In the expansion of \(\left[\mathrm{a}^{(2 / 5)}+\mathrm{b}^{(1 / 3)}\right]^{35} \mathrm{a} \neq \mathrm{b}\), the number of terms in which the power of a and \(\mathrm{b}\) are integers are (a) 1 (b) 2 (c) 3 (d) 4

We are given the expression \(\left[a^{\frac{2}{5}}+b^{\frac{1}{3}}\right]^{35}\), and a requirement that \(a\neq b\). We will be working with the expansion of this expression, so using the Binomial theorem is key for this task.

To expand the given expression, we can use the binomial theorem, which states that for any non-negative integer n: \[(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k}y^k\] Applying the binomial theorem to our expression, we get: \[\left[a^{\frac{2}{5}}+b^{\frac{1}{3}}\right]^{35} = \sum_{k=0}^{35} \binom{35}{k} a^{\frac{2}{5}(35-k)}b^{\frac{1}{3}k}\]

In this step, our goal is to find all the terms in the expansion where both powers of a and b are integers. To do this, we need to find the values of k for which both \(\frac{2}{5}(35-k)\) and \(\frac{1}{3}k\) are integers. For a power to be an integer, the denominator of the exponent should divide the numerator. Therefore, we can express the exponents in the form: 1) \(\frac{2}{5}(35-k)=\frac{2(35-k)}{5} = 2m\), where \(m\) is an integer, and 2) \(\frac{1}{3}k=\frac{k}{3} = n\), where \(n\) is an integer.

Now lets solve these equations to find values of k which make both exponents integers. From equation (1): \(2(35-k)=5m\), and we get \(k = 35-\frac{5m}{2}\) From equation (2): \(k=3n\) Now we need to substitute the value of k from equation (2) into equation (1) to get the relationship between m and n: \(35-\frac{5m}{2} = 3n\) We can rearrange and solve for m: \(35-3n = \frac{5m}{2}\) \(70-6n = 5m\) Let's manually check integer values for n in the range of [1,11] and see if we get integer values for m: 1) n = 1 --> \(m=12\) 2) n = 2 --> \(m=8\) 3) n = 3 --> \(m=4\) 4) n = 4 --> \(m=0\) For all other values of n, m becomes negative. Since m can only be a non-negative integer, these are the only 4 values of n that give us integer m values.

Now, we can obtain the values of k that give us integer exponents using the relationship \(k=3n\): 1) n = 1 --> \(k=3\) 2) n = 2 --> \(k=6\) 3) n = 3 --> \(k=9\) 4) n = 4 --> \(k=12\) These are the 4 values of k that make both exponents in each term integers. Therefore, there are 4 terms in the expansion in which both powers of a and b are integers. So, the correct answer is (d) 4.