The product of two polynomials
Question: The product of two polynomials is a ________. Solution: The product of two polynomials is a polynomials....
Read More →(x + a) (x + b) = x2 + (a + b) x + ________.
Question: (x + a) (x + b) = x2+ (a + b) x + ________. Solution: (x + a) (x + b) = x2+ (a + b) x + ab = (x + a) (x + b) = x (x + b) + a (x + b) = x2+ xb + xa + ab = x2+ x (b + a) + ab...
Read More →Prove the following
Question: (a + b)2 2ab = ___________ + ____________ Solution: (a + b)2 2ab = a2+ b2 = (a + b)2 2ab = a2+ 2ab + b2 2ab = a2+ b2...
Read More →Find the second order derivatives of each of the following functions:
Question: Find the second order derivatives of each of the following functions: $\tan ^{-1} x$ Solution: Basic idea: $\sqrt{S e c o n d}$ order derivative is nothing but derivative of derivative i.e. $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$ $\sqrt{T h e}$ idea of chain rule of differentiation: If $\mathrm{f}$ is any real-valued function which is the composition of two functions $u$ and $v$, i.e. $f=v(u(x))$. Fo...
Read More →(a – b)2 + ____________ = a2 – b2
Question: (a b)2+ ____________ = a2 b2 Solution: (a b)2+ (2ab 2b2) = a2 b2 = (a b)2+ (2ab 2b2) = a2+ b2 2ab + 2ab 2b2 = a2 b2...
Read More →Prove the following
Question: a2 b2= (a + b ) __________. Solution: a2 b2= (a + b) (a b) [from the standard identities]...
Read More →(a – b) _________ = a2 – 2ab + b2
Question: (a b) _________ = a2 2ab + b2 Solution: (a b) (a b) = (a b)2= a2 2ab + b2 (a b) (a b)= a (a b) b (a b) = a2 ab ba + b2 = a2 2ab + b2...
Read More →Find the second order derivatives of each of the following functions:
Question: Find the second order derivatives of each of the following functions: $x^{3} \log x$ Solution: $\sqrt{B a s i c}$ Idea: Second order derivative is nothing but derivative of derivative i.e. $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$ $\sqrt{T h e}$ idea of chain rule of differentiation: If $\mathrm{f}$ is any real-valued function which is the composition of two functions $u$ and $v$, i.e. $f=v(u(x))$. For the sake of simplicity just assume $t=u(x)$ Then $f=v(t) ....
Read More →a (b + c) = a × ____ + a × _____.
Question: a (b + c) = a ____ + a _____. Solution: a (b + c) = a b + a c. [by using left distributive law] = ab + ac...
Read More →The product of two terms
Question: The product of two terms with unlike signs is a term. Solution: The product of two terms with unlike signs is a negative term. Let us assume two unlike terms are, 3p and 2q = -3p 2q = 6pq...
Read More →The product of two terms with
Question: The product of two terms with like signs is a term. Solution: The product of two terms with like signs is a positive term. Let us assume two like terms are, 3p and 2q = 3p 2q = 6pq...
Read More →Prove the following
Question: The value of (a + b)2 (a b)2is (a) 4ab (b) 4ab (c) 2a2+ 2b2 (d) 2a2 2b2 Solution: (a) 4ab The value of (a + b)2 (a b)2= (a2+ b2+ 2ab) (a2+ b2 2ab) = a2 a2+ b2 b2+ 2ab + 2ab = 4ab...
Read More →The value of
Question: The value of (a + b)2+ (a b)2is (a) 2a + 2b (b) 2a 2b (c) 2a2+ 2b2 (d) 2a2 2b2 Solution: (c) 2a2+ 2b2 (a + b)2+ (a b)2= (a2+ b2+ 2ab) + (a2+ b2 2ab) = (a2+ a2) + (b2+ b2) + (2ab 2ab) = 2a2+ 2b2...
Read More →The value of
Question: The value of (3x3+9x2+ 27x) 3x is (a) x2+9 + 27x (b) 3x3+3x2+ 27x (c) 3x3+9x2+ 9 (d) x2+3x + 9 Solution: (d) x2+3x + 9 The value of (3x3+9x2+ 27x) 3x = (3x3+ 9x2+ 27x)/3x Takeout 3x as common, = 3x (x2+ 3x + 9)/3x = x2+ 3x + 9...
Read More →The value of
Question: The value of (2x2+ 4) 2 is (a) 2x2+ 2 (b) x2 + 2 (c) x2+ 4 (d) 2x2+ 4 Solution: (b) x2+ 2 The value of (2x2+ 4) 2 = (2x2+ 4)/2 = (2(x2+ 2))/2 = x2+ 2...
Read More →The value of
Question: The value of ( 27x2y) ( 9xy) is (a) 3xy (b) 3xy (c) 3x (d) 3x Solution: (d) 3x The value of ( 27x2y) ( 9xy) = (-27 x x y)/(- 9 x y) = (27/9)x [divide both numerator, denominator by 3] = 3x...
Read More →The factors of
Question: The factors of x2 4 are (a) (x 2), (x 2) (b) (x + 2), (x 2) (c) (x + 2), (x + 2) (d) (x 4), (x 4) Solution: (b) (x + 2), (x 2) The factors of x2 4 are, X2 4 = x2 22 = (x + 2) (x 2)...
Read More →The factorised form of
Question: The factorised form of 3x 24 is (a) 3x 24 (b) 3 (x 8) (c) 24 (x 3) (d) 3(x 12) Solution: (b) 3 (x 8) The factorised form of 3x 24 is, Take out 3 as common, = 3 (x 8)...
Read More →Number of factors of
Question: Number of factors of (a + b)2is (a) 4 (b) 3 (c) 2 (d) 1 Solution: (c) 2 Number of factors of (a + b)2is = (a + b) (a + b) no further factorisation is possible....
Read More →If the sums of n terms of two APs are in ratio
Question: If the sums of n terms of two APs are in ratio (2n + 3) : (3n + 2), find the ratio of their 10th terms. Solution: Given: sums of n terms of two APs are in ratio (2n + 3) : (3n + 2) To find: find the ratio of their 10th terms For the sum of n terms of two APs is given by $\mathrm{S}_{1}=\frac{\mathrm{n}}{2}\left(2 \mathrm{a}_{1}+(\mathrm{n}-1) \times \mathrm{d}_{1}\right)$ $\mathrm{S}_{2}=\frac{\mathrm{n}}{2}\left(2 \mathrm{a}_{2}+(\mathrm{n}-1) \times \mathrm{d}_{2}\right)$ $\frac{\mat...
Read More →An irreducible factor
Question: An irreducible factor of 24x2y2is (a) x2 (b) y2 (c) x (d) 24x Solution: (c) x An irreducible factor is a factor which cannot be expressed further as a product of factors. Such a factorisation is called an irreducible factorisation. 24x2y2= 2 2 2 3 x x y y Therefore an irreducible factor is x....
Read More →The sum of n terms of an AP is
Question: The sum of $n$ terms of an $A P$ is $\frac{1}{2}$ an $^{2}+b n$. Find the common difference. Solution: Given: the sum of $n$ terms of an AP is $\frac{1}{2} a n^{2}+b n$ To find: common difference. Put n = 1 we get First term $=\frac{1}{2}+b$ Put n = 2 we get First term + second term = 2 a + 2 b Second term $=\frac{3}{2} a+b$ Therefore common difference will be Second term - first term Common difference $=2 \mathrm{a}$...
Read More →Find the second order derivatives of each of the following functions:
Question: Find the second order derivatives of each of the following functions: $e^{6 x} \cos 3 x$ Solution: $\sqrt{B a s i c}$ Idea: Second order derivative is nothing but derivative of derivative i.e. $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$ $\sqrt{T h e}$ idea of chain rule of differentiation: If $\mathrm{f}$ is any real-valued function which is the composition of two functions $u$ and $v, i . e . f=v(u(x))$...
Read More →The common factor of
Question: The common factor of 3ab and 2cd is (a) 1 (b) 1 (c) a (d) c Solution: (a) 1 Considering the two monomials 3ab and 2cd there is no common factor except 1....
Read More →On dividing p (4p2 – 16)
Question: On dividing p (4p2 16) by 4p (p 2), we get (a) 2p + 4 (b) 2p 4 (c) p + 2 (d) p 2 Solution: (c) p + 2 On dividing p (4p2 16) by 4p (p 2) = (p((2p)2 (4)2))/ (4p(p 2)) = ((2p 4) (2p + 4))/(4(p 2)) Take out the common factors = ((2(p 2)) (2 (p + 4)))/(4(p -2)) = (4(p 2)(p + 2))/ (4(p 2)) = p + 2...
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