If the solve the problem
Question: If $y=\frac{\log x}{x}$, show that $\frac{d^{2} y}{d x^{2}}=\frac{2 \log x-3}{x^{3}}$ Solution: Basic 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. $f=v(u(x))$. For the sake of...
Read More →Prove that
Question: If $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}(x \neq 0)$ then show that a, b, c, d are in GP. Solution: $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}$ (Given data in the question) $\rightarrow$ (1) Cross multiplying (1) and expanding $(a+b x)(b-c x)=(b+c x)(a-b x)$ $a b-a c x+b^{2} x-b c x^{2}=b a-b^{2} x+a c x-b c x^{2}$ $2 b^{2} x=2 a c x$ $b^{2}=a c \rightarrow(i)$ If three terms are in GP, then the middle term is the Geometric Mean of first term and last term. $\righta...
Read More →Suppose for the principal P, rate R% and time T,
Question: Suppose for the principal P, rate R% and time T, the simple interest is S and compound interest is C. Consider the possibilities (i) CS (ii) C = S (iii) C S Then, (a) only (i) is correct (b) either (i) or (ii) is correct (c) either (ii) or (iii) is correct (d) only (iii) is correct Solution: (a) only (i) is correct. Explanation: Let Principal, P = Rs. 100, Rate = 10% and Time = 1 year Simple interest (SI)= (PRT)/100 = (100101)/100 = Rs.10 Since, Amount = P(1+R/100)T=100(1+10/100)1= 100...
Read More →In a finite GP, prove that the product of the terms equidistant from the
Question: In a finite GP, prove that the product of the terms equidistant from the beginning and end is the product of first and last terms. Solution: We need to prove that the product of the terms equidistant from the beginning and end is the product of first and last terms in a finite GP. Let us first consider a finite GP. $\mathrm{A}, \mathrm{AR}, \mathrm{AR}^{2} \ldots \mathrm{AR}^{\mathrm{n}-1}, \mathrm{AR}^{\mathrm{n}}$ Where n is finite. Product of first and last terms in the given GP $=A...
Read More →If the solve the problem
Question: If $y=2 \sin x+3 \cos x$, show that: $\frac{d^{2} y}{d x^{2}}+y=0$ Solution: Basic idea: $\sqrt{S e c o n d}$ order derivative is nothing but derivative of derivative i.e. $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$ The 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...
Read More →The third term of a GP is 4; Find the product of its five terms.
Question: The third term of a GP is 4; Find the product of its five terms. Solution: Given that the third term of the GP, a3 = 4 Let us assume the GP mentioned in the question be $\frac{\mathrm{A}}{\mathrm{R}^{2}}, \frac{\mathrm{A}}{\mathrm{R}}, \mathrm{A}, \mathrm{AR}, \mathrm{AR}^{2}$ With the first term $\frac{\mathrm{A}}{\mathrm{R}^{2}}$ and common ratio $\mathrm{R}$. Now, the third term in the assumed GP is A. So, A = 4 (given data) Now Product of the first five terms of $G P=\frac{A}{R^{2}...
Read More →The distance between the Sun and the Earth is
Question: The distance between the Sun and the Earth is 1.496 x 108km and : distance between the Earth and the Moon is 3.84 x 108m. During solar eclipse, the Moon comes in between the Earth and the Sun. What is the distance between the Moon and the Sun at that particular time? Solution: The distance between the Sun and the Earth is 1.496 x 10s km = 1.496 x108x103m = 1496 x108m The distance between the Earth and the Moon is 3.84 x108m. The distance between the Moon and the Sun at particular time ...
Read More →Mass of Mars is
Question: Mass of Mars is 6.42 x 1029kg and mass of the Sun is 1.99 x 1030kg. What is the total mass? Solution: Mass of Mars = 6.42 x 1029kg Mass of the Sun = 1.99 x 1030kg Total mass of Mars and Sun together = 6.42 x 1029+ 1.99 x 1030 = 6.42 x 1029+ 19.9 x 1029= 26.32 x 1029kg...
Read More →If the solve the problem
Question: If $y=\log (\sin x)$, prove that: $\frac{d^{3} y}{d x^{2}}=2 \cos x \operatorname{cose}^{3} x$ 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(...
Read More →Find three machines that can be replaced
Question: Find three machines that can be replaced with hook-up of (x 5) machines. Solution: Since, 52= 25, 53= 125, 54= 625 Hence, (x 52), (x 53)and (x 54) machine can replace (x 5) hook-up machine....
Read More →Find two repeater machines
Question: Find two repeater machines that will do the same work as a (x 81) machine. Solution: Two repeater machines that do the same work as (x 81) are (x 34) and (x 92). Since, factor of 81 are.3 and 9....
Read More →There are 86400 sec in a day.
Question: There are 86400 sec in a day. How many days long is a second? Express your answer in scientific notation. Solution: Total seconds in a day = 86400 So, a second is long as 1/86400 = 0.000011574 Scientific notation of 0.000011574= 1.1574 x10-5days...
Read More →One fermi is equal to
Question: One fermi is equal to10-15 metre. The radius of a proton is 1.3 fermi. Write the radius of a proton (in metres) in standard form. Solution: The radius of a proton is 1.3 fermi. One fermi is equal to10-15m. So, the radius of the proton is 1.3 x10-15m.Hence, standard form of radius of the proton is 1.3 x 10-15m....
Read More →An inch is approximately equal
Question: An inch is approximately equal to 0.02543 metres. Write this distance in standard form. Solution: Standard form of 0.02543 m = 0.2543 x10-1m = 2.543 x 10-2m Hence, standard form of 0.025434s 2.543 x 10-2m....
Read More →The planet Uranus is approximately 2,896,819,200,000
Question: The planet Uranus is approximately 2,896,819,200,000 metres away from the Sun. What is this distance in standard form? Solution: Distance between the planet Uranus and the Sun is 2896819200000 m. Standard form of 2896819200000 = 28968192 x 10 x 10 x 10 x 10 x 10 = 28968192 x 105= 2.8968192 x 1012m...
Read More →Planet A is at a distance of
Question: Planet A is at a distance of 9.35 x 106km from Earth and planet B is 6.27 x 107 km from Earth. Which planet is nearer to Earth? Solution: Distance between planetA and Earth = 9.35 x 106km Distance between planet B and Earth = 6.27 x 107km For finding difference between above two distances, we have to change both in same exponent of 10, i.e. 9.35 x.106= 0.935 x 107, clearly 6.27 x 107is greater. So, planet A is nearer to Earth....
Read More →A new born bear weights 4 kg.
Question: A new born bear weights 4 kg. How many kilograms might a five year old bear weight if its weight increases by the power of 2 in 5 yr? Solution: Weight of new born bear = 4 kg Weight increases by the power of 2 in 5 yr. Weight of bear in 5 yr = (4)2= 16 kg...
Read More →The number of red blood cells
Question: The number of red blood cells per cubic millimetre of blood is approximately mm3) Solution: The average body contain 5 L of blood. Also, the number of red blood cells per cubic millimetre of blood is approximately 5.5 million. Blood contained by body = 5 L = 5 x 100000 mm3 Red blood cells = 5 x 100000 mm3 Blood = 5.5 x 1000000 x 5 x 100000= 55 x 5 x105 + 5 = 275 x1010= 2.75 x 1010x 102= 2.75 x 1012...
Read More →Write 0.000005678 in the standard form.
Question: Write 0.000005678 in the standard form. Solution: For standard form, 0.000005678 = 0.5678 x 10-5= 5.678 x 10-5x 10-1= 5.678 x 10-6Hence, 5.678 x 10-6is the standard form of 0.000005678....
Read More →Solve this
Question: If $a, b, c$ are the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of a GP, show that $(q-r) \log a+(r-p) \log b+(p-q) \log c=0$ Solution: As per the question, $a, b$ and $c$ are the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ term of GP. Let us assume the required $\mathrm{GP}$ as $\mathrm{A}, \mathrm{AR}, \mathrm{AR}^{2}, \mathrm{AR}^{3} \ldots$ Now, the $\mathrm{n}^{\text {th }}$ term in the GP, $\mathrm{a}_{\mathrm{n}}=\mathrm{AR}^{\mathrm{n}-1}$ $p^{\...
Read More →Large numbers can be expressed
Question: Large numbers can be expressed in the standard form by using positive exponents. Solution: True e.g. 2360000 = 236 x 10 x 10 x 10 x 10= 236 x 104 = 2.36 x104x 102=2.36 x106...
Read More →The standard form for
Question: The standard form for 203000 is 2.03 x 105. Solution: True For standard form, 203000 = 203 x 10 x 10 x 10 = 203 x 103 = 2.03 x 102x 103= 2.03 x105...
Read More →The standard form for
Question: The standard form for 0.000037 is 3.7 x 10-5 Solution: True For standard form, 0.000037 = 0.37 x 10-4= 3.7 x 10-5...
Read More →Prove the following
Question: 24.58 = 2 x 10 + 4 x 1+5 x 10 + 8 x 100 Solution: FalseR H S = 2 x 10+ 4 x 1+ 5 x 10+ 8 x 100=20+ 4 + 50+ 800=874 L H S R H S...
Read More →The usual form for
Question: The usual form for 2.3 x10-10is_______. Solution: For usual form, 2.3 x10-10= 0.23 x 10-11= 0.00000000023Hence, the usual form for 2.3 x 10-10is 0.00000000023....
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