If log32, log3 (2^x – 5) and log^3 (2^x – 7/2) are in A.P., then what is the value of x ? -Maths 9th

If log32, log3 (2^x – 5) and log^3 (2^x – 7/2) are in A.P., then what is the value of x ? -Maths 9th
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If 1/3 log3 M + 3 log3 N = 1 + log 0.008 5, then -Maths 9th

Description : If 1/3 log3 M + 3 log3 N = 1 + log 0.008 5, then -Maths 9th

Last Answer : (b) \(N^9=rac{9}{M}\)\(rac{1}{3}\) log3 M + 3 log3 N = 1 + log0.008 5⇒ log3 M1/3 + log3N3 = 1 + log0.0085 ⇒ log3 M1/3 N3 = 1 + log0.0085 ⇒ M1/3 N3 = 3(1 + log0.0085) ⇒ M1/3 N3 = 31 . 3log0.0085⇒ \(N^9=rac{27 ... ^9=rac{27}{M}\big(3^{log_{rac{1}{5}}5}\big)\) = \(rac{1}{M}\) (27)(3-1) = \(rac{9}{M}.\)

If 1, log9 (3^(1–x) + 2) and log3 (4.3^x –1) are in A.P., then x is equal to -Maths 9th

Description : If 1, log9 (3^(1–x) + 2) and log3 (4.3^x –1) are in A.P., then x is equal to -Maths 9th

Last Answer : (d) log3\(\big(rac{3}{4}\big)\) 1, log9 (31 - x + 2), log3 (4.3x -1) are in A.P. ⇒ log33 , log3 (31- x + 2)1/2, log3 (4.3x - 1) are in A.P. ⇒ 3, (31 - x + 2)1/2, (4.3x - 1) are in G.P.\ ... (rac{1}{3}\), \(rac{3}{4}\)∴ Rejecting the negative value, we have 3x = \(rac{3}{4}\)⇒ x = log3\(rac{3}{4}.\)

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Description : Evaluate x if log3 (3 + x) + log3 (8 – x) – log3 (9x – 8) = 2 – log39 -Maths 9th

Last Answer : (c) 4log3 (3 + x) + log3 (8 - x) - log3 (9x - 8) = 2 - log39 ⇒ log3 (3 + x) + log3 (8 - x) - log3 (9x - 8) + log39 = 2⇒ log3 \(\bigg[rac{(3+x)(8-x)(9)}{(9x-8)}\bigg]=2\)⇒ \(rac{9(24+8x-3x-x^2)}{(9x- ... = 9x - 8 ⇒ x2 + 4x - 32 = 0 ⇒ (x + 8) (x - 4) = 0 ⇒ x = - 8, 4. Taking the positive value x = 4.

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Last Answer : Let l+m−2nlogx​=m+n−2llogy​=n+l−2mlogz​=k(say) So, we get logx=k(l+m−2n) ....... (i) logy=k(m+n−2l) ....... (ii) logz=k(n+l−2m) ....... (iii) ∴logx+logy+logz=k(l+m−2n)+k(m+n−2l)+k(n+l−2m) ⇒logx+logy+logz=kl+km−2kn+km+kn−2kl+kn+kl−2km ⇒log(xyz)=0 ⇒logxyz=log1 ⇒xyz=1

If (log x)/(a^2+ab+b^2) = (log y)/(b^2+bc+c^2) = (log z)/(c^2+ca+a^2), then x^(a-b). y^(b-c). z^(c-a) = -Maths 9th

Description : If (log x)/(a^2+ab+b^2) = (log y)/(b^2+bc+c^2) = (log z)/(c^2+ca+a^2), then x^(a-b). y^(b-c). z^(c-a) = -Maths 9th

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Last Answer : (b) 1Let \(rac{ ext{log}\,x}{l+m-2n}\) = \(rac{ ext{log}\,y}{m+n-2l}\) = \(rac{ ext{log}\,z}{n+l-2m}\) = k. Thenlog x = k(l + m – 2n), log y = k(m + n – 2l); log z = k(n + l – 2m) ⇒ log x + log y + log z = k(l + m – 2n) + k(m + n – 2l) + k(n + l – 2m)⇒ log(xyz) = 0 ⇒ log(xyz) = log 1 ⇒ xyz = 1.

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Description : The value of 6 + log(3/2) -Maths 9th

Last Answer : (d) 4Let A = 6 + \( ext{log}_{rac{3}{2}}\)\(\bigg[rac{1}{3\sqrt2}\sqrt{4-rac{1}{3\sqrt2}\sqrt{4-rac{1}{3\sqrt2}\sqrt{4-rac{1}{3\sqrt2}\sqrt{4-rac{1}{3\sqrt2}}}}}...\) Let P = \(\sqrt{4-rac{1}{3\sqrt2}\sqrt{4-rac ... (rac{3}{2}\big)^{-2}\) = 6 - 2 \( ext{log}_{rac{3}{2}}\) \(rac{3}{2}\) = 6 - 2 = 4.

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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

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Description : If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Last Answer : Given that the following polynomials leave the same remainder when divided by (x - 4) : We are to find the value of a. Remainder theorem: When (x - b) divides a polynomial p(x), then the remainder is p(b). So, from (i) and (ii), we get Thus, the required value of a is 1.

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Description : If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. -Maths 9th

Last Answer : p(x) = (k2 – 14) x2 – 2x – 12 Here a = k2 – 14, b = -2, c = -12 Sum of the zeroes, (α + β) = 1 …[Given] ⇒ − = 1 ⇒ −(−2)2−14 = 1 ⇒ k2 – 14 = 2 ⇒ k2 = 16 ⇒ k = ±4

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Description : If the mode of the data 5, 8, 4,5,5,8, 4, 7, 8, x is 5, then find the value of x. -Maths 9th

Last Answer : The value of x = 5.

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Description : if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Last Answer : x = 1 y = -2 2x-y = p Therefore, p = 2(1)-(-2) = 2 + 2 = 4

if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Description : if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Last Answer : 2x-y=p put x-1,y=-2 =2(1)-(-2)=p p=4

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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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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

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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Last Answer : (b) 1 - log25 Given, log2 (5.2x + 1), log4 (21- x + 1), 1 are in A.P. ⇒ log2 (5.2x + 1) + 1 = 2 log4 (21 - x + 1) ⇒ log2 (5.2x + 1) + log22 = 2 log22 (21-x + 1)⇒ log2 (5.2x + 1).2 = 2 x \(rac12\) ... a=-rac{1}{2}\big)\)⇒ log 2x = log \(rac{2}{5}\)⇒ x log2 2 = log2 2 - log2 5 ⇒ \(x\) = 1 - log2 5.

What is the sum, of 'n' terms in the series : log m + log -Maths 9th

Description : What is the sum, of 'n' terms in the series : log m + log -Maths 9th

Last Answer : (d) log \(\bigg[rac{m^{(1+n)}}{n^{(n-1)}}\bigg]^{rac{n}{2}}\) S = log m + log \(rac{m^2}{n}\) + log \(rac{m^3}{n^2}\) + ...........n terms= log \(\bigg[\)m.\(rac{m^2}{n}\).\(rac{m^3}{n^2}\)........ ... 2}}}{n^{rac{n(n-1)}{2}}}\)\(\bigg]\) = log \(\bigg[rac{m^{(1+n)}}{n^{(n-1)}}\bigg]^{rac{n}{2}}\)

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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.

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

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.

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

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

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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

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

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Description : The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

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Description : For what value of k,(x+1) is a factor of p(x) - kx(square) - x - 4 ? -Maths 9th

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Description : Find the value of k if (x-2)is a factor of polynomial p(x) = 2x(cube) - 6x(square) + 5x + k. -Maths 9th

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Description : Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

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Let p and q be the roots of the quadratic equation x^2 – (a – 2)x – a – 1 = 0. What is the minimum possible value of p^2 + q^2 ? -Maths 9th

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For what value of p is the coefficient of x^2 in the product (2x – 1) (x – k) (px + 1) equal to 0 and the constant term equal to 2 ? -Maths 9th

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

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

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

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

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

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

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

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

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

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

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

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.

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

If p(x) is a common multiple of degree 6 of the polynomials f(x) = x^3 + x^2 – x – 1 and g(x) = x^3 – x^2 + x – 1, then which -Maths 9th

Description : If p(x) is a common multiple of degree 6 of the polynomials f(x) = x^3 + x^2 – x – 1 and g(x) = x^3 – x^2 + x – 1, then which -Maths 9th

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If x + (1/x) = p, then x^6+(1/x^6) equals to : -Maths 9th

Description : If x + (1/x) = p, then x^6+(1/x^6) equals to : -Maths 9th

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If x^3 + px + q and x^3 + qx + p have a common factor, then which of the following is correct ? -Maths 9th

Description : If x^3 + px + q and x^3 + qx + p have a common factor, then which of the following is correct ? -Maths 9th

Last Answer : x3+px+q 3x2+p have common root Let common root =a a3+ap+q=0 a2=3−p​a=3−p​​4p3+27q2=? a3(a2+p)+q=0 ⇒3−p​​[(3−p​)+p]+q=0 3−p​​(3+2p​)=−q ⇒27−p(4p2)​=q2

If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Description : If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Last Answer : Since, the point (x = 3, y = 4) lies on the equation 3y = ax + 7, then the equation will be , satisfied by the point. Now, put x = 3 and y = 4 in given equation, we get 3(4) = a (3)+7 ⇒ 12 = 3a+7 ⇒ 3a = 12 – 7 ⇒ 3a = 5 Hence, the value of a is 5/3.

If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Description : If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Last Answer : Since, the point (x = 3, y = 4) lies on the equation 3y = ax + 7, then the equation will be , satisfied by the point. Now, put x = 3 and y = 4 in given equation, we get 3(4) = a (3)+7 ⇒ 12 = 3a+7 ⇒ 3a = 12 – 7 ⇒ 3a = 5 Hence, the value of a is 5/3.

If vertices of a triangles are (1, k), (4, -3) and (-9, 7) and its area is 15 sq. units then find then the value of k. -Maths 9th

Description : If vertices of a triangles are (1, k), (4, -3) and (-9, 7) and its area is 15 sq. units then find then the value of k. -Maths 9th

Last Answer : hope it helps if the vertices of a triangle are (1,k),(4,−3)(−9,7) area = 15 sq.units. find the value of k. Area of △ 21 [x1 (y2 −y3 )+x2 (y3 −y1 )+x3 (y1 −y2 )]=15 21 [1(−3−7)+ ... k+3)]=15 21 [(−10+28−4k−9k−27)]=15 −10+28−4k−9k−27=30 −10+28−13k−27=30 −13k=30+10+27−28 −13k=39 k=1339 k=−3 thank u