       = 32 - 24 + 8 + 4 - 1

       = 32 + 8 + 4 - 24 - 1

       = 44 - 25

       = 19

∵ Remainder₁ = 19

$\rule{200}2$

q(x) = 3x³ - 2x² + x - 1 is divided by (x - 1)

→ x = 1

So,

q(1) = 3(1)³ - 2(1)² + (1) - 1

     = 3(1) - 2(1) + 1 - 1

     = 3 - 2 + 1 - 1

     = 3 + 1 - 2 - 1

     = 4 - 3

     = 1

∵ Remainder₂ = 1

Now,

$\blue{\underline{\underline{\tt{According\:\: to\:\: the \:\:question}}}}$

Twice the sum of the remainders is

Sum of remainders = Remainder₁ + Remainder₂

                                = 19 + 1

                                = 20

Now,

Twice of sum of remainders = 2 × (20) = $\boxed{\boxed{\red{\sf{40}}}}$

Hence,

Option $\boxed{\bf{D) 40}}$ is the correct answer