JEE Main 202227 Jun, Evening shiftMathematicsMediumMatrices and Determinants245× since 2002Let f(x)=∣a−10axa−1ax2axa∣, a∈Rf(x) = \left| {\matrix{ a & { - 1} & 0 \cr {ax} & a & { - 1} \cr {a{x^2}} & {ax} & a \cr } } \right|,\,a \in Rf(x)=aaxax2−1aax0−1a,a∈R. Then the sum of the squares of all the values of a, for which 2f′(10)−f′(5)+100=02f'(10) - f'(5) + 100 = 02f′(10)−f′(5)+100=0, isAOption A. 117BOption B. 106COption C. 125DOption D. 136CheckSkip · show solution
