A) -1
B) 1
C) \[log\text{ }2\]
D) \[-log\text{ }2\]
Correct Answer: A
Solution :
[a] \[{{x}^{2x}}-2{{x}^{x}}\cot y-1=0\] (i) Now at x=1, \[1-2\cot y-1=0\Rightarrow \cot y=0\Rightarrow y=\frac{\pi }{2}\] Now differentiating (i) w.r.t,x, we get \[2{{x}^{2x}}(1+logx)-2\left[ {{x}^{x}}(-cose{{c}^{2}}y)\frac{dy}{dx}+\cot y{{x}^{x}}(1+logx) \right]=0\] Now at \[(1,\pi /2)\], \[2(1+log1)-2\left[ 1(-1){{\left( \frac{dy}{dx} \right)}_{1,\pi /2}}+0 \right]=0\] \[\Rightarrow 2+2{{\left( \frac{dy}{dx} \right)}_{(1,\pi /2)}}=0\] \[\Rightarrow {{\left( \frac{dy}{dx} \right)}_{(1,\pi /2)}}=-1\]
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