NVAMediumJEE 2025Nature of Roots & Formation of Equations

JEE Mathematics 2025 Question with Solution

If for some α,β\alpha, \beta; αβ\alpha \leq \beta, α+β=8\alpha + \beta = 8 and sec2(tan1α)+csc2(cot1β)=36,\sec^2(\tan^{-1} \alpha) + \csc^2(\cot^{-1} \beta) = 36, then α2+β\alpha^2 + \beta is:

Answer

Correct answer:14

Step-by-step solution

Standard Method

Given: αβ\alpha \leq \beta, α+β=8\alpha + \beta = 8, and

sec2(tan1α)+csc2(cot1β)=36\sec^2(\tan^{-1} \alpha) + \csc^2(\cot^{-1} \beta) = 36

Find: α2+β\alpha^2 + \beta

Use the identities

sec2(tan1α)=1+α2\sec^2(\tan^{-1} \alpha) = 1 + \alpha^2

and

csc2(cot1β)=1+β2\csc^2(\cot^{-1} \beta) = 1 + \beta^2

So the given equation becomes

1+α2+1+β2=361 + \alpha^2 + 1 + \beta^2 = 36

Hence,

α2+β2=34\alpha^2 + \beta^2 = 34

Now square α+β=8\alpha + \beta = 8:

(α+β)2=64(\alpha + \beta)^2 = 64

Therefore,

α2+2αβ+β2=64\alpha^2 + 2\alpha\beta + \beta^2 = 64

Substituting α2+β2=34\alpha^2 + \beta^2 = 34,

34+2αβ=6434 + 2\alpha\beta = 64

So,

αβ=15\alpha\beta = 15

Now α\alpha and β\beta are roots of

t28t+15=0t^2 - 8t + 15 = 0

Factoring,

(t3)(t5)=0(t-3)(t-5)=0

Thus the values are 33 and 55. Since αβ\alpha \leq \beta, we get α=3\alpha = 3 and β=5\beta = 5. Finally,

α2+β=32+5=14\alpha^2 + \beta = 3^2 + 5 = 14

Therefore, the value of α2+β\alpha^2 + \beta is 1414.

Using sum and product of roots

Given:

sec2(tan1α)+csc2(cot1β)=36\sec^2(\tan^{-1} \alpha) + \csc^2(\cot^{-1} \beta) = 36

and α+β=8\alpha + \beta = 8

Find: α2+β\alpha^2 + \beta

Let A=tan1αA = \tan^{-1}\alpha and B=cot1βB = \cot^{-1}\beta. Then tanA=α\tan A = \alpha and cotB=β\cot B = \beta. Using standard identities,

sec2A=1+tan2A,csc2B=1+cot2B\sec^2 A = 1 + \tan^2 A, \qquad \csc^2 B = 1 + \cot^2 B

Hence,

1+α2+1+β2=361 + \alpha^2 + 1 + \beta^2 = 36

which gives

α2+β2=34\alpha^2 + \beta^2 = 34

Also,

(α+β)2=α2+β2+2αβ(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta

Substitute α+β=8\alpha + \beta = 8 and α2+β2=34\alpha^2 + \beta^2 = 34:

64=34+2αβ64 = 34 + 2\alpha\beta

So,

2αβ=302\alpha\beta = 30

and therefore

αβ=15\alpha\beta = 15

Now form the quadratic equation with roots α\alpha and β\beta:

x2(α+β)x+αβ=0x^2 - (\alpha+\beta)x + \alpha\beta = 0

Therefore,

x28x+15=0x^2 - 8x + 15 = 0

Solving,

(x3)(x5)=0(x-3)(x-5)=0

Thus x=3x=3 or x=5x=5. Since αβ\alpha \leq \beta,

α=3,β=5\alpha = 3, \qquad \beta = 5

Now compute

α2+β=9+5=14\alpha^2 + \beta = 9 + 5 = 14

So the required numerical value is 1414.

Note: The second provided approach ends with α2+β2=14\alpha^2 + \beta^2 = 14, which is inconsistent with its own working. The correct quantity asked and evaluated from the valid working is α2+β=14\alpha^2 + \beta = 14.

Common mistakes

  • Using the wrong identities for inverse trigonometric forms. Here, sec2(tan1x)=1+x2\sec^2(\tan^{-1}x)=1+x^2 and csc2(cot1x)=1+x2\csc^2(\cot^{-1}x)=1+x^2. Replacing them incorrectly changes the entire algebra. Always convert the trigonometric expressions first.

  • Forgetting to use αβ\alpha \leq \beta after solving the quadratic. The roots are 33 and 55, but assigning α=5\alpha=5 and β=3\beta=3 would give the wrong final value. Use the order condition to assign the roots correctly.

  • Confusing the required expression α2+β\alpha^2 + \beta with α2+β2\alpha^2 + \beta^2. From the relation you get α2+β2=34\alpha^2 + \beta^2 = 34, not the asked quantity. After finding α\alpha and β\beta, substitute into the exact expression asked.

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