1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Assertion (A) If $p{x}^{2}-2x+2=0$ has real roots, then $p\le \frac{1}{2}.$ Reason (R) The equation $\left({a}^{2}{+}^{b}\right){x}^{2}+2\left(ac+bd\right)x+\left({c}^{2}+{d}^{2}\right)=0$ has no real root, if ad ≠ bc. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A). (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A). (c) Assertion (A) is true and Reason (R) is false. (d) Assertion (A) is false and Reason (R) is true.

Open in App
Solution

## (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A). $\text{For no real roots, we have:}\phantom{\rule{0ex}{0ex}}\left({b}^{2}-4ac\right)<0\phantom{\rule{0ex}{0ex}}\text{Now,}\phantom{\rule{0ex}{0ex}}\text{4}{\left(ac+bd\right)}^{2}\text{}-4\left({a}^{2}+{b}^{2}\right)\left({c}^{2}+{d}^{2}\right)\phantom{\rule{0ex}{0ex}}=4{a}^{2}{c}^{2}+4{b}^{2}{d}^{2}+8abcd-4{a}^{2}{c}^{2}-4{a}^{2}{d}^{2}-4{b}^{2}{c}^{2}-4{b}^{2}{d}^{2}\phantom{\rule{0ex}{0ex}}=-4\left({a}^{2}{d}^{2}-2abcd+{b}^{2}{c}^{2}\right)\phantom{\rule{0ex}{0ex}}=-4{\left(ad-bc\right)}^{2}<0,\text{when}\left(ad-bc\right)\text{}\ne 0⇒ad\ne bc\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\text{Reason}\left(\text{R}\right)\text{is true}\text{.}\phantom{\rule{0ex}{0ex}}\text{Now,}\phantom{\rule{0ex}{0ex}}p{x}^{2}-2x+2=0has realroots;\mathrm{therefore},\mathrm{we}\mathrm{have}:\phantom{\rule{0ex}{0ex}}\left({b}^{2}-4ac\right)\ge 0\phantom{\rule{0ex}{0ex}}⇒4-4×p×2\ge 0\phantom{\rule{0ex}{0ex}}⇒4-8p\ge 0\phantom{\rule{0ex}{0ex}}⇒8p\le 4\phantom{\rule{0ex}{0ex}}⇒p\le \frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\text{Assertion}\left(\text{A}\right)\text{is true}\text{.}\phantom{\rule{0ex}{0ex}}\text{Thus, both Reason}\left(\text{R}\right)\text{and Assertion}\left(\text{A}\right)\text{are true but Reason}\left(\text{R}\right)\text{is not a correct}\phantom{\rule{0ex}{0ex}}\text{explanation of Assertion}\left(\text{A}\right).\phantom{\rule{0ex}{0ex}}$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Animalia
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program