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Quadratic Equation and Inequalities question

2018 · 15 Apr · Shift 1 · Q35
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  5. /2018 · 15 Apr · Shift 1 · Q35

Quadratic Equation and Inequalities question

2018 · 15 Apr · Shift 1 · Q35

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If tanA and tanB are the roots of the quadratic equation, 3x2 −-− 10x −-− 25 = 0, then the value of 3 sin2(A + B) −-− 10 sin(A + B).cos(A + B) −-− 25 cos2(A + B) is :
  1. A
    −-− 10
  2. B
    10
  3. C
    −-− 25
  4. D
    25
View written solutionFree

Correct answer: C

  1. Let the roots of the quadratic equation 3x2−10x−25=03x^2-10x-25=03x2−10x−25=0 be x=tan⁡A,x=tan⁡B.x=\tan A,\quad x=\tan B.x=tanA,x=tanB.

So, by Vieta's formulas, tan⁡A+tan⁡B=103,\tan A+\tan B=\frac{10}{3},tanA+tanB=310​, tan⁡Atan⁡B=−253.\tan A\tan B=\frac{-25}{3}.tanAtanB=3−25​.

  1. We need to find 3sin⁡2(A+B)−10sin⁡(A+B)cos⁡(A+B)−25cos⁡2(A+B).3\sin^2(A+B)-10\sin(A+B)\cos(A+B)-25\cos^2(A+B).3sin2(A+B)−10sin(A+B)cos(A+B)−25cos2(A+B).

Let θ=A+B.\theta=A+B.θ=A+B. Then the expression becomes 3sin⁡2θ−10sin⁡θcos⁡θ−25cos⁡2θ.3\sin^2\theta-10\sin\theta\cos\theta-25\cos^2\theta.3sin2θ−10sinθcosθ−25cos2θ.

  1. Use tan⁡θ=tan⁡A+tan⁡B1−tan⁡Atan⁡B.\tan\theta=\frac{\tan A+\tan B}{1-\tan A\tan B}.tanθ=1−tanAtanBtanA+tanB​. Substituting the values,
=\frac{\frac{10}{3}}{\frac{28}{3}}=\frac{10}{28}=\frac{5}{14}.$$ So, $$\tan\theta=\frac{5}{14}.$$ 4. Write the required expression in terms of $\tan\theta$. Since $$\sin^2\theta=\tan^2\theta\cos^2\theta, \qquad \sin\theta\cos\theta=\tan\theta\cos^2\theta,$$ we get $$3\sin^2\theta-10\sin\theta\cos\theta-25\cos^2\theta =\cos^2\theta\left(3\tan^2\theta-10\tan\theta-25\right).$$ Now substitute $\tan\theta=\frac{5}{14}$: $$3\tan^2\theta-10\tan\theta-25 =3\left(\frac{25}{196}\right)-10\left(\frac{5}{14}\right)-25.$$ Taking LCM $196$, $$=\frac{75}{196}-\frac{700}{196}-\frac{4900}{196} =\frac{-5525}{196}.$$ Also, $$\cos^2\theta=\frac{1}{1+\tan^2\theta} =\frac{1}{1+\frac{25}{196}}= rac{1}{\frac{221}{196}}=\frac{196}{221}.$$ Therefore, $$\cos^2\theta\left(3\tan^2\theta-10\tan\theta-25\right) =\frac{196}{221}\cdot \frac{-5525}{196} =\frac{-5525}{221}=-25.$$ 5. Hence, $$3\sin^2(A+B)-10\sin(A+B)\cos(A+B)-25\cos^2(A+B)=-25.$$ So the correct option is: $$\boxed{\text{C: }-25}$$ 6. Verification with stored answer: Stored correct answer is C, which matches our result.
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