Question

For real number $a,b,c$ and $d$ , if $a^2+b^2=4$ and $c^2+d^2=1$, then possible value of $ac+bd$ is / are

• A. $2$
• B. $3$
• C. $1$
• D. $\frac{1}{4}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $2$
C $1$
D $\frac{1}{4}$

According to Cauchy-Schwarz inequality:

$(ac+bd{)}^2\le (a^2+b^2\left)\right(c^2+d^2)$ $(\forall a,b,c,d\in R)$

Substituting the values here gives:

$(ac+bd{)}^2\le 4$

$(ac+bd{)}^{\le }2$

Therefore, it can take all values given in the options except 3.