Which one of the following is divisible by (1 + a + a^5) and (1 + a^4 + a^5) individually ? -Maths 9th

Which one of the following is divisible by (1 + a + a^5) and (1 + a^4 + a^5) individually ? -Maths 9th
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Find the probability that a two digit number formed by the digit 1, 2, 3, 4 and 5 is divisible by 4. -Maths 9th

Description : Find the probability that a two digit number formed by the digit 1, 2, 3, 4 and 5 is divisible by 4. -Maths 9th

Last Answer : The two digit numbers can be formed by putting any of 5 digits at the one 's place and also one of the 5 digits at ten's place. So, Total number of 2-digit numbers that can be formed using these 5-digits = 5 5 = ... 52}, i.e, 5 in number. ∴ Required probability = \(rac{5}{25}\) = \(rac{1}{5}.\)

If x^5 – 9x^2 + 12x – 14 is divisible by (x – 3), what is the remainder ? -Maths 9th

Description : If x^5 – 9x^2 + 12x – 14 is divisible by (x – 3), what is the remainder ? -Maths 9th

Last Answer : answer:

If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Description : If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Last Answer : The divisor is x4−1=(x−1)(x+1)(x2+1) By factor theorem, f(1)=f(−1)=0 Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1 i.e., p+q=4 and p−q=−2 Adding the two, 2p=2 i.e. p=1 and ∴ q=3. ∴ p2+q2=1+9=10

What should be subtracted from 27x^3 – 9x^2 – 6x – 5 to make it exactly divisible by (3x – 1) -Maths 9th

Description : What should be subtracted from 27x^3 – 9x^2 – 6x – 5 to make it exactly divisible by (3x – 1) -Maths 9th

Last Answer : answer:

For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Description : For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Last Answer : Let p(x) = x3 -2mx2 +16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2)3 -2m(-2)2 + 16=0 ⇒ -8-8m+16=0 ⇒ 8 = 8 m m = 1 Hence, the value of m is 1 .

Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Description : Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Last Answer : Let p(x) = 2x4 - 5x3 + 2x2 - x+ 2 firstly, factorise x2-3x+2. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] = x(x-2)-1 (x-2)= (x-1)(x-2) Hence, 0 of x2-3x+2 are land 2. We have to prove that, 2x4 ... )2 - 2 + 2 = 2x16-5x8+2x4+ 0 = 32 - 40 + 8 = 40 - 40 =0 Hence, p(x) is divisible by x2-3x+2.

For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Description : For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Last Answer : Let p(x) = x3 -2mx2 +16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2)3 -2m(-2)2 + 16=0 ⇒ -8-8m+16=0 ⇒ 8 = 8 m m = 1 Hence, the value of m is 1 .

Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Description : Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Last Answer : Let p(x) = 2x4 - 5x3 + 2x2 - x+ 2 firstly, factorise x2-3x+2. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] = x(x-2)-1 (x-2)= (x-1)(x-2) Hence, 0 of x2-3x+2 are land 2. We have to prove that, 2x4 ... )2 - 2 + 2 = 2x16-5x8+2x4+ 0 = 32 - 40 + 8 = 40 - 40 =0 Hence, p(x) is divisible by x2-3x+2.

FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Description : FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Last Answer : As the given polynomial divisible by x-2 means the polynomial satisfies for the value x=2 So putting x=2 in x²+(4-k)x+2 yields 0 ⇒2²+(4-k)2+2=0 ⇒4+8-2k+2=0 ⇒ 2k=14 ⇒ k= ... ;-3x+2 if factorized yields (x-1)(x-2). Thus is divisible by x-2 as well as divisible by x-1.

FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Description : FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Last Answer : The value of 'k' is 4

An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

Description : An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

Last Answer : As there are 200 integers, total number of exhaustive, mutually exclusive and equally likely cases, i.e, n(S) = 200 Let A : Event of integer chosen from 1 to 200 being divisible by 6⇒ n(A) = 33 \(\bigg(rac{200}{6}=33rac{1}{3}\ ... (rac{25}{200}\) - \(rac{8}{200}\) = \(rac{50}{200}\) = \(rac{1}{4}\).

A five digit number is formed by the digits 0, 1, 2, 3, 4 (without repetition). Find the probability that the number formed is divisible by 4 ? -Maths 9th

Description : A five digit number is formed by the digits 0, 1, 2, 3, 4 (without repetition). Find the probability that the number formed is divisible by 4 ? -Maths 9th

Last Answer : Without repetition, a five -digit number can be formed using the five digits in 5! ways (5 4 3 2 1) Out of these 5! numbers, 4! numbers will be starting with digit 0. (0 (fixed) 4 3 2 1) ∴ Total ... + 6 + 6 + 4 + 4 + 4 = 30∴ Required probability = \(rac{30}{96}\) = \(rac{5}{16}.\)

If three natural numbersfrom 1 to 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is -Maths 9th

Description : If three natural numbersfrom 1 to 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is -Maths 9th

Last Answer : (c) \(rac{4}{1155}\)Let n(S) = Number of ways of selecting 3 numbers from 100 numbers = 100C3 Let E : Event of selecting three numbers divisible by both 2 and 3 from numbers 1 to 100 = Event of selecting three ... C_3}{^{100}C_3}\) = \(rac{16 imes15 imes14}{100 imes99 imes98}\) = \(rac{4}{1155}\).

For what value of m will the expression 3x^3 + mx^2 + 4x – 4m be divisible by x + 2 ? -Maths 9th

Description : For what value of m will the expression 3x^3 + mx^2 + 4x – 4m be divisible by x + 2 ? -Maths 9th

Last Answer : f(x) = 3x3 + mx2 + 4x – 4m f(x) is divisible by (x + 2) if f(–2) = 0 Now f(–2) = 3(–2)3 + m(–2)2 + 4(–2) – 4m = – 24 + 4m – ... ; 4m = – 32 ≠ 0 ∴ No such value of m exists for which (x + 2) is a factor of the given expression

Without actual division show that 2x^4 – 6x^3 + 3x^2 + 3x – 2 is exactly divisible by x^2 – 3x + 2 -Maths 9th

Description : Without actual division show that 2x^4 – 6x^3 + 3x^2 + 3x – 2 is exactly divisible by x^2 – 3x + 2 -Maths 9th

Last Answer : answer:

Without actual division, show that (x – 1)^2n – x^2n + 2x – 1 is divisible by 2x^3 – 3x^2 + x. -Maths 9th

Description : Without actual division, show that (x – 1)^2n – x^2n + 2x – 1 is divisible by 2x^3 – 3x^2 + x. -Maths 9th

Last Answer : answer:

For what value of k, will the expression (3x^3 – kx^2 + 4x + 16) be divisible by (x – k/2) ? -Maths 9th

Description : For what value of k, will the expression (3x^3 – kx^2 + 4x + 16) be divisible by (x – k/2) ? -Maths 9th

Last Answer : Given f(x) = 3x³ - kx² + 4x + 16. Since (x - k/2) is a factor of polynomial. This means x = k/2 is the zero of the given polynomial. ⇒ f(k/2) = 3(k/2)³ - k(k/2)² + 4(k/2) + 16 ⇒ 0 ... - 4k + 32) + 4(k² - 4k + 32) ⇒ 0 = (k + 4)(k² - 4k + 32) ⇒ k = -4.

x^4 + xy^3 + x^3y + xz^3 + y^4 + yz^3 is divisible by : -Maths 9th

Description : x^4 + xy^3 + x^3y + xz^3 + y^4 + yz^3 is divisible by : -Maths 9th

Last Answer : answer:

If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Description : If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Last Answer : Let f(x)=px3+x2−2x−q Since f(x) is divisible by (x−1) and (x+1) so x=1 and −1 must make f(x)=0. Therefore, p+1−2−q=0, i.e., p−q=1; and −p+1+2−q=0, i.e., p+q=3 Thus p=2 and q=1

Consider the following statements : 1. a^n + b^n is divisible by a + b if n = 2k + 1, where k is a positive integer. -Maths 9th

Description : Consider the following statements : 1. a^n + b^n is divisible by a + b if n = 2k + 1, where k is a positive integer. -Maths 9th

Last Answer : Statement (1) is correct as for k = 1, n = 2 × 1 + 1 = 3. ∴ a3 + b3 = (a + b) (a2 + b2 – ab) which is divisible by (a + b), statement (2) is also correct as for k = 1, n = 2, ∴ a2 – b2 = (a – b) (a + b) which is divisible by (a – b).

(x^n – a^n) is divisible by (x – a) -Maths 9th

Description : (x^n – a^n) is divisible by (x – a) -Maths 9th

Last Answer : answer:

If 4x^2 – 6x + m is divisible by x – 3, which one of the following is the greatest divisor of m ? -Maths 9th

Description : If 4x^2 – 6x + m is divisible by x – 3, which one of the following is the greatest divisor of m ? -Maths 9th

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