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Question : How to solve following up pgt mathematics question paper solution Solution : Let \[z=i\left(\frac{3+i}{2-i}+\frac{3-i}{2+i}\right)\] \[=i\left(1+i+1-i\right)\] \[z=2i\] \[=2\left(\cos{\left(\frac{\pi}{2}\right)}+i\sin{\left(\frac{\pi}{2}\right)}\right) \] So \[Amplitude(z)=\frac{\pi}{2}\] Hence correct option is (1)

Question : How to calculate amplitude of a complex number in the following problem Solution : Let the complex number as \[z=i\left(\frac{3+i}{2-i}+\frac{3-i}{2+i}\right)\] simplyfying we get \[z=i\left(1+i+1-i\right)\] or \[z=2i\] writing it…

Question : Find solution of the question of up pgt mathematics previous years question paper Solution : By the given expression \[limit_{h\rightarrow0}\frac{f\left(c+h\right)+f\left(c-h\right)-2f\left(c\right)}{h}\] we shall write this as \[=limit_{h\rightarrow0}\frac{f\left(c+h\right)-f(c)+f\left(c-h\right)-f(c)}{h}\] which becomes…

Question : Solve following up pgt mathematics previous years question Solution : given function \[y=\tan^{-1}{\left(\frac{3a^2x-x^3}{a\left(a^2-3x^2\right)}\right)\ }\] to find derivative of y \[\frac{dy}{dx}=\frac{1}{1+\frac{\left(3a^2x-x^3\right)^2}{a^2\left(a^2-3x^2\right)^2}\ }\frac{d}{dx}\left(\frac{3a^2x-x^3}{a\left(a^2-3x^2\right)}\right)\] simplifying this we get \[=\frac{1}{1+\frac{\left(3a^2x-x^3\right)^2}{a^2\left(a^2-3x^2\right)^2}\ }\left(\frac{3a^2-3x^2}{a(a^2-3x^2)}+\frac{6x(3a^2x-x^3)}{a{(a^2-3x^2)}^2}\right)\] after…

Question : Find solution of the following question of pgt mathematics question paper Solution : From continuity \[f\left(1\right)=-\frac{1}{\left|1\right|}=-1\] \[limit_{h\rightarrow0}f\left(1+h\right)=limit_{h\rightarrow0}-\frac{1}{\left|1+h\right|}=-1\] \[limit_{h\rightarrow0}f\left(1-h\right)=limit_{h\rightarrow0}a\left(1-h\right)^2-b =a-b\] f is continuous at x =1 if \[a-b=-1\] Now…

Question : Solve following up pgt math previous years question Solution: This function is discontinuous at 3 points at x=-1,0,1. Since \[\log{\left|x\right|}\] is not defined at x=0, so the function…

Question : Find solution of UP PGT mathematics previous years question Solution : To evaluate \[limit_{x\rightarrow\infty}\left(\frac{x}{1+x}\right)^{2x}=\left(limit_{x\rightarrow\infty}\left(\frac{1+x-1}{1+x}\right)^x\right)^2\] We can write it as \[=\left(limit_{1+x\rightarrow\infty}\left(1-\frac{1}{1+x}\right)^{1+x}\right)^2 \] and this becomes \[=\left(e^{-1}\right)^2\] which is equal…

Question : Find solution of the following UP PGT mathematics previous years question paper Solution : First we shall check whether f is One -One \[f\left(x_1\right)=f\left(x_2\right)\] \[x_1^2\ =x_2^2\] \[x_1^2-x_2^2=0\] \[\left(x_1-x_2\right)\left(x_1+x_2\right)=0\]…

Question : Find solution of the following UP PGT math previous years question papers solution Solution : given sets \[A=\left\{1,2,3,4,5,6,7,8,9\right\}\] and \[B=\left\{2,3,5,7\right\}\] Number of elements in A and B are…

Question : Find solution of the following UP PGT math previous years Question paper solution Solution : Since \(a^2\) and \(a^2\) are not prime to each other hence R is…