# If the term free from x in the expansion of (x-kx2)10 is 405, find the value of k. - Mathematics

Sum

If the term free from x in the expansion of (sqrt(x) - k/x^2)^10 is 405, find the value of k.

#### Solution

The given expression is (sqrt(x) - k/x^2)^10

General term "T"_(r + 1) = ""^n"C"_r x^(n - r) y^r

= ""^10"C"_r (sqrt(x))^(10 - r) ((-k)/x^2)^r

= ""^10"C"_r (x)^((10 - r)/2) (-k)^r (1/x^(2r))

= ""^10"C"_r (x)^((10 - r)/2 - 2r) (-k)^r

= ""^10"C"_r (x)^((10 - r - 4r)/2) (- k)^r

= ""^10"C"_r (x)^((10 - 5r)/2) (- k)^r

For getting term free from x

(10 - 5r)/2 = 0

⇒ r = 2

On putting the value of r in the above expression

We get ""^10"C"_2  (-k)^2

According to the condition of the question, we have

""^10"C"_2 k^2 = 405

⇒ (10*9)/(2*1) k^2 = 405

⇒ 45k2 = 405

⇒ k2 = 405/45 = 9

∴ k = +-  3

Hence, the value of k = ±3

Concept: General and Middle Terms
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Chapter 8: Binomial Theorem - Exercise [Page 142]

#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Exercise | Q 2 | Page 142

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