Question

Assertion (A)
If $p{x}^2-2x+2=0$ has real roots, then $p\le \frac{1}{2}.$

Reason (R)
The equation $\left(a^2{+}^b\right)x^2+2(ac+bd)x+(c^2+d^2)=0$ 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.

Answer 1

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

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