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यदि व्यंजक `(x^(2))/(y^(2))+tx+(y^(2))/4` एक पूर्ण वगै है तो `t` का मान होगा?A. `+-1`B. `+-2`C. `0`D. `+-3` |
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Answer» Correct Answer - A `(x^(2))/(y^(2))+tx+(y^(2))/4` (given) To make it a perfect square it should be in the form. `A^(2) +-2AB+B^(2)=(A+-B)^(2)` ltbr `=(x/y)^(2)+-tx+(y/2)^(2)` `= A^(2) +-2AB+B^(2)` `A=x/y,B=y/2` & `2AB=tx` So `tx=+-2xx x/y xx y/2` `tx=+-x` `t=+-1` |
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