If the solve the problem
Question: If $y=x^{3} \log x$, prove that $\frac{d^{4} y}{d x^{4}}=\frac{6}{x}$ Solution: Basic idea: $\sqrt{S e c o n d}$ 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 $f=v(t)$. By chain rule...
Read More →The standard form for
Question: The standard form for 0.000000008 is_______. Solution: For standard form, 0.000000008 = 0.8 x10-8= 8 x 10-9=8.0 x10-9 Hence, the standard form for 0.000000008 is 8.0 x 10-9...
Read More →The standard form for
Question: The standard form for 32500000000 is_______. Solution: For standard form, 32500000000 = 3250 x 102x 102x 103 = 3250 x 107= 3.250 x 1010or 3.25 x 1010 Hence, the standard form for 32500000000 is 3.25 x 1010....
Read More →To add the numbers given in standard form,
Question: To add the numbers given in standard form, we first convert them into number with_______exponents. Solution: To add the numbers given in standard form, we first convert them into numbers with equal exponents. e.g. 2.46 x 106 + 24.6 x 105 = 2.46 x 105+ 2.46 x 106= 4.92 x 106...
Read More →The value of
Question: The value of 3 x 10-7 is equal to_______ Solution: Given, 3 x 10-7 = 3.0 x 10-7 Now, placing decimal seven place towards left of original position, we get 0.0000003. Hence, the value of 3 x 10-7is equal to 0.0000003....
Read More →Find the 4th term from the end of the GP
Question: Find the $4^{\text {th }}$ term from the end of the GP $\frac{2}{27} \cdot \frac{2}{9}, \frac{2}{3}, \ldots, 162$ Solution: The given GP is $\frac{2}{27} \cdot \frac{2}{9}, \frac{2}{3}, \ldots, 162 . \rightarrow(1)$ The first term in the GP, $\mathrm{a}_{1}=\mathrm{a}=\frac{2}{27}$ The second term in the GP, $\mathrm{a}_{2}=\frac{2}{9}$ The common ratio, $r=3$ The last term in the given GP is $a_{n}=162$. Second last term in the GP $=a_{n-1}=a r^{n-2}$ Starting from the end, the series...
Read More →Prove the following
Question: If 36 = 6 6 = 62, then 1/36 expressed as a power with the base 6 is 6-2. Solution: Explanation: 36 = 6 6 = 62 1/36 = 1/62= 6-2...
Read More →The usual form of
Question: The usual form of 2.39461 106is 2394610. Solution: Explanation: 2.39461 106= 2.39461 10 10 10 10 10 10 = 239461 10 = 2394610...
Read More →The usual form of
Question: The usual form of 3.41 106is 3410000. Solution: Explanation: 3.41 106= 3.41 10 10 10 10 10 10 = 341 10 10 10 10 = 3410000...
Read More →The standard form of
Question: The standard form of 12340000 is 1.234 107 Solution: Explanation: 12340000 = 1234 104= 1.234 103 104= 1.234 107...
Read More →The standard form of
Question: The standard form of (1/100000000) is 1.0 10-8 Solution: Explanation: (1/100000000) = 1/1108= 1.0 10-8...
Read More →Find the
Question: Find the $6^{\text {th }}$ term from the end of GP $8,4,2 \ldots \frac{1}{1024} .$ Solution: The given GP is $8,4,2 \ldots \frac{1}{1024} \cdot \rightarrow(1)$ First term in the GP, $a_{1}=a=8$ Second term in the GP, $a_{2}=a r=4$ The common ratio, $r=\frac{4}{8}=\frac{1}{2}$ The last term in the given GP is $\frac{1}{1024}$. Second last term in the GP $=a_{n-1}=a r^{n-2}$ Starting from the end, the series forms another GP in the form, $a r^{n-1}, a r^{n-2}, a r^{n-3} \ldots a r^{3}, a...
Read More →Find the value
Question: Find the value [4-1+3-1+ 6-2]-1 Solution: [4-1+3-1+ 6-2]-1 = (1/4+1/3+1/62)-1 = [(9+12+1)/36]-1 = (22/36)-1 = (36/22)...
Read More →If the solve the problem
Question: If $y=x+\tan x$, show that: $\cos ^{2} x \frac{d^{2} y}{d x^{2}}-2 y-2 x=0$ 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 $f$ 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 simp...
Read More →Prove the following
Question: [(2/13)-6(2/13)3]3 (2/13)-9= (2/13)-36 Solution: Explanation: [(2/13)-6(2/13)3]3 (2/13)-9 = [(2/13)-6-3]3 (2/13)-9 = [(2/13)-9]3 (2/13)-9 = (2/13)-93 (2/13)-9 = (2/13)-27 (2/13)-9 = (2/13)-27-9 = (2/13)-36...
Read More →By multiplying
Question: By multiplying (10)5by (10)-10we get 10-5 Solution: Explanation: (10)5 (10)-10= 105+(-10)= 105-10= 10-5...
Read More →Very large numbers can be expressed
Question: Very large numbers can be expressed in standard form by usingexponents. Solution: Very large numbers can be expressed in standard form by using positive exponents, i.e. 23000 = 23 x 103=2.3 x103x 101=2.3 x 104...
Read More →The first term of a GP is -3 and the square of the second term is equal to its
Question: The first term of a GP is -3 and the square of the second term is equal to its $4^{\text {th }}$ term. Find its $7^{\text {th }}$ term. Solution: It is given that the first term of GP is -3. So, a = -3 It is also given that the square of the second term is equal to its 4th term. $\therefore\left(a_{2}\right)^{2}=a_{4}$ $n^{\text {th }}$ term of GP, $a_{n}=a r^{n-1}$ So, $a_{2}=a r ; a_{4}=a r^{3}$ $(a r)^{2}=a r^{3} \rightarrow a=r=-3$ Now, the $7^{\text {th }}$ term in the GP, a7 $=a ...
Read More →The 5th, 8th and 11th terms of a GP are a, b, c respectively
Question: The $5^{\text {th }}, 8^{\text {th }}$ and $11^{\text {th }}$ terms of a GP are $a, b, c$ respectively. Show that $b^{2}=a c$ Solution: It is given in the question that $5^{\text {th }}, 8^{\text {th }}$ and $11^{\text {th }}$ terms of GP are $a, b$ and $c$ respectively. Let us assume the GP is A, AR, AR $^{2}$, and AR3.... So, the $n^{\text {th }}$ term of this GP is $a_{n}=A R^{n-1}$ Now, $5^{\text {th }}$ term, $a_{5}=A R^{4}=a \rightarrow(1)$ $8^{\text {th }}$ term, $a_{8}=A R^{7}=...
Read More →Which of the following is not
Question: Which of the following is not the reciprocal of (2/3)4? (a) (3/2)4 (b) (3/2)-4 (c) (2/3)-4 (d) 34/24 Solution: (b) (3/2)-4 Explanation: (2/3)4= 1/(2/3)-4= (3/2)-4...
Read More →Cube of -1/2 is
Question: Cube of -1/2 is (a) 1/8 (b) 1/16 (c) -1/8 (d) -1/16 Solution: (c) -1/8 Explanation: Cube of -1/2 = (-1/2)3 = (-1/2) (-1/2) (-1/2) = -1/8...
Read More →If the solve the problem
Question: If $y=e^{-x} \cos x$, show that $: \frac{d^{2} y}{d x^{2}}=2 e^{-x} \sin x$ Solution: Basic idea: $\sqrt{S e c o n d}$ order derivative is nothing but derivative of derivative i.e. $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}\right)$ $\sqrt{T h e}$ 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. $f=v(u(x))$. F...
Read More →For a non-zero rational number z,
Question: For a non-zero rational number z, (z-2)3equal to (a) z6 (b) z-6 (c)z1 (d) z4 Solution: (b) z-6 (By the law of exponents: (am)n=amn)...
Read More →Find the GP whose 4th and 7th terms are
Question: Find the GP whose $4^{\text {th }}$ and $7^{\text {th }}$ terms are $\frac{1}{18}$ and $\frac{-1}{486}$ respectively. Solution: The $n^{\text {th }}$ term of a GP is $a_{n}=a r^{n-1}$ It's given in the question that $4^{\text {th }}$ term of the GP is $\frac{1}{18}$ and $7^{\text {th }}$ term of GP is $-\frac{1}{486}$. So, $a_{4}=a r^{3}=\frac{1}{18} \rightarrow(1)$ $a^{7}=a r^{6}=-\frac{1}{486} \longrightarrow(2)$ $\frac{(2)}{(1)} \rightarrow \frac{\operatorname{ar}^{6}}{\operatorname...
Read More →For a non-zero rational number p,
Question: For a non-zero rational number p, p13 p8is equal to (a) p5 (b) p21 (c) p-5 (d) p-19 Solution: (a) p5 (By law of exponent: (a)m(a)n= (a)m-n)...
Read More →