IF p(x)=10x-4x -Maths 9th

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Description : IF p(x)=10x-4x -Maths 9th

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Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

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Description : p(x)=x3-2x2-4x-1, g(x)=x- -Maths 9th

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Description : Find the value of m, so that 2x -1 be a factor of 8x4 +4x3 -16x2 +10x+07. -Maths 9th

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Description : Solve for x: 5(4x + 3) = 3(x -2) -Maths 9th

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Description : If sin^4x + sin^2x = 1 then what is 1 are the value of cot^4 x + cot^2 x? -Maths 9th

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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

Description : If x^2 – 4x + 1 = 0, then what is the value of x^3 + 1/x^3? -Maths 9th

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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.

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Description : When x^3 + 2x^2 + 4x + b is divided by (x + 1), the quotient is x^2 + ax + 3 and the remainder -Maths 9th

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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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Description : Find any four solutions of the equation 4x+3y=12. -Maths 9th

Last Answer : Given equation is 4x + 3y =12 On putting x = 0 in Eq. (i), we get 4(0) +3y =12 ⇒ 3y =12 ⇒ y = 12 / 3 = 4 So, (0, 4) is a solution of given equation. On putting y = 0 in Eq. (i), we ... given equation. Hence the four solutions of given equation are (0, 4), (3, 0), (1, 8 / 3) and (2,4 / 3).

Description : Find any four solutions of the equation 4x+3y=12. -Maths 9th

Last Answer : Given equation is 4x + 3y =12 On putting x = 0 in Eq. (i), we get 4(0) +3y =12 ⇒ 3y =12 ⇒ y = 12 / 3 = 4 So, (0, 4) is a solution of given equation. On putting y = 0 in Eq. (i), we ... given equation. Hence the four solutions of given equation are (0, 4), (3, 0), (1, 8 / 3) and (2,4 / 3).

Description : factorise the following 4x^2+20x+25 -Maths 9th

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Description : factorise the following 4x^2+20x+25 -Maths 9th

Last Answer : 4x^2+20x+25 4x^2+10x+10x+25 2x(2x+5)+5(2x+5) (2x+5)(2x+5) 2x^2+2(2x)(5)+25 2x^2+20x+25

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Description : Using factor theorem, factorise the polynomial x3 + x2 - 4x - 4. -Maths 9th

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Description : Find the coordinate where the linear equation 4x - 23 y = 7 meets at y-axis. -Maths 9th

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Description : The solution of 4x^2 + 4x + 1 > 0 is -Maths 9th

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Description : What is the equation of the line joining the origin with the point of intersection of the lines 4x + 3y = 12 and 3x + 4y = 12 ? -Maths 9th

Last Answer : (b) (5, 6)Let the foot of the perpendicular be M(x1, y1) Slope of line AB, i.e., y = -x + 11 = -1 Slope of line PM = \(rac{y_1-3}{x_1-2}\)Now, PM ⊥ AB⇒ \(\bigg(rac{y_1-3}{x_1-2}\bigg)\) x - ... get 2x1 = 10 ⇒ x1 = 5 Putting x1 in (ii), we get y1 = 6. ∴ Required foot of the perpendicular M is (5, 6).

Description : The image of the origin with reference to the line 4x + 3y – 25 = 0 is -Maths 9th

Last Answer : (c) 25x + 25y - 4 = 0 The point of intersection of the given lines can be obtained by solving the equations of the two lines simultaneously. 100x + 50y = 1 ...(i) 75x + 25y = -3 ... = \(rac{4}{25}\)∴ Eqn of required line: x + y = \(rac{4}{25}\) ⇒ 25 + 25y - 4 = 0.

Description : Find the following products: (i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx) (ii) (4x -3y + 2z) (16x2 + 9y2+ 4z2 + 12xy + 6yz – 8zx) (iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca) (iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz) -Maths 9th

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Description : Find the following products: (i) (3x + 2y) (9X2 – 6xy + Ay2) (ii) (4x – 5y) (16x2 + 20xy + 25y2) (iii) (7p4 + q) (49p8 – 7p4q + q2) -Maths 9th

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Description : Find the value of 64x3 – 125z3, if 4x – 5z = 16 and xz = 12. -Maths 9th

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Description : p(x)=x3+3x2+3x+1, g(x) = x+2 -Maths 9th

Last Answer : p(x) = x3+3x2+3x+1, g(x) = x+2 g(x) = 0 ⇒ x+2 = 0 ⇒ x = −2 ∴ Zero of g(x) is -2. Now, p(−2) = (−2)3+3(−2)2+3(−2)+1 = −8+12−6+1 = −1 ≠ 0 ∴By factor theorem, g(x) is not a factor of p(x

Description : Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Last Answer : Solution: p(x) = 2x3+x2–2x–1, g(x) = x+1 g(x) = 0 ⇒ x+1 = 0 ⇒ x = −1 ∴Zero of g(x) is -1. Now, p(−1) = 2(−1)3+(−1)2–2(−1)–1 = −2+1+2−1 = 0 ∴By factor theorem, g(x) is a factor of p(x

Description : 2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

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Description : Determine the remainder when polynomial p(x) is divided by x - 2 . -Maths 9th

Last Answer : p(x) = x4 - 3x2 + 2x - 5 According to remainder theorem, the required remainder will be = p(2) p(x) = x4 - 3x2 + 2x - 5 ∴ p(2) = 24 - 3(2)2 + 2(2) - 5 =16 - 12 + 4 - 5 = 3

Description : If p (x) = x + 3, then p(x)+ p (- x) is equal to -Maths 9th

Last Answer : (d) Given p(x) = x+3, put x = -x in the given equation, we get p(-x) = -x+3 Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6

Description : Zero of the polynomial p(x)=2x+5 is -Maths 9th

Last Answer : (b) Given, p(x) = 2x+5 For zero of the polynomial, put p(x) = 0 ∴ 2x + 5 = 0 ⇒ -5/2 Hence, zero of the polynomial p(x) is -5/2.

Description : Find the zeroes of the polynomial p(x)= (x – 2)2 – (x+ 2)2. -Maths 9th

Last Answer : Given, polynomial is p(x) = (x – 2)2 – (x+ 2)2 For zeroes of polynomial, put p(x) = 0 (x – 2)2 – (x+ 2)2 = 0 (x-2 + x+2)(x-2-x-2) = 0 [using identity, a2-b2 =(a-b)(a + b)] ⇒ (2x)(-4) = 0

Description : Check whether p(x) is a multiple of g(x) or not -Maths 9th

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Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Last Answer : Show that p = r.

Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Last Answer : (d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.