41 The derivative 43 Example 49 Define f R → R by f(x) = x2 sin(1/x) if x ̸= 0, 0 if x = 0 Then f is differentiable on R (See Figure 1) It follows from the product and chain rules proved below that f is differentiable at x ̸= 0 with derivative f′(x) = 2xsin 1 x −cos 1 x Moreover, f is differentiable at 0 with f′(0) = 0, since limClick here👆to get an answer to your question ️ If y = f (2x 1/x^2 1 ) and f^'(x) = sin x^2 , then dy/dx = (IIT JEE, 19)Essentially correct, but more complicated than required You make some confusion about k and j, though Suppose x and y are rational solutions of x^2y^2=3^k, with k an odd integer
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If y=f(x)=x+2/x-1 then x is equal to-I=1 A i for some integer n If n = 2, then X = A 1 ∪A 2, A 1 and A 2 are closed in X, fA 1 A 1 → Y and fA 2 A 2 → Y are continuous and fA 1(x) = fA 2(x) for every x ∈ A 1∩A 2 Hence, by the pasting lemma, we can construct continuous f0 X → Y such that f0(x) = fA 1(x) if x ∈ A 1 and f0(x) = fA 2(x) if x ∈ A 2 It is clearIf x = 0, y =0 Stepbystep explanation Hi, to obtain the values of the variable y in each case, we have to replace the variable x
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If f0 is a homeomorphism of X with Z, then f X → Y is a topological imbedding (or simply imbedding) of X in Y Example 5 Consider F (−1,1) → R defined by F(x) = x/(1 − x2) Then F is continuous and one to one (since F0(x) = (1x2)/(1−x2)2 ≥ 0) and is continuous on R So F is a homeomorphism (where R has the standard topologyClick here👆to get an answer to your question ️ Let f N→ X f(x) = 4x^2 12x 15 Then, f^1(y) = ?(b) f(x,y) = 2x2 xy2 −2, D= {(x,y) ∈ R2;x2 y2 64} Solution (a) Calculating the partial derivatives gives f x = y 2 √ x − 1, f y = √ x−2y 6 Finding critical points f x = 0, f y = 0 =⇒ (x,y) = (4,4) Note that (4,4) ∈ Dand that f(4,4) = 12 The boundary of Dconsists of
If xy=2 and x−y=1, then what is xy?Solutionput (xy)=v then differentiate both sides with respect to 'x' we get 1dy/dx=dv/dx or, dy/dx=dv/dx—1 this value put in to equation (I), first arranging equation (I) dy/dx= (xy1) (xy—2)/ (xy2) (xy—1) or, dv/dx—1= (v1) (v—2)/ (v2) (v—1) or,dv/dx= (v^2—2vv—2)/ (v^22v—v—2) 1X 1 6=x 2 Then f(x;y) = x1 2Fhas di erent values on P 1 and P 2, so Fseparates points Since Ris compact, the StoneWeierstrass theorem implies that Fis dense in C(R) Alternatively, note that Fincludes all polynomials in (x;y), and the polynomials are dense in C(R), so Fis dense in C(R) 7
ProofLet fK g 2A be a family of convex sets, and let K = \ 2AK Then, for any x;y2 K by de nition of the intersection of a family of sets, x;y2 K for all 2 nd each of these sets is convex Hence for any 2 A;and 2 0;1;(1 )x y2 KDivide \frac{f1}{f}, the coefficient of the x term, by 2 to get \frac{1}{2}\frac{1}{2f} Then add the square of \frac{1}{2}\frac{1}{2f} to both sides of the equationThis is the Solution of Question From RD SHARMA book of CLASS 11 CHAPTER RELATIONS AND FUNCTIONS This Question is also available in R S AGGARWAL book of CLAS
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R X f X(x) dx= 1 Alternately, X may be described by its cumulative distribution function (CDF) The CDF of Xis the function F X(x) that gives, for any specified number x∈X, the probability that theIf x = −1, y =1 ;X y = 2 (equation 1) x y = 1 (equation 2) Add the equations for 2x = 3, meaning that x = 3/2 Using x = 3/2 in equation 1 gives y = 1/2 Using x = 3/2 in equation 2 also gives y = 1/2 (3/2)*(1/2) = 3/4 Answer xy = 3/4
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Given function f(x) =x/(x1) Then f(1/x) = (1/x)/(1/x) 1 f(1/x) = (1/x)/(1 x)/x f(1/x) = (x/x)/(1 x) f(1/x) = 1/(1 x) Hence, f(x) =x/(x1) then f(1/x) = 12 (a) Define uniform continuity on R for a function f R → R (b) Suppose that f,g R → R are uniformly continuous on R (i) Prove that f g is uniformly continuous on R (ii) Give an example to show that fg need not be uniformly continuous on R Solution • (a) A function f R → R is uniformly continuous if for every ϵ > 0 there exists δ > 0 such that f(x)−f(y) < ϵ for all xThe total area under f X(x) is 1;
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94 7 Metric Spaces Then d is a metric on R Nearly all the concepts we discuss for metric spaces are natural generalizations of the corresponding concepts for R with this absolutevalue metric Example 74 Define d R2 ×R2 → R by d(x,y) = √ (x1 −y1)2 (x2 −y2)2 x = (x1,x2), y = (y1,y2)Then d is a metric on R2, called the Euclidean, or ℓ2, metricIt corresponds toBASIC STATISTICS 3 We can then write 9 as i=1 (xi − a)2 = i=1 (xi −x¯)2 i=1 (¯x− a)2 (11) Equation 11 is clearly minimized when a =¯xNow consider part b of theorem 1 Expand the second expression in part b and simplifyY1 n 1 f X(y 1 n) where, f X() is the pdf of X which is given Here are some more examples Example 1 Suppose Xfollows the exponential distribution with = 1 If Y = p X nd the pdf of Y Example 2 Let X ˘N(0;1) If Y = eX nd the pdf of Y Note Y it is said to have a lognormal distribution Example 3 Let Xbe a continuous random variable with
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The partition theorem says that if Bn is a partition of the sample space then EX = X n EXjBnP(Bn) Now suppose that X and Y are discrete RV's If y is in the range of Y then Y = y is a event with nonzero probability, so we can use it as the B in the aboveM 2, and M 3 be metric spaces Let gbe a uniformly continuous function from M 1 into M 2, and let fbe a uniformlyAnswer to If f(x)= g(h(x)), then f'(4) = 5 9 8 4 7 6 3 5 yvalues yvalues 4 h(x) 2 3 8(x) 2 1 1 1 4 5 1 4 5 2 3 xvalues
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5k views asked in Class XI Maths by vijay Premium (539 points) If (1i) 2 / 2i = xiy, then find the value of xy complex numbers and quadratic equationsAssuming that f is nonzero, we can rewrite this equation as 1 f(x) 1 f(1 / x) = 1 Making the substitution g(y) = − 1 2 1 f ( ey), we get the functional equation g(y) g( − y) = 0, which simply says that g is odd This equation can be solved, with the initial condition, by setting g(y) = cy for a suitable constant c Answer If x = −2, y =8 ;
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1 Let y = x 1 x, try now to express x as a function of y We have x 2 − x y 1 = 0 x = y ± y 2 − 4 2 Substitute this value for x in your expression for f f ( y) = ( y ± y 2 − 4 2) 2 ( 2 y ± y 2 − 4) 2 f ( y) = y 2 − 2Figure 1 A plot of the function y = x2 sin(1=x) and a detail near the origin with the parabolas y = x2 shown in red Then fis di erentiable on R (See Figure 1) It follows from the product and chain rules proved below that fis di erentiable at x6= 0 with derivative f0(x) = 2xsin 1 x cos 1 x Moreover, fis di erentiable at 0 with f0(0) = 0F X(x) ≥0 for all values of x∈X 2The rule of total probability holds;
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Clearly f(x) =x^2 for x>/=0, and f(x)= x^2 for x0 For y/=0, g(f(x))=sqrt(x^2)=x and f(g(y)) =(sqrt(y))^2 =y, as x>/=0 => y>/=0 If 8 f(x) 6f(1/x) = x 5 and y = x2 f(x) then (dy/dx) at x = – 1 is (a) {(– 1)/(14)} (b) 0 (c) (1/14) (d) none of there Welcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get solutions to their queriesSelect a few x x values, and plug them into the equation to find the corresponding y y values The x x values should be selected around the vertex Tap for more steps Replace the variable x x with − 2 2 in the expression f ( − 2) = ( − 2) 2 2 ( − 2) − 1 f ( 2) = ( 2) 2 2 ( 2) 1
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Domain RR{1,3} Range RR # To Find the domain Equate the denominator(x^22x3) to zero, then solve the equation for x rarr x^22x3=0 rarr x=((2)sqrt((2)^24*(1)*(3)))/(2*1) rarr x= 12 => x= 1 and x=3 This means that, when x=1 or 3, we have the x^22x3=0 Implying that f(x)=color(red)(x/0) which is undefined Hence, the domain is all real numbers except 11The PDF f X(x) is positivevalued;(b) Pick = 1 Given any >0, pick x>0 such that 3 x2 2 >1 Then d(x 2;x) < but we have d(f(x 2);f(x)) = j(x 2)3 x3j= j 3 x2 2 3 2x 22 3 23 j 3 x2 2 >1 This shows that f(x) = x3 is not uniformly continuous on R 445 Let M 1;
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If the eccentricity of the hyperbola $\frac {x^2}{a^2}\frac {y^2}{b^2}=1$ is $\frac {5}{4}$ and $2x 3y 6 = 0$ is a focal chord of the hyperbola, then theGiven a function g with this property, we can easily construct a suitable f Just let f ( x) = { g ( x) x ≥ 0 g ( − x) x < 0 If g is additionally continuous then so is f We can find a lot of continuous g Pick a 1 ∈ ( 0, 1), let a 0 = 0 and recursively a n = a n − 2 2 1 for n ≥ 22x2Thenwecanpluginfory to get f(x,y)=f(x)= ±x p 2x2 The boundary's critical points are precisely those values of x for which 0=f0(x)=⌥ 2(x2 1) p 2x2 This is only true when x = ±1 We then find the corresponding values of y and find the extreme points on the boundary are (1,1),(1,1),(1,1), and (1,1) • Alternatively, we could
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To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW If `2f (x1)f((1x)/x)=x , then f(x)` is If y = sin^1 x/√(1 x^2), then show that (1 x^2)d^2y/dx^2 3xdy/dx y = 0 asked in Mathematics by Samantha ( 3k points) continuity and differntiabilitySolution for If f(x, y) ha then 2y (11) = (1,1) = 3 Select one True O False Q Find the period of f(x) = tan(4x) 2 and sketch its graphChose a scale for the axes that show at A The objective is to find the period of the given trigonometric function and sketch the graph
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If f(x y) = f(x) × f(y) ∀ x, y and f(5) = 2, f '(0) = 3, then f '(5) is (a) 0 (b) 1 (c) 6 (d) 2 Find the set of values of x satisfying sin x cos x = 1, ∀ x ∈ 0, 2πX−y2 − x 6y, D= {(x,y) ∈ R2;0 6x69,0 6y65};Can take y = 2 Then if f(x) = 2x1 = 2, then x = 1/2 ∈/ Z Example 24 Define f Z → Z by f(n) = n5 Show that f is onto Suppose that y ∈ Z is an arbitrary integer We need to show that there exists x ∈ Z such that f(x) = y However take x = y − 5 ∈ Z Then f(x) = x5 = y −55 = y It follows that f is onto 3 OnetoOne Correspondence
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Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack ExchangeDefinition X is a continuous random variable if there is a function f(x) so that for any constants a and b, with −∞ ≤ a ≤ b ≤ ∞, P(a ≤ X ≤ b) =Z b a f(x) dx (1) • For δ small, P(a ≤ X ≤ a δ) ≈ f(a) δ • The function f(x) is called the probability density function (pdf) • For any a, P(X = a) = P(a ≤ X ≤ a) =R a a f(x) dx = 0 • A discrete randomPutting x=1y in first equation f (1y)2f (1 1y)= (1y)^22 eq3 Multiply by 2 in eq3 we get 2f (1y)4f (y)=2× (1y)^2 2×2 eq4 Adding eq4 and eq2 we get 3f (y)=y^2 2 2× (1y)^2 4 f (y)= — (3×y^2–4y8)/3 Then put y=x again to have f (x) f (x)= — (3×x^2–4x8)/3
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