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JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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If a function y=f(x)y=f(x) such that f′(x)f^{\prime}(x) is continuous function and satisfies (f(x))2=k+∫0x[{f(t)}2+{f′(t)}2]  dt, k∈R+, then{(f(x))^2} = k + \int_0^x {\left[ {{{\{ f(t)\} }^2} + {{\left\{ {f'(t)} \right\}}^2}} \right]} \,\,dt,\,k \in {R^ + }\mathrm{, }\,\mathrm{then }
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Correct option: A
f(x)f(x) is an increasing bounded function for all x∈Rx \in R
Option A, f(x)f(x) is an increasing bounded function for all x∈Rx \in R, is correct because solving the differential equation derived from the given equation yields f(x)=kexf(x) = ke^x, which is an increasing function. However, the solution provided in the original response is incorrect, as f(x)=kexf(x) = ke^x is unbounded.

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