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If a<0, b>0, then sqrta. sqrtb is equal ...

If `a<0, b>0,` then `sqrta. sqrtb` is equal to :

A

`isqrt(|a|b)`

B

`isqrt(|a||b|)`

C

`isqrt(|a||b|)`

D

`-sqrt(|a||b|)`

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The correct Answer is:
To solve the problem, we need to analyze the expression \(\sqrt{a} \cdot \sqrt{b}\) given the conditions \(a < 0\) and \(b > 0\). ### Step-by-Step Solution: 1. **Identify the Values of a and b**: - We know that \(a < 0\) means \(a\) is a negative number. - We know that \(b > 0\) means \(b\) is a positive number. 2. **Understanding the Square Roots**: - The square root of a negative number is not a real number. Instead, it is expressed in terms of imaginary numbers. - Specifically, \(\sqrt{a}\) can be written as \(\sqrt{-|a|} = i\sqrt{|a|}\) since \(a\) is negative. 3. **Calculating \(\sqrt{a}\)**: - Since \(a < 0\), we can express \(\sqrt{a}\) as: \[ \sqrt{a} = \sqrt{-|a|} = i\sqrt{|a|} \] 4. **Calculating \(\sqrt{b}\)**: - Since \(b > 0\), the square root of \(b\) is simply: \[ \sqrt{b} = \sqrt{b} \] 5. **Multiplying the Square Roots**: - Now, we multiply \(\sqrt{a}\) and \(\sqrt{b}\): \[ \sqrt{a} \cdot \sqrt{b} = (i\sqrt{|a|}) \cdot (\sqrt{b}) = i\sqrt{|a|b} \] 6. **Final Expression**: - Therefore, the final expression for \(\sqrt{a} \cdot \sqrt{b}\) is: \[ \sqrt{a} \cdot \sqrt{b} = i\sqrt{|a|b} \] ### Conclusion: Thus, if \(a < 0\) and \(b > 0\), then \(\sqrt{a} \cdot \sqrt{b} = i\sqrt{|a|b}\).

To solve the problem, we need to analyze the expression \(\sqrt{a} \cdot \sqrt{b}\) given the conditions \(a < 0\) and \(b > 0\). ### Step-by-Step Solution: 1. **Identify the Values of a and b**: - We know that \(a < 0\) means \(a\) is a negative number. - We know that \(b > 0\) means \(b\) is a positive number. ...
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OBJECTIVE RD SHARMA ENGLISH-COMPLEX NUMBERS -Chapter Test
  1. If a<0, b>0, then sqrta. sqrtb is equal to :

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  2. The locus of the center of a circle which touches the circles |z-z1|=a...

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  3. Prove that for positive integers n(1) and n(2), the value of express...

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  4. The value of abs(sqrt( 2i) - sqrt(2i)) is :

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  5. Prove that the triangle formed by the points 1,(1+i)/(sqrt(2)),a n di ...

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  6. The value of ((1+ i sqrt(3))/(1-isqrt(3)))+ ((1-isqrt(3))/(1+isqrt(3)...

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  7. If alpha+ibeta=tan^(-1) (z), z=x+iy and alpha is constant, the locus o...

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  8. If cosA+cosB+cosC=0,sinA+sinB+sinC=0andA+B+C=180^(@) then the value of...

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  9. Find the sum 1xx(2-omega)xx(2-omega^(2))+2xx(-3-omega)xx(3-omega^(2))+...

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  10. The value of the expression (1+(1)/(omega))+(1+(1)/(omega^(2)))+(2+(1)...

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  11. The condition that x^(n+1)-x^(n)+1 shall be divisible by x^(2)-x+1 is ...

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  12. The expression (1+i)^(n1)+(1+i^(3))^(n2) is real iff

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  13. If |{:(6i,3i,1),(4,3i,-1),(20,3,i):}|=x+iy, then (x, y) is equal to

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  14. If cosalpha+2cosbeta+3cosgamma=sinalpha+2sinbeta+3singamma=0,t h e nt ...

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  15. If cosalpha+2cosbeta+3cosgamma=sinalpha+2sinbeta+3singamma=0,t h e nt ...

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  16. Sum of the series sum(r=0)^n (-1)^r ^nCr[i^(5r)+i^(6r)+i^(7r)+i^(8r)] ...

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  17. If az(1)+bz(2)+cz(3)=0 for complex numbers z(1),z(2),z(3) and real num...

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  18. If 2z1-3z2 + z3=0, then z1, z2 and z3 are represented by

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  19. If Re((z+4)/(2z-1)) = 1/2 then z is represented by a point lying on

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  20. The vertices of a square are z(1),z(2),z(3) and z(4) taken in the anti...

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  21. Let lambda in R . If the origin and the non-real roots of 2z^2+2z+lam...

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