| A continuous function f(x) satisfying |
| (i) \[{{x}^{4}}-4{{x}^{2}}\le f(x)\le 2{{x}^{2}}-{{x}^{3}}\] |
| (il) area bounded by\[y=f(x),\text{ }y={{x}^{4}}-4{{x}^{2}},\], y - axis and line \[x=t(0\,\,\le \,\,t\,\,\le \,\,2)\] is k times the area bounded by\[y=f(x),\,\,y=2{{x}^{2}}-{{x}^{3}}\], y - axis and line \[x=t(0\le t\le 2)\], is given as |
A) \[f(x)=\frac{{{x}^{4}}-k{{x}^{3}}+(2k-4){{x}^{2}}}{1+k}\]
B) \[f(x)=\frac{{{x}^{4}}+{{x}^{3}}+k{{x}^{2}}}{k+1}\]
C) \[f(x)=\frac{k{{x}^{4}}-{{x}^{3}}-2k{{x}^{2}}}{k+1}\]
D) \[f(x)=\frac{k}{k+1}({{x}^{4}}-{{x}^{2}})\]
Correct Answer: A
Solution :
| \[\int\limits_{0}^{t}{(f(x)-({{x}^{4}}-4{{x}^{2}}))dx=k\int\limits_{0}^{t}{(2{{x}^{2}}-{{x}^{3}}-f(x))dx}}\] |
| Differentiating w.r.t. t |
| \[f(t)=\frac{{{t}^{4}}-k{{t}^{3}}+(2k-4){{t}^{2}}}{1+k}\] |
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